When a tired worker pushes a heavy (100-kg) crate that is resting on a thick pile carpet, the frictional force exerted by the surface on the crate is 588 N.
When a tired worker pushes a heavy (100-kg) crate that is resting on a thick pile carpet, the frictional force exerted by the surface can be calculated as follows:
The weight of the crate = m × g = 100 kg × 9.8 m/s² = 980 N
Force applied by the worker = F = 600 N
The force of friction acting on the crate is given by the following formula:
Ff = μF
Where, μ is the coefficient of friction, F is the normal force acting on the crate.
Notes: The normal force is equal and opposite to the weight of the crate. i.e., N = 980 N1. The frictional force exerted by the surface on the crate is the static frictional force initially. Hence, we use the coefficient of static friction for our calculation.
2. If the force applied by the worker is not enough to overcome the static frictional force, then the crate will not move and the frictional force will remain static friction.
3. Once the crate starts moving, the static friction will convert to kinetic friction. Hence, we will use the coefficient of kinetic friction if the force applied by the worker is greater than the force of static friction. Initially, the force applied by the worker is less than the force of static friction, hence the frictional force exerted on the crate will be the static frictional force.
Frictional force = Ff = μN
The normal force acting on the crate = Weight of the crate = 980 N
Frictional force =
Ff = μN
= 0.6 × 980 N
= 588 N
Therefore, the frictional force exerted by the surface on the crate is 588 N.
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uppose you hit a 0.058- kg tennis ball so that the ball then moves with an acceleration of 10 m/s2 . if you were to hit a basketball of mass 0.58 kg with the same force, what would the acceleration a of the basketball be? express your answer in meters per second squared.
The acceleration of the basketball would be the same as the tennis ball, 10 m/s2. This is because in this scenario, the acceleration of the two objects is determined by the same force, and not the mass of the object.
Acceleration is the rate of change of an object's velocity given by the equation a = F/m, where F is the force applied, and m is the mass of the object. Since the force is the same in both cases, the acceleration of the two objects must also be the same. This means that the basketball, which has a mass of 0.58 kg, will still accelerate at 10 m/s2.
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gold has a density of 0.01932 kg/cm^3. what volume in cm^3 would be occupied by a 77.7 g sample of gold
77.7 g sample of gold will occupy a volume of 4 cm3.
A 77.7 g sample of gold will occupy a volume of 4 cm3, as calculated by using the equation D = m/V, where D is the density of gold (0.01932 kg/cm3), m is the mass of gold (77.7 g), and V is the volume of gold (4 cm3).
The mass needs to be converted from grams to kilograms, and the volume needs to be calculated. Divide the mass of gold (77.7 g) by 1,000:
m (in kg) = 77.7 g ÷ 1,000 = 0.0777 kg
V = m/D
V = 0.0777 kg ÷ 0.01932 kg/cm3 = 4 cm3
Therefore, a 77.7 g sample of gold will occupy a volume of 4 cm3.
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if you rub a balloon against your head, then electrons from the atoms that make up your hair get transferred to the balloon. the balloon becomes negatively charged and your hair becomes positively charged. what happens if you place balloon by hair?
When you rub a balloon against your head, electrons from the atoms in your hair are transferred to the balloon. This causes the balloon to become negatively charged, while your hair becomes positively charged. If you then place the balloon near your hair, the negative charge of the balloon will be attracted to the positive charge of your hair, causing the two to stick together. This phenomenon is known as electrostatic attraction.
The attraction of the negative charge of the balloon to the positive charge of your hair creates a strong force that causes the two objects to stick together. This force is known as the electrostatic force of attraction. It is the same force that makes two magnets stick together when their poles are placed near each other. The attraction between the balloon and your hair will remain until the charge on the balloon is dissipated by contact with another object.
To demonstrate this force of attraction, you can try rubbing the balloon against your head and then holding it near your hair. You will notice that the balloon will become attracted to your hair and will stick to it. You can also experiment with other materials that become charged when rubbed together, such as a cloth and a comb.
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a reservoir behind a dam is 15 m deep. what is the pressure a. at the base of the dam? b. 5.0 m from the top of the dam?
a. The pressure at the base of the dam is 147.15 kPa.
b. The pressure 5.0 m from the top of the dam is 98.1 kPa.
The pressure at the base of the dam can be calculated using the formula:
P = ρgh
where P is the pressure, ρ is the density of the liquid, g is the acceleration due to gravity, and h is the depth of the liquid.
