The magnitude of the acceleration of the proton is approximately 2.25 × 10^17 m/s^2. We can use Coulomb's law and Newton's second law.
To calculate the magnitude of the acceleration of a proton due to the electric field created by the charged bead, we can use Coulomb's law and Newton's second law.
The electric force between the charged bead and the proton is given by Coulomb's law:
F = k * |q1| * |q2| / r^2
where F is the electric force, k is the Coulomb's constant (k = 8.99 × 10^9 N m^2/C^2), |q1| and |q2| are the magnitudes of the charges, and r is the distance between them.
The electric force can also be expressed as:
F = m * a
where m is the mass of the proton and a is its acceleration.
Setting these two equations equal to each other, we have:
k * |q1| * |q2| / r^2 = m * a
We can rearrange this equation to solve for the acceleration:
a = (k * |q1| * |q2|) / (m * r^2)
Substituting the given values:
k = 8.99 × 10^9 N m^2/C^2,
|q1| = 11 nC = 11 × 10^-9 C,
|q2| = charge of a proton = 1.6 × 10^-19 C,
m = mass of a proton = 1.67 × 10^-27 kg,
r = 0.60 cm = 0.60 × 10^-2 m,
we can calculate the acceleration:
a = (8.99 × 10^9 N m^2/C^2 * 11 × 10^-9 C * 1.6 × 10^-19 C) / (1.67 × 10^-27 kg * (0.60 × 10^-2 m)^2)
Evaluating this expression, the magnitude of the acceleration (ap) of the proton is approximately:
ap ≈ 2.25 × 10^17 m/s^2
Therefore, the magnitude of the acceleration of the proton is approximately 2.25 × 10^17 m/s^2.
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An infinitely long cylinder with radius R1 and charge density rho has a small cylinder (length L=50 cm, radius R2
r(R₂) ≈ √(L₂ + R₁₂) + 2kρL ln(R₁ / R₂) / √(L₂ + R1₂). The electric field at point P is then: E = kρ / r₂ ≈ kρ / [L₂ + R₁₂ + 2kρL ln(R₁ / R₂)]. The contribution of a small element of the cylinder with length dx, charge density ρ, and radius x to the electric field at point P is : dE = k · ρ · dx / r
The contribution of a small element of the cylinder with length dx, charge density ρ, and radius x to the electric field at point P is : dE = k · ρ · dx / r, where k is Coulomb's constant. We can use the Pythagorean theorem to relate r and x: r₂= L₂ + (R₁ - x)₂
Squaring both sides and differentiating with respect to x yields: 2r · dr / dx = -2(R₁ - x)
Therefore, dr / dx = -(R₁ - x) / r
Integrating this expression from x = 0 to x = R₂,
we obtain: r(R₂) - r(0) = -∫0R₂(R₁ - x) / r dx
We can use the substitution u = r₂ to simplify the integral:∫1r₁ du / √(r₁₂ - u) = -∫R₂₀(R₁ - x) dx / xR₁ > R₂, the integral can be approximated as: ∫R₂₀(R₁ - x) dx / x ≈ 2(R₁ - R₂) ln (R₁ / R₂)
Therefore: r(R₂) ≈ √(L₂ + R₁₂) + 2kρL ln(R₁ / R₂) / √(L₂ + R1₂)
The electric field at point P is then: E = kρ / r₂ ≈ kρ / [L₂ + R₁₂ + 2kρL ln(R₁ / R₂)]
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A700-tum solenoid, 24 cm long has a diameter of 2.7 cm A11-turn coil is wound tightly around the center of the solenoid . If the current in the solenoid increases uniformily from 0 to 42 A in 0 60 s, what will be the induced emf in the short coll during this time? Express your answer to two significant figures and include the appropriate units.
The induced emf in the short coil during this time is -1.12 × 10⁻⁸ V. The formula to calculate the induced emf in the short coil during this time is given by the following formula:ε=−N(ΔΦ/Δt)
The formula to calculate the induced emf in the short coil during this time is given by the following formula:ε=−N(ΔΦ/Δt)where N is the number of turns in the short coil and ΔΦ/Δt is the change in the magnetic flux over time. The change in magnetic flux over time is given by the following formula:
ΔΦ/Δt=μ_0NA(ΔI/Δt)where μ0 is the permeability of free space, A is the cross-sectional area of the solenoid, and ΔI/Δt is the rate of change of current in the solenoid.
Substituting the values given in the question: μ0 = 4π × 10⁻⁷ T·m/A,
N = 11, A = (π/4) × (2.7 × 10⁻² m)²
= 5.73 × 10⁻⁴ m²,
ΔI/Δt = 42 A/60 s
= 0.7 A/s,
we have: ΔΦ/Δt =4π × 10⁻⁷ T·m/A × 11 × 5.73 × 10⁻⁴ m² × 0.7 A/s
= 1.02 × 10⁻⁹ Wb/s (2 SF)
Therefore, the induced emf in the short coil during this time is:
ε=−N(ΔΦ/Δt)
=−11 × 1.02 × 10⁻⁹ V/s
= -1.12 × 10⁻⁸ V (2 SF)
Answer: The induced emf in the short coil during this time is -1.12 × 10⁻⁸ V.
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You go for a walk and travel 27.0 m at an angle 24 East of North. You then walk 35.4 m and an angle 32 South of East. What is the magnitude of your displacement vector in metres?
The magnitude of your displacement vector is approximately 55.10 meters. To find the magnitude of the displacement vector, we need to calculate the resultant vector by adding the two vectors together.
For the first vector (27.0 m at an angle 24° east of north):
27.0 m * sin(24°) = 11.07 m (northward)
27.0 m * cos(24°) = 24.71 m (eastward)
For the second vector (35.4 m at an angle 32° south of east):
The east component is given by:
35.4 m * cos(32°) = 29.83 m (eastward)
The south component is given by:
35.4 m * sin(32°) = 18.60 m (southward)
11.07 m (northward) - 18.60 m (southward) = -7.53 m (southward)
And let's add the east components together:
24.71 m (eastward) + 29.83 m (eastward) = 54.54 m (eastward)
So, the resultant vector is 54.54 m eastward and -7.53 m southward.
To find the magnitude of the displacement vector, we can use the Pythagorean theorem:
magnitude = sqrt((eastward)^2 + (southward)^2)
magnitude = sqrt((54.54 m)^2 + (-7.53 m)^2)
magnitude ≈ 55.10 m
Therefore, the magnitude of your displacement vector is approximately 55.10 meters.