Assuming the density of water is 1000 kg/m³ and acceleration due to gravity is 9.81 m/s², the pressure at the base of the dam is:
P = 1000 x 9.81 x 15
P = 147,150 Pa or 147.15 kPa
Therefore, the pressure at the base of the dam is 147.15 kPa.
b. To calculate the pressure 5.0 m from the top of the dam, we can use the formula:
P = ρgh
where h is the depth of the liquid from the surface to the point where we want to calculate the pressure. In this case, h = 15 - 5 = 10 m.
Using the same values for density and acceleration due to gravity, the pressure at 5.0 m from the top of the dam is:
P = 1000 x 9.81 x 10
P = 98,100 Pa or 98.1 kPa
Therefore, the pressure 5.0 m from the top of the dam is 98.1 kPa.
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1) For the diagram below, you are looking down from above at a puck sliding on a frictionless
tabletop. The puck is sliding at a constant speed along a circular fence and just leaves
contact with the fence at position B. The fence and puck are also frictionless.
B
B
a) On the top drawing draw vectors and label them
for
any forces acting parallel to the table top on
the puck at positions A and B.
b) On the side views below, draw vectors and label
them for any forces acting perpendicular to the
tabletop.
side view at A
side view at B
c) Which of the forces above are inward forces?
d) Which of the forces above are outward forces?
e) What would be the equation for the net force on
the puck?
f) On the top drawing at position B draw a line at least
an inch long showing the path the puck will travel
after point B. How do you know?
g) On the bottom drawing, draw and label a velocity
and an acceleration vector on the puck at positions
A and B.
h) At position A is the net force on the puck zero?
How do you know?
i) At position B is the net force on the puck zero?
How do you know?
c the forces that are inward forces are ;F3 and f4
d. The centrifugal force is the outward force
e. equation is force = mu²/r
F. there would be no restriction at B
How to solve the problemsd. Along with circular motion, there will be centrifugal force (pseudo force) in the outer direction. It will exert an opposite force to that of inward forces.
(f) The path will not be constrained after point B. As a result, until a force acts on the puck, it will move in a straight line in accordance with Newton's first law. It will follow the straight line you have shown in the image since it is moving at a constant speed, v.
(g) The velocity vector will always move in the circular path's tangential direction (tangent to the circular fence). The circular path's center is always where the centripetal acceleration will act (in the normal direction of the circular path).
(h & I). The puck is moving in a circular motion while just altering its speed's direction and not its magnitude. As a result, the puck will experience a net force and acceleration. The net force is referred to as a centripetal or inward force. Without such an internal force, an object would always move in the same direction and stay in a straight line. This will clarify the two situations where the puck was in position A and B.
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an electron starts from rest a distance of 42 cm from a fixed point charge of 0.128 how fast will electron be moving when it is very far away
The speed of the electron when it is very far away from the point charge of 0.128 depends on the amount of energy it has gained from the electric field. As the electron moves closer to the charge, the electric field gets stronger and the electron accelerates. By the time the electron reaches a distance of 42 cm from the point charge, it has gained enough energy from the electric field to reach a velocity of 8.97 x 106 m/s.
As the electron moves away from the point charge, the strength of the electric field decreases and the electron starts to decelerate. Eventually, the electric field will become so weak that the electron reaches a point where its speed stops decreasing and stabilizes. This point is referred to as the “asymptote”, and the speed of the electron at this point is known as the “asymptotic velocity”.
The asymptotic velocity of the electron can be calculated using the formula: V asymptotic = (2q/m)1/2, where q is the charge of the electron and m is its mass.
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Do any of the force pairs suggested in Question 5 not produce an acceleration? If so which one(s).
A. A skier uses her ski poles to start moving downhill
B. A boat propeller spins rapidly in the water
C. A baseball player hits a pitched ball with a bat
D. A party balloon contains rapidly moving helium atoms
All of the given options produce an acceleration that are force pairs suggested in Question 5.
When a skier uses her ski poles to start moving downhill then the ski poles exert a backward force on the ground while the ground exerts a forward force on poles and produces acceleration.
Similarly in case B. when a boat propeller spins rapidly in the water the propeller exert a backward force on the water while the water exerts a forward force on propeller and produces acceleration.
In case C. when a baseball player hits a pitched ball with a bat the bat exert a backward force on the ball while the ball exerts a force away from bat and produces acceleration.