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On a frictionless surface, an 80 gram meter stick lies at rest on a frictionless surface. The origin lies at the 60-cm mark and is along x axis. At the 100 cm mark, there is an 80 gram lump of clay. Also, there is another 80 gram lump of clay moving 2.50 m/s in positive y direction. This second lump of clay collides and sticks at the 12 cm mark. What is angular momentum around center of stick?
What quantities are conserved in collision accounting for angular momentum, energy, momentum, and rotational energy? Give an explanation for each.
2. Calculate the moment of inertia for the two lumps of clay + stick after collision.
3. Calculate the velocity of the center of mass of the meter stick after the collision?
4. Calculate the angular velocity of the stick after collision.
5. Calculate where the center of the stick is after it has completed one rotation?
A friction less surface, an 80 gram meter stick lies at rest on a friction less surface. The origin lies at the 60-cm mark and is along x axis. At the 100 cm mark, there is an 80 gram lump of clay.( 1)) The angular momentum around the center of the stick is zero.(2)The moment of inertia for the two lumps of clay + stick after collision is 0.08 kg×m^2.(3)The velocity of the center of mass of the meter stick after the collision is 0 m/s.(4) The angular velocity of the stick after collision is 4.3 rad/s.(5)The center of the stick will be at the 60 cm mark after it has completed one rotation
The following solution are :
1. This is because the initial angular momentum of the system is zero, and there are no external torques acting on the system after the collision.
2)The quantities conserved in the collision are angular momentum, energy, and momentum. Angular momentum is conserved because there are no external torques acting on the system. Energy is conserved because the collision is elastic. Momentum is conserved because the collision is head-on and there is no net external force acting on the system.
The moment of inertia for the two lumps of clay + stick after collision is 0.08 kg×m^2. This is calculated using the equation I = mr^2, where m is the mass of the system (160 g) and r is the distance from the center of mass to the axis of rotation (58 cm).
3) The velocity of the center of mass of the meter stick after the collision is 0 m/s. This is because the center of mass of the system does not move in a collision.
4) The angular velocity of the stick after collision is 4.3 rad/s. This is calculated using the equation ω = L/I, where L is the angular momentum of the system (0.16 kg.m^2rad/s) and I is the moment of inertia of the system (0.08 kg×m^2).
5) The center of the stick will be at the 60 cm mark after it has completed one rotation. This is because the center of mass of the system does not move in a collision.
Here are the steps in more detail:
The initial angular momentum of the system is zero. This is because the first lump of clay is not rotating, and the second lump of clay has no angular momentum because it is moving in a straight line. There are no external torques acting on the system after the collision. This is because the surface is friction less, so there is no frictional force acting on the system. There are also no other forces acting on the system, so the net torque is zero. The angular momentum of the system is conserved because there are no external torques acting on the system. This means that the angular momentum of the system after the collision must be equal to the angular momentum of the system before the collision, which is zero. The energy of the system is conserved because the collision is elastic. This means that the total kinetic energy of the system before the collision is equal to the total kinetic energy of the system after the collision. The momentum of the system is conserved because the collision is head-on and there is no net external force acting on the system. This means that the total momentum of the system before the collision is equal to the total momentum of the system after the collision. The center of mass of the meter stick does not move in a collision. This is because the collision is perfectly elastic, and there are no external forces acting on the system. The angular velocity of the stick after collision is calculated using the equation ω = L/I, where L is the angular momentum of the system (0.16 kgm^2rad/s) and I is the moment of inertia of the system (0.08 kg×m^2). This gives us an angular velocity of 4.3 rad/s. The center of the stick will be at the 60 cm mark after it has completed one rotation. This is because the center of mass of the system does not move in a collision.To learn more about angular momentum visit: https://brainly.com/question/4126751
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Electrons from the main beam at the Stanford Linear Accelerator Center can reach speeds as large as 0.9999999997 c. Let these electrons enter a detector 1 m long. Calculate the length of the detector in the rest frame of one of the particles.
The length of the detector in the rest frame of one of the particles is 0.010129 m.
Stanford Linear Accelerator Center is a research institute that has developed an accelerator to generate high-energy electron and positron beams. These beams are then collided with each other or a fixed target to investigate subatomic particles and their properties. The electrons at this facility can reach a velocity of 0.9999999997 c.
The length of the detector in the rest frame of one of the particles is calculated as follows:Let’s start by calculating the velocity of the electrons. V= 0.9999999997 c.
Velocity can be defined as distance traveled per unit time. Hence, it is necessary to use the Lorentz factor to calculate the length of the detector in the rest frame of one of the particles.
Lorentz factor γ is given byγ = 1 / √(1 – v²/c²)where v is the velocity of the particle and c is the speed of light.γ = 1 / √(1 – (0.9999999997c)²/c²)γ = 98.7887
Now that we have the value of γ, we can calculate the length of the detector in the rest frame of one of the particles.The length of the detector as seen by an observer at rest is L = 1 m.
So, the length of the detector in the rest frame of one of the particles is given byL' = L / γL' = 1 m / 98.7887L' = 0.010129 m
Therefore, the length of the detector in the rest frame of one of the particles is 0.010129 m.
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An object with mass 3.2 kg is moving in one dimension subject to a time-dependent force given by the function F (1) = 3.172
At t = 1.0 s, the object is moving in the -x direction at a speed of 8.8 m/s.
What is the object's velocity at t = 2.1 s?
An object with mass 3.2 kg is moving in one dimension subject to a time-dependent force given by the function F (1) = 3.172. At t = 2.1 s, its velocity is -18.8 m/s in the -x direction.
To solve this problem, we can use the following equation:
F = ma
where
F is the force acting on the object
m is the mass of the object
a is the acceleration of the object
We know that the force acting on the object is given by the function F(t) = 3.172. We also know that the mass of the object is 3.2 kg. We can use these values to find the acceleration of the object:
a = F/m = 3.172 N/kg = 0.988 m/s²
We know that the object is moving in the -x direction at a speed of 8.8 m/s at t = 1.0 s. We can use this information to find the object's velocity at t = 2.1 s:
v = u + at
where
v is the object's velocity at t = 2.1 s
u is the object's velocity at t = 1.0 s
a is the acceleration of the object
Substituting the known values, we get:
v = -8.8 m/s + 0.988 m/s² * 2.1 s = -18.8 m/s
Therefore, the object's velocity at t = 2.1 s is -18.8 m/s.