In case D. when a party balloon contains rapidly moving helium atoms the helium atoms exert an outward force on the balloon while the balloon exerts an inward force on helium atoms and produces acceleration.
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find an expression for the magnitude of the force on the lower book by the upper book in the elevator situation. is it equal to the weight of the upper book? should it be?
Yes, the magnitude of the force on the lower book by the upper book in the elevator situation is equal to the weight of the upper book. This can be expressed as F = mg,
What is magnitude expression?The expression for the magnitude of the force on the lower book by the upper book in the elevator situation is;
F(lower on upper) = (m(lower) + m(upper))g
Where F(lower on upper) is the magnitude of the force on the lower book by the upper book,
m(lower) is the mass of the lower book,
m(upper) is the mass of the upper book and
g is the acceleration due to gravity.
Therefore, the magnitude of the force on the lower book by the upper book is not equal to the weight of the upper book since it takes into account the mass of the lower book as well.
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a .1 kg ball traveling 20 m/s is caught by a catcher. in stopping the ball, the mitt recoils for .01 second. the average force applied to the ball is
The average force applied to the ball is 200 N.
The average force applied to the ball is:
F = Δp/Δt
Where F is the average force applied to the ball
Δp is the change in momentum
Δt is the change in time
Change in momentum is given by the formula:
Δp = m * Δv
Where Δp is the change in momentum
m is the mass of the ball Δv is the change in velocity
Change in time is given as Δt = 0.01 s
The mass of the ball is 0.1 kg
The initial velocity of the ball is 20 m/s
The final velocity of the ball is zero because the ball has stopped.
Δv = -20 m/s
Substitute the values in the formula,
Δp = m * ΔvΔp = 0.1 * (-20)Δp = -2 Ns
F = Δp/ΔtF
= (-2 Ns) / (0.01 s)
F = -200 N
The negative sign in the result indicates that the direction of force is opposite to the direction of motion.
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water flows through a pipe with a cross-sectional area of 0.002 m2 at a mass flow rate of 4 kg/s. the density of water is 1 000 kg/m3. determine its average velocity. multiple choice question. 20 m/s 200 m/s 0.02 m/s 2 m/s 0.2 m/s
Option D: 2 m/s is the average velocity of the water flowing through a pipe with a cross-sectional area of 0.002 m2 at a mass flow rate of 4 kg/s.
According to the question:
cross-sectional area of the pipe = 0.002m²
Mass flowrate = 4 kg/s
Density of water = 1000 kg/m³
We are asked to find, average velocity =?
Average velocity is the net or total displacement covered by a body in a given time. The mass flow rate divided by the pipe's cross-sectional area and density ratio is the formula for calculating a fluid's average velocity.
As a result, the water's average flow rate through the pipe is provided by:
v = m / (ρ × A)
where, v is the average velocity, m is the mass flow rate, ρ is the density of water, and A is the cross-sectional area of the pipe. Substituting the values in the above equation we get:
v = 4 / (1000 × 0.002)
v = 2m/s
Therefore, the average velocity of water flowing through a pipe of cross-sectional area of 0.002m² is 2m/s.
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Correct question is:
Water flows through a pipe with a cross-sectional area of 0.002 m2 at a mass flow rate of 4 kg/s. The density of water is 1 000 kg/m3. Determine its average velocity. Multiple choice question.
20 m/s
200 m/s
0.02 m/s
2 m/s
0.2 m/s
1. A string of length 1.23 m vibrates in such a way that a standing wave with wavelength 0.820 m forms. What harmonic is the string vibrating at?
Answer:
1. A string of length 1.23 m vibrates in such a way that a standing wave with wavelength 0.820 m forms. What harmonic is the string vibrating at?
what is the relationship between velocity of moving body and the force acting on it?
Answer:
The relation between the momentum of a body and the force acting on it is that the rate of change of momentum of a body is directly proportional to the applied force and takes place in the direction of the force acting.
Blocks A (mass 3.50 kg) and B (mass 10.00 kg) move on a frictionless, horizontal surface. Initially, block B is at rest and block A is moving toward it at 9.00 m/s. The blocks are equipped with ideal spring bumpers. The collision is head-on, so all motion before and after the collision is along a straight line. Let +x be the direction of the initial motion of A. Find the maximum energy stored in the spring bumpers and the velocity of each block at the time of the collision
The total energy that can be stored in the spring bumpers is 43.8 J, or KE = 43.8.