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In a demolition derby, a car known as 'slippery Pete' has a mass of 1520 kg is moving east with a speed of 15.79 m/s. the driver small truck named 'vindicator" has a mass of 1350 kg is driving north at 17.4 m/s. when the two cars collide their bodies stick together.
what is the common speed of the cars after the collision?
a) 11.5 m/s
b) 11.7 m/s
c) 11.1 m/s
d) 11.3 m/s
Counties fairs and international events frequently feature demolition derbies.
Thus, The traditional demolition derby event features five or more drivers compete by purposefully smashing their automobiles into one another, though restrictions vary depending on the event. The winner is the last driver whose car is still in working order.
The United States is where demolition derbies first appeared, and other Western countries swiftly caught on. For instance, the country of Australia hosted its inaugural demolition derby in January 1963. Demolition derbies—also known as "destruction derbies"—are frequently held in the UK and other parts of Europe after a long day of banger racing.
Whiplash and other major injuries are uncommon in demolition derbies, although they do occur.
Thus, Counties fairs and international events frequently feature demolition derbies.
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The correct option is (none of the above). The given masses of the cars involved in the collision are:
Mass of 'slippery Pete' = 1520 kg
Mass of 'vindicator' = 1350 kg
The given velocities of the cars involved in the collision are:
Velocity of 'slippery Pete' = 15.79 m/s
Velocity of 'vindicator' = 17.4 m/s
The initial momentum of the system is given by: P(initial) = m1v1 + m2v2
where m1 and v1 are the mass and velocity of car 1, and m2 and v2 are the mass and velocity of car 2. Substituting the given values, we get:
P(initial) = (1520 kg) (15.79 m/s) + (1350 kg) (17.4 m/s)P(initial) = 23969 + 23490P(initial) = 47459 kg m/s
Since the two cars stick together after the collision, they can be considered as a single body. The final momentum of the system is given by:P(final) = (m1 + m2) vf
where m1 and m2 are the masses of the two cars, and vf is the final velocity of the combined cars. Substituting the given values, we get:
P(final) = (1520 kg + 1350 kg) vfP(final) = 2870 kg vf
Since momentum is conserved in the system, we can equate P(initial) to P(final) and solve for vf. So:
P(initial) = P(final)47459 kg m/s = 2870 kg vf vf = 47459 kg m/s ÷ 2870 kg vf = 16.51 m/s
The common speed of the cars after the collision is 16.51 m/s, which when rounded off to one decimal place, is 16.5 m/s.Therefore, the correct option is (none of the above).
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A raindrop of mass 3.35× 10⁻⁵ kg falls vertically at constant speed under the influence of gravity and air resis-tance. Model the drop as a particle. As it falls 100m, what is the work done on the raindrop(b) by air resistance?
The work done by gravity is equal to the work done by air resistance, the work done on the raindrop by air resistance is also 3.27×10⁻² J.
This means that the work done by gravity is equal to the work done by air resistance.
The work done by gravity can be calculated using the formula: Work = force x distance. The force of gravity acting on the raindrop is given by the equation: F = mg, where m is the mass of the raindrop and g is the acceleration due to gravity (9.8 m/s²).
First, we need to calculate the force of gravity acting on the raindrop. The mass of the raindrop is given as 3.35×10⁻⁵ kg. Therefore, the force of gravity can be calculated as:
F = mg
F = (3.35×10⁻⁵ kg) x (9.8 m/s²)
F = 3.27×10⁻⁴ N
Next, we calculate the work done by gravity over a distance of 100 m:
Work = force x distance
Work = (3.27×10⁻⁴ N) x (100 m)
Work = 3.27×10⁻² J
Since the work done by gravity is equal to the work done by air resistance, the work done on the raindrop by air resistance is also 3.27×10⁻² J.
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5cm, qA = 2μC at the origin x = 0, qß = 1µC at x = : 4 cm, 2 1 cm - = X1 What is the potential difference Vx1 Vx2? Again, note the sign change on the charge. (2 points per case) Also in each case please provide the integral you are doing and then also provide your answer with units.
The potential difference between Vx1 Vx2 when x1 = 4cm, and x2 = 2 cm . The formula for potential difference is given by V = VB - VA Where VB is the potential at point B, and VA is the potential at point A.
Integral formula: Potential difference is defined as the work done per unit charge to move a charge from one point to another, and is represented mathematically as the line integral of the electric field between the two points in question, as shown below:
V = - ∫E.ds
Where, E is the electric field, ds is an infinitesimal element of the path taken by the charge, and the integral is taken along the path between the two points in question. Here, E can be determined using Coulomb's law, given as:
F = k.q1.q2/r^2
Here, r is the distance between the two charges and k is the Coulomb's constant which is equal to 1/4πε_0. Where ε_0 is the permittivity of free space, which is equal to 8.85 x 10^-12 C^2/(N.m^2).
When x1 = 4 cm, q1 = 1 µC, q2 = - 2 µC, and x2 = 2 cm, The distance between the two charges, r = (4 - 2) cm = 2 cm = 0.02 m.
Therefore,
F = k.q1.q2/r^2 = (1/4πε_0).(1 x 10^-6) x (-2 x 10^-6)/(0.02)^2 = - 0.225 N
Using the formula for electric potential,
Vx1 - Vx2 = ∫E.dx = (- 0.225) x 10^3 x ∫(2 - 4)/100 dx = (0.225) x 10^3 x ∫2/100 - 4/100 dx= (0.225) x 10^3 x (- 2/100) = -4.5V
Therefore, the potential difference Vx1 Vx2 is equal to - 4.5 V.
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What is the minimum stopping distance for the same car traveling at a speed of 36 m/s ?
The minimum stopping distance for the car traveling at a speed of 36 m/s is 117 meters.
The minimum stopping distance for a car can be calculated using the formula:
Stopping Distance = Thinking Distance + Braking Distance
The thinking distance is the distance the car travels while the driver reacts to a situation and applies the brakes. The braking distance is the distance the car travels while braking to a stop.
To calculate the thinking distance, we can use the formula: Thinking Distance = Speed x Reaction Time.
Given that the car is traveling at a speed of 36 m/s, we need to know the reaction time of the driver to calculate the thinking distance. Let's assume a typical reaction time of 1 second for this example.
Thinking Distance = 36 m/s x 1 s = 36 m
To calculate the braking distance, we need to use the formula: Braking Distance = (Speed 2) / (2 x Deceleration)
Deceleration is the rate at which the car slows down. Let's assume a deceleration of 8 m/s^2 for this example.
Braking Distance = (36 m/s) 2 / (2 x 8 m/s 2) = 81 m
Therefore, the minimum stopping distance for the same car traveling at a speed of 36 m/s is the sum of the thinking distance and the braking distance:
Stopping Distance = 36 m + 81 m = 117 m
The minimum stopping distance for the car traveling at a speed of 36 m/s is 117 meters.
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A projectile is projected from the origin with a velocity of 45.0 m/s at an angle of 30.0 degrees above the horizontal. What is the time it takes for the projectile to hit the ground?
The time it takes the projectile to hit the ground is 4.59 s.
What is time of flight?The time taken by the projectile to complete its trajectory is called time of flight.
To calculate the time of flight of the projectile to hit the ground,we used the formula below
Formula:
T = 2usin∅/g.................. Equation 1Where:
T = Time of flight of the projectile∅ = Angle of the projectileg = Acceleration due to gravityu = Intial velocityFrom the question,
Given:
u = 45 m/s∅ = 30°g = 9.8 m/s²Substitute these values into equation 1
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The time it takes for the projectile to hit the ground is 4.59 seconds.
A projectile is projected from the origin with a velocity of 45.0 m/s at an angle of 30.0 degrees above the horizontal.
The horizontal and vertical motions of a projectile are independent of one another. As a result, the horizontal motion is constant velocity motion, whereas the vertical motion is free-fall motion.
Let's calculate the time it takes for the projectile to hit the ground:
First, we will calculate the time it takes for the projectile to reach the maximum height. Using the formula:v_y = v_iy + a_ytFinal velocity = 0 (since the projectile stops at the top)
v_iy = 45 sin 30° = 22.5 m/st = ?a_y = - 9.8 m/s² (negative acceleration since it is directed downwards) 0 = 22.5 - 9.8tt = 22.5 / 9.8t = 2.3 s
The time taken for the projectile to reach its highest point is 2.3 s.
Next, we can calculate the time taken for the projectile to reach the ground. Using the formula:y = v_iyt + (1/2) a_yt²y = 0 (since the projectile hits the ground)
v_iy = 22.5 m/s (from above)t = ?a_y = - 9.8 m/s² (negative since it is directed downwards) 0 = 22.5t - 4.9t²t(4.9t - 22.5) = 0t = 0 s (initially)t = 4.59 s (when the projectile hits the ground)
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A mass of 26 kg is suspended by two cords from a ceiling. The cords have lengths of 17 cm and 21 cm, and the distance between the points where they are attached on the ceiling is 29 cm. Determine the tension in each of the two cords. Include a clear diagram
The tension in the 17 cm cord is 156.3 N and the tension in the 21 cm cord is 110.3 N.
The mass of 26 kg is suspended by two cords from a ceiling. The cords have lengths of 17 cm and 21 cm, and the distance between the points where they are attached to the ceiling is 29 cm.
To determine the tension in each of the two cords, we first sketch the diagram of the system of the two cords and the mass that is being suspended from the cords.From the diagram, we can see that the forces acting on the mass are the weight of the mass and the tensions in the cords. Thus we have two equations of equilibrium as follows:Equation (1) resolves forces in the vertical direction: `T1 sin θ1 + T2 sin θ2 = Fg
For the 17 cm cord, the vertical component of tension T1 is T1 sin(θ1), and for the 21 cm cord, the vertical component of tension T2 is T2 sin(θ2).
Since the mass is in equilibrium, the sum of the vertical forces must be zero:
T1 sin(θ1) + T2 sin(θ2) = mg
We can also consider the horizontal components of tension T1 and T2. The horizontal component of T1 is T1 cos(θ1), and the horizontal component of T2 is T2 cos(θ2). The horizontal components must cancel out each other since there is no horizontal acceleration:
T1 cos(θ1) = T2 cos(θ2)
Using these two equations, we can solve for the tensions T1 and T2
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:
A child bounces a super ball on the sidewalk, the linear impulse delivered by the sidewalk is 2N.s during the 1/800 s of contact, what is the magnitude of the average force exerted on the ball by the sidewalk.
The magnitude of the average force exerted on the ball by the sidewalk is 1600 N.
The impulse momentum theorem states that the change in momentum of an object is equal to the impulse acting on it. When a child bounces a super ball on the sidewalk, the linear impulse delivered by the sidewalk is 2N.s during the 1/800 s of contact.
This means that the impulse acting on the ball is 2 N.s, and it occurs over a time of 1/800 s. We can use this information to determine the magnitude of the average force exerted on the ball by the sidewalk. The impulse momentum theorem is expressed as:
I = Δp where I is the impulse and Δp is the change in momentum. We can rearrange this equation to solve for the change in momentum: Δp = I
momentum is expressed as: p = mv where p is momentum, m is mass, and v is velocity. Since the mass of the ball remains constant, we can simplify this equation to: p = mv = mΔv where Δv is the change in velocity. We can now substitute this expression for momentum into the impulse momentum theorem equation: Δp = I = mΔv
Solving for Δv, we get: Δv = I/m
We know that the impulse acting on the ball is 2 N.s and that it occurs over a time of 1/800 s. To determine the average force exerted on the ball by the sidewalk, we need to calculate the change in velocity. However, we do not know the mass of the ball. Therefore, we will assume a mass of 1 kg, which is reasonable for a super ball. Using this assumption, we can calculate the change in velocity:
Δv = I/m
= 2 N.s / 1 kg
= 2 m/s
The average force exerted on the ball by the sidewalk is equal to the rate of change of momentum, which is given by:F = Δp / t where t is the time over which the force is applied. Since the force is applied over a time of 1/800 s, we can substitute this value into the equation:
F = Δp / t = mΔv / t
= (1 kg)(2 m/s) / (1/800 s)
= 1600 N
The magnitude of the average force exerted on the ball by the sidewalk is 1600 N. This means that the sidewalk exerts a strong force on the ball to change its direction. It also means that the ball exerts an equal and opposite force on the sidewalk, as required by Newton's third law of motion.
Answer: The magnitude of the average force exerted on the ball by the sidewalk is 1600 N.
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The magnitude of the average force exerted on the ball by the sidewalk is 1600 N.
We have been given the following:Linear impulse delivered by the sidewalk = 2 N
The impulse delivered by the sidewalk can be calculated using the formula:
Impulse = Force * Time
Given that the impulse delivered is 2 N·s and the contact time is 1/800 s, we can rearrange the equation to solve for the average force:
Force = Impulse / Time
Substituting the values:
Force = 2 N·s / (1/800 s)
Force = 2 N·s * (800 s)
Force = 1600 N
Therefore, the magnitude of the average force exerted on the ball by the sidewalk is 1600 N.