What is the formula for energy capacity?The battery's power capacity is the amount of energy it can hold. Its power is commonly stated in Watt-hours (the symbol Wh) (the symbol Wh). A Watt-hour is equal to the voltage (V) and current (Amps) that a battery can produce for a specific period of time (generally in hours). Voltage * Amps * hours = Wh.
Block A's momentum before to the impact can be calculated using the formula p1 = m1v1 = (3.50 kg)(9.00 m/s) = 31.5 kgm/s.
Block B's initial momentum is p2 = m2v2 = 0, indicating that it is at rest.
Prior to the collision, the system's total momentum was equal to 31.5 kgm/p1 + p2.
[tex]p1 + p2 = (m1 + m2)v[/tex]
[tex]31.5 kgm/s = (3.50 kg + 10.00 kg) * v[/tex]
[tex]31.5 kgm/s = 13.50 kg * v[/tex]
[tex]v = 31.5 kg*m/s / 13.50 kg = 2.33 m/s[/tex]
The kinetic energy of block A before the collision is given by KE1 = ([tex]1/2)m1v1^2 = (1/2)(3.50 kg)(9.00 m/s)^2 = 141.8[/tex] J
The kinetic energy of block B before the collision is KE2 = [tex](1/2)m2v2^2 = 0[/tex]
The total kinetic energy before the collision is KE1 + KE2 = 141.8 J
[tex]ΔKE = KEf - KEi = (1/2)(m1 + m2)v^2 - KE1 - KE2[/tex]
[tex]ΔKE = (1/2)(3.50 kg + 10.00 kg)(2.33 m/s)^2 - 141.8 J - 0[/tex]
ΔKE = 43.8 J[tex]31.5 kgm/s = 13.50 kg * v[/tex]
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gravitational potential energy when potential energy0.25 kg ball is suspended from a light spring.find gravitaional potential energy.g
The gravitational potential energy of a 0.25 kg ball suspended from a light spring is 0.6125 J.
Calculate the gravitational potential energy below.
The gravitational potential energy formula is as follows:
GPE = mgh
where m is the mass of the object, g is the acceleration due to gravity, and h is the height of the object from a reference point.
Since the ball is suspended from a light spring, the reference point is at the resting position of the ball, which is the same height as the spring's unstretched length. As a result, h is the length of the stretched spring.
Assume the stretched length of the light spring is 0.25 m, and the acceleration due to gravity is [tex]9.8 m/s^2[/tex].
GPE = mgh
= 0.25 × 9.8 × 0.25
= 0.6125 J
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pry on the power steering reservoir to adjust the tension of the power steering belt. true or false?
The statement "pry on the power steering reservoir to adjust the tension of the power steering belt" is: false.
The tension of the power steering belt is adjusted by adjusting the position of the power steering pump. There is a tension adjustment bolt on the power steering pump that is used to adjust the tension of the power steering belt. The adjustment bolt should be turned clockwise or counterclockwise to adjust the tension of the belt.
A belt tension gauge may be used to ensure that the belt is properly tensioned. A pry bar should not be used on the power steering reservoir to adjust the tension of the power steering belt. This could cause damage to the reservoir or other components of the power steering system. The reservoir should be inspected for damage or leaks, but it should not be used to adjust the tension of the belt.
In summary, the tension of the power steering belt should be adjusted by adjusting the position of the power steering pump, not by prying on the power steering reservoir.
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air is compressed in a piston-cylinder device from 5 m3 to 3 m3 at a constant pressure of 1000 kpa. determine the amount of boundary work for the process.
The amount of boundary work for the given process is −2000 kJ.
For the piston-cylinder device, the values are as follows:
The initial volume, V₁ = 5m³, Final volume, V₂ = 3m³, Pressure, P = 1000 kPa.
We need to determine the amount of boundary work for the given process. The boundary work is represented as Wb.
Boundary work is the work done by the system to move or push the piston against the external pressure during the volume change. Boundary work,
Wb = P × (V₂ − V₁)
Here, P is the pressure, and (V₂ − V₁) is the change in volume.
Substituting the given values in the formula,
Wb = 1000 kPa × (3 m³ − 5 m³)
Wb = 1000 kPa × (−2 m³)
Wb = −2000 kJ.
Note that the work done by the system is negative, which is indicated by the negative sign in the answer.
Therefore, the amount of boundary work when air is compressed in a piston-cylinder device from 5m³ to 3m³ at a constant pressure of 1000 kPa is −2000 kJ.