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Four point charges are located at the comers of a square. Fach charge has magnitude 1 , a0 nc and the square has sides of length 3.00 om. Find the magnitude of the ei of the square of all of the charges are positive and three of illve charges are pesittve and one is negative. (a) sis the charges are positive N/C (b) three of the charges are pesitive and one is negative Nre (a) ill the tharges are jotitive Nye (b) three of the charges are Dettive aref ene is negative N'C
The problem is related to Coulomb's law, which describes the interaction of charges with one another. It is necessary to consider four point charges located at the corners of a square. Each charge has a magnitude of 1 and is positioned a0 nc away from the square, which has sides of length 3.00 om.
The task is to determine the magnitude of the electric field generated by the square of charges if all charges are positive and three are positive, and one is negative. (a) is the charges are positive N/C (b) three of the charges are positive and one is negative Nre (a) ill the charges are positive Nye (b) three of the charges are Dettive aref ene is negative N'C.
Electric field is a vector quantity that is denoted by E. The formula of electric field is E = F / q. The electric field is the force per unit charge acting on a charge placed in the electric field, where F is the force acting on the charge and q is the magnitude of the charge.In the case where all four charges are positive, the magnitude of the electric field generated by the square of charges isE = k * Q / r²The total electric field due to four charges of magnitude q is the vector sum of the individual fields created by each of the charges.E = E1 + E2 + E3 + E4.
We know that the charges at opposite corners of the square have a net electric field of zero because they lie on the same diagonal line. So, we only need to consider the fields created by the two charges along the same diagonal line. Let's say that the charges on this diagonal line are q1 and q2. The distance between them is a, and the distance from each charge to the midpoint of the line is b.
The electric field generated by each of the charges isE = k * q / r²E1 = k * q1 / b²E2 = k * q2 / b²The net electric field at the midpoint of the line isE = E1 - E2 = k * (q1 - q2) / b²The magnitude of the electric field isE = k * (q1 - q2) / b²The distance b is equal to half the length of the diagonal of the square, which isL = √(3² + 3²) = 3√2.
The magnitude of the electric field at the midpoint of the diagonal isE = k * (q1 - q2) / (3√2)²E = k * (q1 - q2) / 18. The electric field at the midpoint of the opposite diagonal is the same magnitude and in the opposite direction. So, the net electric field at the center of the square is zero. So, in this case, the answer is (c) all charges are positive Nye.
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Here are some questions about toast. for a total of 5.0 minutes? (a) How many kWh of energy does a 550 W toaster use in the morning if it is in operation (b) At a cost of 9.44 cents/kWh, how much (in s) would this appliance add to your monthly electric bill if you made toast four mornings per week? (Assume that 1 month = 4 weeks.
(a) A 550 W toaster operates for 5.0 minutes in the morning. To calculate the energy usage in kilowatt-hours (kWh), we need to convert the power from watts to kilowatts and then multiply it by the time in hours.
Since 1 kilowatt is equal to 1000 watts, the toaster's power can be expressed as 0.55 kW (550 W ÷ 1000). Multiplying the power by the time gives us the energy usage: 0.55 kW × 5.0 min ÷ 60 min/hour = 0.0458 kWh.
(b) Assuming four mornings per week, we can calculate the monthly energy consumption of the toaster. Since 1 month is equal to 4 weeks, the number of mornings in a month is 4 × 4 = 16.
Multiplying the energy usage per morning (0.0458 kWh) by the number of mornings in a month (16) gives us the total energy consumption per month: 0.0458 kWh/morning × 16 mornings = 0.7328 kWh/month.
To determine the cost, we multiply the energy consumption by the cost per kilowatt-hour (9.44 cents/kWh).
Converting cents to dollars (1 dollar = 100 cents), the cost can be calculated as follows: 0.7328 kWh/month × $0.0944/kWh = $0.0696/month.
Therefore, if you made toast four mornings per week, the toaster would add approximately $0.0696 to your monthly electric bill.
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An LED lamp has a 700 lx intensity at a distance of 1.0 m.
What is the intensity of the
lamp at 6.5 m away?
The intensity of the LED lamp at a distance of 6.5 m away is approximately 16.59 lx.
To calculate the intensity of the LED lamp at a distance of 6.5 m away, we can use the inverse square law, which states that the intensity of light decreases inversely proportional to the square of the distance.
Given:
Initial intensity (I1) = 700 lx
Initial distance (d1) = 1.0 m
Target distance (d2) = 6.5 m
The formula to calculate the intensity at the target distance is:
I2 = I1 * (d1 / d2)^2
Substituting the given values:
I2 = 700 lx * (1.0 m / 6.5 m)^2
Calculating the value:
I2 = 700 lx * (0.1538)^2
I2 ≈ 700 lx * 0.0237
I2 ≈ 16.59 lx
Therefore, the intensity of the LED lamp at a distance of 6.5 m away is approximately 16.59 lx.
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A block, W 180 lbs rests on a rough level plane. The coefficient of friction is 0.42, what horizontal push will cause the block to move? What inclined push making 45° with the horizontal will cause the block to move?
The inclined push making a 45° angle with the horizontal should satisfy the equation: Horizontal component = inclined push × cos(45°) ≥ Frictional force
To determine the horizontal push required to make the block move, we need to consider the force of friction acting on the block. The force of friction can be calculated using the formula:
Frictional force = coefficient of friction × normal force
The normal force is equal to the weight of the block, which is 180 lbs. Therefore, the normal force is 180 lbs × acceleration due to gravity.
To find the horizontal push, we need to overcome the force of friction. The force of friction is given by the equation:
Frictional force = coefficient of friction × normal force
Let's calculate the force of friction:
Frictional force = 0.42 × (180 lbs × acceleration due to gravity)
Now we can calculate the horizontal push:
Horizontal push = Frictional force
To Know the inclined push making a 45° angle with the horizontal, we need to consider the force components acting on the block. The horizontal component of the inclined push will contribute to overcoming the force of friction, while the vertical component will assist in counteracting the weight of the block.
Since the inclined push makes a 45° angle with the horizontal, the horizontal component can be calculated using the formula:
Horizontal component = inclined push × cos(45°)
To make the block move, the horizontal component of the inclined push should be equal to or greater than the force of friction calculated previously.
Therefore, the inclined push making a 45° angle with the horizontal should satisfy the equation:
Horizontal component = inclined push × cos(45°) ≥ Frictional force
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1. Two equal-mass hockey pucks undergo a glancing collision. Puck 1 is initially at rest and is struck by puck 2 travelling at a velocity of 13 m/s [E). After the collision Puck 1 travels at an angle of [E 18° N] with a velocity of 20m/s, what is he velocity and direction of Puck 2 [41] [4A
After the glancing collision between two equal-mass hockey pucks, Puck 1 moves at an angle of 18° north of east with a velocity of 20 m/s. To determine the velocity and direction of Puck 2, we need to use the principles of conservation of momentum and analyze the vector components of the velocities before and after the collision.