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Please help. Due at Midnight!
The magnitude and direction of the net force on the center charge is 3.929 x 10⁻⁴ N.
What is unit of charge?The unit of charge is the Coulomb (C). It is named after French physicist Charles-Augustin de Coulomb and is defined as the amount of electric charge that flows through a circuit when a current of one ampere flows for one second. One Coulomb is also equivalent to the charge on approximately 6.24 x 10¹⁸ electrons. The Coulomb is one of the seven base SI units (International System of Units) and is used to measure electric charge in physics and engineering.
So, the magnitude of the net force on the center charge is 3.929 x 10⁻⁴ N. Since F12 is directed towards the left, and F23 is directed towards the right, the net force is also directed towards the left. Therefore, the direction of the net force on the center charge is to the left.
According to Coulomb's law to calculate the force exerted by each of the other charges on the center charge, and then add them vectorially.
Let's call the left charge Q1, the center charge Q2, and the right charge Q3.
The force exerted on Q2 by Q1 is given by:
F₁₂ = k * |Q1| * |Q2| / r₁₂²
where k is Coulomb's constant, |Q1| and |Q2| are the magnitudes of the charges, and r₁₂ is the distance between them. Since Q1 is positive and Q2 is negative, the force F₁₂ is attractive and directed towards Q1. Because the distance between them is 2m, we can say:
F₁₂ = 9 x 10⁹ Nm²/C² * |52 x 10⁻⁶ C| * |3.10 x 10⁻⁶ C| / (2m)²
= 3.468 x 10⁻⁴ N (attractive)
The force exerted on Q2 by Q3 is given by:
F₂₃ = k * |Q2| * |Q3| / r₂₃²
where |Q3| is positive, and |Q2| is negative, so the force F23 is repulsive and directed away from Q3. The distance between them is also 2m, so:
F₂₃ = 9 x 10⁹ Nm²/C² * |3.10 x 10⁻⁶ C| * |68 x 10⁻⁶ C| / (2m)²
= 5.383 x 10⁻⁵ N (repulsive)
To find the net force on Q2, we need to add these two forces vectorially. Since they act along the same line, we can simply subtract their magnitudes:
Fnet = |F₁₂| - |F₂₃|
= 3.468 x 10⁻⁴ N - 5.383 x 10⁻⁵N
= 3.929 x 10⁻⁴ N.
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an object moves around a circular path at a constant speed and makes ten complete revolutions in 5 seconds. what is the frequency of rotation?
Answer:
circular path, 10 rotation in 5 sec
frequency of rotation = 10/5 = 2
Explanation:
The frequency of rotation for an object moving around a circular path at a constant speed and making ten complete revolutions in 5 seconds is 2 Hz (Hertz). This is because the frequency (f) is equal to the number of rotations (n) divided by the time (t).
In this case, n = 10 and t = 5, so the frequency is f = n/t = 10/5 = 2 Hz.
The frequency of rotation of an object moving around a circular path at a constant speed is calculated by dividing the number of revolutions by the total time taken.
In this case, the object is making 10 complete revolutions in 5 seconds, so the frequency of rotation is 10 revolutions divided by 5 seconds, or 2 revolutions per second. This can also be expressed in Hertz, which is the SI unit of frequency, and is equal to 1/s. In this case, the frequency is 2 Hertz. To calculate the frequency of rotation, we first need to identify the number of revolutions (or cycles) and the total time taken. Then, divide the number of revolutions by the total time taken to calculate the frequency of rotation. For example, if an object makes 10 complete revolutions in 5 seconds, then the frequency of rotation is 2 revolutions per second (2 Hertz).
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a 23.9 a current flows in a long, straight wire. find the strength of the resulting magnetic field at a distance of 58.3 cm from the wire.
The magnetic field at a distance of 58.3 cm from a long, straight wire carrying a 23.9 A current, the strength of the resulting magnetic field can be found using the equation B = μ0*I/2π*r, where B is the magnetic field strength, μ0 is the permeability of free space, I is current, and r is the distance.
Therefore, the strength of the magnetic field at 58.3 cm from the wire is B = 4π * 10-7 * 23.9/2π * 58.3 = 0.0067 N/Amp.
The magnetic field strength due to the current in the wire is caused by the current producing a magnetic field, which is a result of moving electric charges (electrons) in the wire. The strength of the magnetic field depends on the magnitude of the current and the distance from the wire.
As the current increases, the magnetic field strength increases; likewise, as the distance from the wire increases, the magnetic field strength decreases. The direction of the magnetic field can be determined using the right-hand rule.
The strength of the magnetic field can be used to calculate the force on a moving charged particle, F = q * v * B, where q is the charge of the particle, v is its velocity, and B is the magnetic field strength. By using this equation, the force acting on a charged particle due to the magnetic field at 58.3 cm from the wire can be found.
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(1) a father racing his son has 1/4 the kinetic energy of the son, who has 1/2 the mass of the father. the father speeds up by 1.7 m/s and then has the same kinetic energy as the son. what are the original speeds of (a) the father and (b) the son?
The original speeds of the father and the son are 1.9 m/s and 5.374 m/s.
Mass of the son = m1
Mass of the father = m2
The son has 1/2 the mass of the father.
m_2 = 2m_1
Original speed of the son = V1
Original speed of the father = V2
Initial kinetic energy of the son = E_1 = m_1[tex]V_1^2[/tex]/2
Initial kinetic energy of the father = E_2 = m_2[tex]V_2^2[/tex] /2
The father has 1/4 the kinetic energy of the son.
E_1 = 4E_2
m_1[tex]V_1^2[/tex]/2 = 4(m_2[tex]V_2^2[/tex] /2)
m_1[tex]V_1^2[/tex]/2 = 2m_2[tex]V_2^2[/tex]
New speed of the father = V_3 = V_2 + 1.9
New kinetic energy of the father = E_3 = m_2 [tex]V_3^2[/tex]/2
The new kinetic energy of the father is equal to the kinetic energy of the son.
E_1 = E_3
m1[tex]V_1^2[/tex]/2 = m2 [tex]V_3^2[/tex]/2
2m2[tex]V_2^2[/tex] = m2 [tex]V_3^2[/tex]/2
4[tex]V_2^2[/tex] = [tex]V_3^2[/tex]
2[tex]V_2[/tex] = [tex]V_3[/tex]
2[tex]V_2[/tex] = [tex]V_2[/tex] + 1.9
[tex]V_2[/tex] = 1.9 m/s
m_1[tex]V_1^2[/tex]/2 = 2m2[tex]V_2^2[/tex]
m_1[tex]V_1^2[/tex]/2 = 2(2m1)(1.9)2
V_1 = 5.374 m/s
a) Original speed of the father = 1.9 m/s
b) Original speed of the son = 5.374 m/s
Speed refers to the rate at which something moves or operates. It can be described as the distance traveled by an object over a certain period of time, or the number of events or operations that occur within a given time frame. Speed is a fundamental concept in physics and is often measured in units such as meters per second, miles per hour, or revolutions per minute.
In everyday life, speed is important in a variety of contexts. For example, the speed of a vehicle can affect its safety and efficiency. The speed at which a computer operates can impact its performance. The speed of communication, such as the transfer of data or the sending of messages, can impact the effectiveness of communication.
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Complete Question: -
A father racing his son has 1/4 the kinetic energy of the son, who has 1/2 the mass of the father. The father speeds up by 1.9 m/s and then has the same kinetic energy as the son. What are the original speeds of (a) the father and (b) the son?
calculate the speed of the second ship with respect to earth if it is fired in the same direction the first spaceship is already moving.
The speed of the second ship is fired in the same direction as the first ship, and the relative velocity of the second ship with respect to the first ship is zero.
To calculate the speed of the second ship with respect to Earth if it is fired in the same direction as the first spaceship is already moving, the formula of relative velocity is used.
The relative velocity formula is V₂ = V₁ + V, where V₂ is the velocity of the second ship, V₁ is the velocity of the first ship, and V is the velocity of the second ship relative to the first ship.
Since the second ship is fired in the same direction as the first ship, the relative velocity is just the difference between the two velocities. The velocity of the first ship is not given, so the answer will be given in terms of relative velocity only.
The speed of the second ship with respect to Earth is the velocity of the second ship plus the velocity of the first ship relative to Earth.
The speed of the second ship with respect to Earth is just the speed of the first ship plus the speed of the second ship.
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a 5100-pound vehicle is driven at a speed of 30 miles per hour on a circular interchange of radius 100 feet. to keep the vehicle from skidding off course, what frictional force must the road surface exert on the tires?
The frictional force must the road surface exert on the tires is 58.667 ft / s.