The principle of conservation of momentum states that the total momentum of a system remains constant before and after a collision, assuming no external forces act on the system. Since the masses of Puck 1 and Puck 2 are equal, their initial momenta are also equal and opposite in direction.
Let's consider the x-axis as east-west and the y-axis as north-south. Before the collision, Puck 2 travels at 13 m/s east (positive x-direction), and Puck 1 is at rest (0 m/s). After the collision, Puck 1 moves at an angle of 18° north of east with a velocity of 20 m/s.
To determine the velocity and direction of Puck 2, we can use vector components. We can break down the velocity of Puck 2 into its x and y components. The x-component of Puck 2's velocity is equal to the initial x-component of Puck 1's velocity (since momentum is conserved). Therefore, Puck 2's x-velocity remains 13 m/s east.
To find Puck 2's y-velocity, we need to consider the conservation of momentum in the y-direction. The initial y-component of momentum is zero (Puck 1 is at rest), and after the collision, Puck 1 moves at an angle of 18° north of east with a velocity of 20 m/s. Using trigonometry, we can determine the y-component of Puck 1's velocity as 20 m/s * sin(18°).
Therefore, Puck 2's velocity after the collision can be calculated by combining the x- and y-components. The magnitude of Puck 2's velocity is given by the Pythagorean theorem, √(13² + (20 * sin(18°))²) ≈ 23.4 m/s. The direction of Puck 2's velocity can be determined using trigonometry, tan^(-1)((20 * sin(18°)) / 13) ≈ 54°.
Hence, after the collision, Puck 2 has a velocity of approximately 23.4 m/s at an angle of 54° north of east.
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Q/C S A glider of mass m is free to slide along a horizontal air track. It is pushed against a launcher at one end of the track. Model the launcher as a light spring of force constant k compressed by a distance x. The glider is released from rest. (c) Is more work done on a cart with a large or a small mass?
More work is done on a cart with a small mass. This relationship arises from the work-energy principle, which states that the work done on an object is equal to the change in its kinetic energy.
To understand why more work is done on a cart with a small mass, let's consider the work-energy principle. According to this principle, the work done on an object is equal to the change in its kinetic energy.
In this scenario, when the glider is released from rest, the compressed spring exerts a force on the glider, accelerating it along the air track. The work done by the spring force is given by the formula:
Work = (1/2) kx²
where k is the force constant of the spring and x is the distance the spring is compressed.
Now, the change in kinetic energy of the glider can be calculated using the formula:
ΔKE = (1/2) mv²
where m is the mass of the glider and v is its final velocity.
From the work-energy principle, we can equate the work done by the spring force to the change in kinetic energy:
(1/2) kx² = (1/2) mv²
Since the initial velocity of the glider is zero, the final velocity v is equal to the square root of (2kx²/m).
Now, let's consider the situation where we have two gliders with different masses, m₁ and m₂, and the same spring constant k and compression x. Using the above equation, we can see that the final velocity of the glider is inversely proportional to the square root of its mass:
v ∝ 1/√m
As a result, a glider with a smaller mass will have a larger final velocity compared to a glider with a larger mass. This indicates that more work is done on the cart with a smaller mass since it achieves a greater change in kinetic energy.
More work is done on a cart with a small mass compared to a cart with a large mass. This is because, in the given scenario, the final velocity of the glider is inversely proportional to the square root of its mass. Therefore, a glider with a smaller mass will experience a larger change in kinetic energy and, consequently, more work will be done on it.
This relationship arises from the work-energy principle, which states that the work done on an object is equal to the change in its kinetic energy. Understanding this concept helps in analyzing the energy transfer and mechanical behavior of objects in systems involving springs and masses.
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ydro Electrical Funda COURSES SCHOOL OF ACCESS AND CONTINUING EDUCA PHYSICS: A REVIEW OF THE PHYSICS YOU WILL NEED TO CO Calculate the capacitive reactance of a capacitor through which 6A flows when 12VAC is applied. Select one: a. 2 ohms b. 0.7 ohms of is page nit 3 Oc. 4 ohms d. 2.7 ohms Jump to... · Next page Unit 4 ► : 7
Calculating the capacitive reactance of a capacitor through which 6A flows when 12VAC is applied.
The capacitive reactance can be calculated as follows: XC = V / I
Where, V = Voltage applied
I = Current flowing
XC = Capacitive reactance
Therefore, substituting the given values,V = 12VACI = 6AXC = V / IXC = 12VAC / 6A = 2 Ω
Thus, the capacitive reactance of a capacitor through which 6A flows when 12VAC is applied is 2 Ω.
The capacitive reactance of a capacitor can be calculated using the formula XC = V / I, where V is the voltage applied, I is the current flowing, and XC is the capacitive reactance. When 12VAC is applied to a capacitor through which 6A flows, the capacitive reactance is 2 Ω.
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Explain it pleaseTwo particles of charge Q are located inside a box. One is at the box center while the other is halfway to one of the corners. Determine the electric flux through the box.
Answer: charge enclosed over epsilon not gives
The electric flux through the box is determined by the charge enclosed within the box divided by the permittivity of free space (ε₀). In this scenario, we have two particles of charge Q, with one located at the center of the box and the other halfway to one of the corners.
Since the charge at the center of the box is equidistant from all sides, it will produce an equal flux through each face of the box. On the other hand, the charge halfway to one of the corners will only contribute to the flux through one face of the box.
Therefore, the total electric flux through the box is given by the charge enclosed, which is the sum of the charges of both particles (2Q), divided by the permittivity of free space (ε₀). Mathematically, it can be expressed as:
Electric Flux = (2Q) / ε₀.
This equation signifies that the electric flux through the box is directly proportional to the total charge enclosed within it. The permittivity of free space (ε₀) is a constant that relates to the ability of the electric field to propagate through a vacuum.
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A stiff wire 41.0 cm long is bent at a right angle in the middle. One section lies along the z axis and the other is along the line y = 2x in the xy plane. A current of 23.5 A flows in the wire-down the z axis and out the line in the xy plane. The wire passes through a uniform magnetic field given by B = (0.318i) T. Determine the magnitude and direction of the total force on the wire.
The magnitude of the total force on the wire is 0.968 N and it is directed along the negative y axis.
What is force?A force is a pull or push upon an object resulting from the object's interaction with another object. Forces can cause an object to change its motion or velocity.