Weight of the vehicle W = 5600 lb
Speed v = 40 miles/h
Radius of circular interchanger = 100 feet.
mass of the vehicle m= w/g = 5600 lb / 32 ft/s2
= m = 175 lb s2 / ft.
Speed of the vehicle V = ds/dt.
V = 40 miles/h 1mile = 5280ft
=40 x 5280 ft / 3600 S
V = 58.667 ft / s
Also curvature k = 1/r = 100ft.
when a vehicle in moving along a circular track, the tyres have a tendancy to slip outwards So to avoid skidding the surface exerts frictional force on the times towards the cente
frictional force F = m x normal component of acceleration
= m x an.
where a_N = k (ds/dt)^2 = kv^2.
F = mk v^2.
Frictional force is a force that opposes the relative motion or tendency of motion between two surfaces in contact. It arises due to the roughness and irregularities present on the surfaces in contact.Static frictional force is the force that prevents two objects from moving relative to each other when a force is applied to them. It is always equal and opposite to the applied force until the maximum value of static frictional force is reached.
Kinetic frictional force is the force that opposes the motion of two surfaces sliding over each other. It is generally less than the maximum static frictional force. The magnitude of frictional force depends on various factors such as the nature of the surfaces in contact, the normal force acting between them, the temperature, and the presence of any lubricants.
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Complete Question: -
A 5600-pound vehicle is driven at a speed of 40 miles per hour on a circular interchange of radius 100 feet. To keep the vehicle from skidding off course, what frictional force must the road surface exert on the tires? (Round your answer to one decimal place.)
1st attempt rolling-circle replication of plasmids proceeds choose one: in opposite directions from multiple origin sites. in one direction from multiple origin sites. in one direction from a single fixed origin. in opposite directions from a single fixed origin.
Based on the given options, the correct answer is: "in one direction from a single fixed origin."
The 1st attempt rolling-circle replication of plasmids proceeds in one direction from a single fixed origin.
This process involves the initiation of DNA replication from a specific origin site on the plasmid.
The replication then proceeds in a circular direction, generating multiple copies of the plasmid.
Overall, plasmids are small, circular pieces of DNA that are separate from the chromosome.
They replicate independently of the chromosome and can carry genes that provide a selective advantage to the cell, such as antibiotic resistance.
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x < If a heater is used for 2 hours and an electric motor for 4 hours, they consume 25 kJ of energy. If the heater is used for 3 hours and the electric motor for 2 hours, they consume 18 kJ of energy. Calculate the energy consumption per hour of the heater and of the electric motor
The energy consumption per hour of the heater is 9 kJ/hour and the energy consumption per hour of the electric motor is 3 kJ/hour.
What is the energy consumption rate?Let's denote the energy consumption per hour of the heater as "h" and the energy consumption per hour of the electric motor as "m".
From the first piece of information, we can set up the equation:
2h + 4m = 25 (equation 1)
Similarly, from the second piece of information, we can set up another equation:
3h + 2m = 18 (equation 2)
We now have two equations with two unknowns, which we can solve using algebraic methods. Multiplying equation 2 by 2 and subtracting it from equation 1 multiplied by 3, we get:
(3h + 6m) - 2(3h + 2m) = 25(3) - 18(2)
Simplifying this expression, we get:
h = 9
Substituting this value of h into equation 2, we get:
3(9) + 2m = 18
Simplifying this expression, we get:
m = 3
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what is the maximum speed with which a 1200- kg car can round a turn of radius 85.0 m on a flat road if the coefficient of static friction between tires and road is 0.60?
When you want to find the maximum speed with which a 1200- kg car can round a turn of radius 85.0 m on a flat road if the coefficient of static friction between tires and road is 0.60, you can use the following formula: v = sqrt(μrg).
Where:v represents the maximum speed with which the car can round a turn r is the radius of the turn g is the acceleration due to gravity, andμ is the coefficient of frictionIn this case, the mass of the car is 1200 kg and the radius of the turn is 85.0 m, while the coefficient of friction is 0.60.
To find the acceleration due to gravity, we can use the value 9.81 m/s². Therefore:v = sqrt(0.60 * 9.81 m/s² * 85.0 m) = 23.7 m/sTherefore, the maximum speed with which the 1200-kg car can round a turn of radius 85.0 m on a flat road if the coefficient of static friction between tires and road is 0.60 is approximately 23.7 m/s.