In this case, the wire is experiencing a magnetic force due to the current in the wire and a magnetic field acting on it. To calculate the magnitude and direction of the total force on the wire, we can use the right-hand rule for magnetic forces. According to this rule, if the thumb of the right hand points in the direction of the current, and the fingers point in the direction of the magnetic field, then the palm will point in the direction of the force.
Let's begin by determining the magnitude of the magnetic force on each section of the wire.
Magnetic force on the section of the wire that lies along the z-axis:
Magnetic force on the section of the wire that lies along the line y = 2x in the xy plane:
Now, we need to calculate the total force on the wire by adding up the forces on each section of the wire. Since the forces are at right angles to each other, we can use the Pythagorean theorem to find the magnitude of the total force.
Ftotal² = Fz² + Fy²Ftotal² = (0.288 N)² + (0.792 N)²F
total = 0.849 N
Now, we need to find the direction of the total force. According to the right-hand rule for magnetic forces, the force on the section of the wire that lies along the line y = 2x in the xy plane is directed along the negative y-axis. Therefore, the total force on the wire is also directed along the negative y-axis.
Thus, the magnitude of the total force on the wire is 0.849 N, and it is directed along the negative y-axis.
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A conducting sphere of radius a, having a total charge Q, is
situated in an electric field initially
uniform, E0. Determine the potential at all points outside the
sphere.
The potential at all points outside a conducting sphere of radius a, with a total charge Q, situated in an initially uniform electric field E0, is the same as the potential due to a point charge Q located at the center of the sphere.
The potential is given by the equation V = kQ/r, where V is the potential, k is the electrostatic constant, Q is the charge, and r is the distance from the center of the sphere to the point.
When a conducting sphere is placed in an electric field, the charges on the surface of the sphere redistribute themselves in such a way that the electric field inside the sphere becomes zero.
Therefore, the electric field outside the sphere is the same as the initial uniform electric field E0.
Since the electric field outside the sphere is uniform, the potential at any point outside the sphere can be determined using the formula for the potential due to a point charge.
The conducting sphere can be considered as a point charge located at its center, with charge Q.
The potential V at a point outside the sphere is given by the equation V = kQ/r, where k is the electrostatic constant ([tex]k = 1/4πε0[/tex]), Q is the total charge on the sphere, and r is the distance from the center of the sphere to the point.
Therefore, the potential at all points outside the conducting sphere is the same as the potential due to a point charge Q located at the center of the sphere, and it can be calculated using the equation V = kQ/r.
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The magnetic field of a plane EM wave is given by B = B0 cos(kz
− ωt)i.
Indicate:
a) The direction of propagation of the wave
b) The direction of E.
Given magnetic field of a plane EM wave is: B = B0cos(kz − ωt)i and we need to find the direction of propagation of the wave and the direction of E.
Let’s discuss this one by one.Direction of propagation of the wave: We can find the direction of propagation of the wave from the magnetic field.
The plane EM wave is propagating along the x-axis as ‘i’ is the unit vector along x-axis. The wave is traveling along the positive x-axis because the cosine function is positive
when kz − ωt = 0 at some x > 0.
Thus, we can say the direction of propagation of the wave is in the positive x-axis.Direction of E: The electric field can be obtained by applying Faraday's Law of Electromagnetic Induction.
We know that E = −dB/dt, where dB/dt is the rate of change of magnetic field w.r.t time. We differentiate the given magnetic field w.r.t time to find the
E.E = - d/dt(B0cos(kz − ωt)i) = B0w*sin(kz − ωt)j
Here, j is the unit vector along the y-axis. As we can see from the equation of electric field, the direction of E is along the positive y-axis. Answer:a) The direction of propagation of the wave is in the positive x-axis.b) The direction of E is along the positive y-axis.
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Light is travelling from medium A(refractive index
1.4) to medium B ( refractive index 1.5). If the incident angle is
44.3⁰, what would be the refracted angle in medium B? Express
answer in degrees
The refracted angle in medium B, when light travels from medium A to medium B, is approximately 41.3 degrees.
To find the refracted angle in medium B when light travels from medium A to medium B, we can use Snell's Law. Snell's Law states that the ratio of the sines of the angles of incidence (θ₁) and refraction (θ₂) is equal to the ratio of the refractive indices (n₁ and n₂) of the two mediums:
n₁ * sin(θ₁) = n₂ * sin(θ₂)
In this case, the incident angle (θ₁) is given as 44.3 degrees, and the refractive indices of medium A and medium B are 1.4 and 1.5, respectively.
Let's plug in the values and solve for the refracted angle (θ₂):
1.4 * sin(44.3°) = 1.5 * sin(θ₂)
θ₂ = arcsin((1.4 * sin(44.3°)) / 1.5)
Evaluating the equation, we find that the refracted angle in medium B is approximately 41.3 degrees. Therefore, the refracted angle in medium B is 41.3° (rounded to one decimal place).
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(10%) Problem 2: The image shows a rocket sled, In the top image all four forward thrusters are engaged, creating a total forward thrust of magnitude 47, where T =519 N. In the bottom image, in addition to the four forward thrusters, one reverse thruster is engaged, creating a reverse thrust of magnitude 7. In both cases a backward force (friction and air drag) of magnitude f = 20 Nacts on the sled. 7 What is the ratio of the greater acceleration to the lesser acceleration?
The ratio of the greater acceleration to the lesser acceleration is approximately 0.985.
In the top image where all four forward thrusters are engaged, the total forward thrust exerted on the sled is 519 N. The backward force due to friction and air drag is 20 N. Using Newton's second law, we can calculate the acceleration in this case:
Forward thrust - Backward force = Mass * Acceleration
519 N - 20 N = Mass * Acceleration₁
In the bottom image, in addition to the four forward thrusters, one reverse thruster is engaged, creating a reverse thrust of magnitude 7 N. The backward force of friction and air drag remains the same at 20 N. The total forward thrust can be calculated as:
Total forward thrust = Forward thrust - Reverse thrust
Total forward thrust = 519 N - 7 N = 512 N
Again, using Newton's second law, we can calculate the acceleration this case:
Total forward thrust - Backward force = Mass * Acceleration
512 N - 20 N = Mass * Acceleration₂
To find the ratio of the greater acceleration (Acceleration₂) to the lesser acceleration (Acceleration₁), we can divide the equations:
(Acceleration₂) / (Acceleration₁) = (512 N - 20 N) / (519 N - 20 N)
Simplifying the expression, we get:
(Acceleration₂) / (Acceleration₁) = 492 N / 499 N
(Acceleration₂) / (Acceleration₁) ≈ 0.985
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In conservation of energy experiment, the relation between the hanging mass (m) and the increase in the length (x) is given by: mg =kx Where (k) is the spring constant and (g) is the acceleration due to gravity (g = 9.81 m/s2). The graph below shows (m vs. x) for three different springs. Which spring has the lowest spring constant (ks)? Spring 2 Spring 3 Spring 1 All the springs have the same Tidliging mass (m) and the increase in the length (x) is given mg =kx Spring (slope - 2km) Spring 2 (slope 1.7km) Spring 3 slope 2.5km) g) is the acceleration due to gravity (g = 9.81 m/s2). The graph below shows (m vs.