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the generator of a car idling at 1300 rpm produces 14.9 v . part a what will the output be at a rotation speed of 2100 rpm , assuming nothing else changes? express your answer to three significant figures and include the appropriate units.
The output of the car generator at a rotation speed of 2100 rpm would be 24.1 V, assuming that nothing else changes.
To calculate the output of the generator, we can use the formula:-
Output Voltage = (RPM/1300) x 14.9 V
We know that the generator is rotating at 2100 rpm, so the output voltage is:-
Output Voltage = (2100/1300) x 14.9 V= 24.09 V (rounded to three significant figures)
Therefore, the output voltage of the car generator at a rotation speed of 2100 rpm is 24.1 V (rounded to one decimal place), assuming nothing else changes. The appropriate unit of voltage is volts (V).
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the force of gravity on the side of the earth facing the moon is the force of gravity acting on the center of the earth group of answer choices greater than smaller than equal to
The force of gravity on the side of the earth facing the moon is greater than the force of gravity acting on the center of the earth.
This is because of the gravitational attraction between the earth and the moon.
The moon’s gravity pulls on the side of the earth that is closer to it, resulting in a larger gravitational force on that side than on the center of the earth. The size of the force on the side of the earth is slightly more than double that at the center, due to the inverse square law.
Thus, the force of gravity at the side of the earth facing the moon is greater than the force of gravity acting on the center of the earth.
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Complete question:
The force of gravity on the side of the earth facing the moon is the force of gravity acting on the center of the earth
greater than
smaller than
equal to
What evidence is there to explain how the temperature of the blocks can be measured?
Answer:
in addition infrared thermometers don't measure metal surfaces particularly well anyway (metals typically have a low emissivity). Measuring electrical resistance is better.
A load of 46 N attached to a spring hanging
vertically stretches the spring 5.5 cm. The
spring is now placed horizontally on a table
and stretched 13 cm.
What force is required to stretch it by this
amount?
Answer in units of N.
Answer:
108.8N
Explanation:
We can use Hooke's Law to solve for the force required to stretch the spring. Hooke's Law states that the force (F) applied to a spring is directly proportional to the amount it is stretched or compressed (x), as long as the spring does not exceed its limit of proportionality. Mathematically, this can be expressed as:
F = kx
where k is the spring constant, which depends on the properties of the spring and is measured in units of N/m (newtons per meter).
To solve for the force required to stretch the spring by 13 cm, we first need to find the spring constant. We can use the information given in the problem to do this. When the load of 46 N was attached to the spring and it was stretched 5.5 cm, we can write:
46 N = k (5.5 cm) (1)
To convert the units of length to meters, we divide both sides by 100:
46 N = k (0.055 m)
Solving for k, we find:
k = 46 N ÷ 0.055 m = 836.36 N/m
Now we can use Hooke's Law again to solve for the force required to stretch the spring by 13 cm. Since the spring is now horizontal, we need to convert the displacement from vertical to horizontal. We can assume that the spring is stretched in a straight line, so the displacement is the same for both orientations. Therefore, we can write:
F = kx
F = (836.36 N/m) (0.13 m)
F ≈ 108.8 N
Therefore, the force required to stretch the spring by 13 cm is approximately 108.8 N.
The cord from an appliance is too short to reach the wall outlet in your room. You have two extension cords to choose from. (a) Find the voltage drop in the first extension cord having a 0.0760 ? resistance and through which 5.60 A is flowing. V (b) The second extension cord is cheaper and utilizes thinner wire. It has a resistance of 0.760 ? and the current flowing through it is 5.60 A. By what amount does the voltage supplied to the appliance change when the first extension cord is replaced by the second?
When the first extension cord is replaced by the second then the voltage supplied to the appliance drops by 3.834 V.
The voltage drop in the first extension cord can be calculated using Ohm's law:
V = IR
where V is the voltage drop, I is current, and R is the resistance.
The voltage drop in the first extension cord is V = IR = (5.60 A) x (0.0760 Ω) = 0.4256 V.
The voltage drop across the second extension cord is also V = IR = (5.60 A) x (0.760 Ω) = 4.256 V.
Therefore, the voltage supplied to the appliance changes by (0.4256 V - 4.256 V) = - 3.8304 V when the first extension cord is replaced by the second.
Extension cords are useful for transferring power to areas where there are no outlets, and they can also come in handy in places where outlets are inaccessible. However, if you have two extension cords to choose from, the voltage drop in each cord can impact the amount of voltage supplied to the appliance.
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