Spring 2 has the lowest spring constant among the three springs in the experiment.
In the given conservation of energy experiment, the relation between the hanging mass (m) and the increase in length (x) is given by mg = kx, where k is the spring constant and g is the acceleration due to gravity (9.81 m/s²).
The graph provided shows the relationship between m and x for three different springs. To determine which spring has the lowest spring constant, we need to compare the slopes of the graph lines. The spring with the lowest slope, which represents the smallest value of k, has the lowest spring constant.
The slope of the graph represents the spring constant (k) in the relation mg = kx. A steeper slope indicates a higher spring constant, while a flatter slope indicates a lower spring constant. Looking at the graph lines for the three springs, we can compare their slopes to determine which one has the lowest spring constant.
If the slope of Spring 1 is 2k, the slope of Spring 2 is 1.7k, and the slope of Spring 3 is 2.5k, we can conclude that Spring 2 has the lowest spring constant (ks). This is because its slope is the smallest among the three, indicating a smaller value for k.
Therefore, Spring 2 has the lowest spring constant among the three springs in the experiment.
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A spaceship, 230-m long for those on board, moves by the Earth at 0.955c. What is its length as measured by an earthbound observer
The length of the spaceship as measured by an earthbound observer is approximately 68.69 meters.
To calculate the length of the spaceship as measured by an earthbound observer, we can use the Lorentz transformation for length contraction:
L' = L × sqrt(1 - (v²/c²))
Where:
L' is the length of the spaceship as measured by the earthbound observer,
L is the proper length of the spaceship (230 m in this case),
v is the velocity of the spaceship relative to the earthbound observer (0.955c),
c is the speed of light.
Substituting the given values:
L' = 230 m × sqrt(1 - (0.955c)²/c²)
To simplify the calculation, we can rewrite (0.955c)² as (0.955)² × c²:
L' = 230 m × sqrt(1 - (0.955)² × c²/c²)
L' = 230 m × sqrt(1 - 0.911025)
L' = 230 m sqrt(0.088975)
L' = 230 m × 0.29828
L' = 68.69 m
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2. A shell is fired from a cliff horizontally with initial velocity of 800 m/s at a target on the ground 150 m below. How far away is the target? ( 2 pts) 3. You are standing 50 feet from a building and throw a ball through a window that is 26 feet above the ground. Your release point is 6 feet off of the ground (hint: you are only concerned with Δ y). You throw the ball at 30ft/sec. At what angle from the horizontal should you throw the ball? (hint: this is your launch angle) (2pts)
Horizontal displacement = 4008 meters
The launch angle should be approximately 20.5°
To find how far away the target is, the horizontal displacement of the shell needs to be found.
This can be done using the formula:
horizontal displacement = initial horizontal velocity x time
The time taken for the shell to reach the ground can be found using the formula:
vertical displacement = initial vertical velocity x time + 0.5 x acceleration x time^2
Since the shell is fired horizontally, its initial vertical velocity is 0. The acceleration due to gravity is 9.8 m/s^2. The vertical displacement is -150 m (since it is below the cliff).
Using these values, we get:-150 = 0 x t + 0.5 x 9.8 x t^2
Solving for t, we get:t = 5.01 seconds
The horizontal displacement is therefore:
horizontal displacement = 800 x 5.01
horizontal displacement = 4008 meters
3. To find the launch angle, we can use the formula:
Δy = (v^2 x sin^2 θ)/2g Where Δy is the vertical displacement (26 ft), v is the initial velocity (30 ft/s), g is the acceleration due to gravity (32 ft/s^2), and θ is the launch angle.
Using these values, we get:26 = (30^2 x sin^2 θ)/2 x 32
Solving for sin^2 θ:sin^2 θ = (2 x 26 x 32)/(30^2)sin^2 θ = 0.12
Taking the square root:sin θ = 0.35θ = sin^-1 (0.35)θ = 20.5°
Therefore, the launch angle should be approximately 20.5°.
Note: The given measurements are in feet, but the calculations are done in fps (feet per second).
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A block of mass 1.30 kg is placed on a frictionless floor and initially pushed northward, whereupon it begins sliding with a constant speed of 5.12 m/s. It eventually collides with a second, stationary block, of mass 4.82 kg, head-on, and rebounds back to the south. The collision is 100% elastic. What will be the speeds of the 1.30-kg and 4.82-kg blocks, respectively, after this collision?
2.05 m/s and 2.56 m/s
1.18 m/s and 2.75 m/s
2.94 m/s and 2.18 m/s
2.18 m/s and 2.94 m/s
To solve this problem, we can use the principle of conservation of momentum and the principle of conservation of kinetic energy.
Before the collision, the total momentum of the system is the sum of the momenta of the two blocks. After the collision, the total momentum remains the same.
Let's denote the initial velocity of the 1.30 kg block as v1i and the initial velocity of the 4.82 kg block as v2i. Since the 1.30 kg block is initially pushed northward, its velocity is positive, while the 4.82 kg block is stationary, so its initial velocity is 0.
Using the conservation of momentum:
(m1 × v1i) + (m2 × v2i) = (m1 × v1f) + (m2 × v2f)
Since the collision is elastic, the total kinetic energy before and after the collision remains the same. The kinetic energy equation can be written as:
0.5 × m1 × (v1i)^2 + 0.5 × m2 × (v2i)^2 = 0.5 × m1 × (v1f)^2 + 0.5 × m2 × (v2f)^2
We can solve these two equations simultaneously to find the final velocities (v1f and v2f) of the blocks after the collision.
Substituting the given masses (m1 = 1.30 kg and m2 = 4.82 kg) and initial velocity values into the equations, we find that the speeds of the 1.30 kg and 4.82 kg blocks after the collision are approximately 2.18 m/s and 2.94 m/s, respectively. Therefore, the correct answer is 2.18 m/s and 2.94 m/s.
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