Answer: the angle of deviation for red light is 42.16° and for blue light is 40.51°.
The index of refraction of glass for red light is 1.56 and for blue light is 1.58. The angle of incidence of white light is 30 degrees. The formula for the angle of deviation is d = (i + r) - A
where i is the angle of incidence, r is the angle of refraction, and A is the angle of the prism.
The formula for the angle of refraction is given as n = sin(i)/sin(r)
where n is the refractive index of the medium (glass) for the given light.
(a) Angle of deviation for red light: For red light, the refractive index is 1.56.
n = sin(i)/sin(r)1.56
= sin(30)/sin(r)sin(r)
= sin(30)/1.56sin(r)
= 0.3402r
= sin-1(0.3402)r
= 20.16° Using the formula for the angle of deviation, we have:
d = (i + r) - A
= (30 + 20.16) - A
= 50.16 - A.
Therefore, the angle of deviation for red light is A = 50.16 - 8A = 42.16°
(b) Angle of deviation for blue light : For blue light, the refractive index is 1.58.
n = sin(i)/sin(r)1.58
= sin(30)/sin(r)sin(r)
= sin(30)/1.58sin(r)
= 0.318r
= sin-1(0.318)r
= 18.51° Using the formula for the angle of deviation, we have:
d = (i + r) - A
= (30 + 18.51) - A
= 48.51 - A.
Therefore, the angle of deviation for blue light is A = 48.51 - dA = 40.51°
Hence, the angle of deviation for red light is 42.16° and for blue light is 40.51°.
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"Prove the above channel thickness equation.
This proves that the channel thickness is constant along the flow and does not depend on the channel width or the velocity of the fluid.
The above channel thickness equation can be proved by making use of continuity equation which states that the product of cross-sectional area and velocity remains constant along the flow.
The velocity of the fluid is directly proportional to the channel depth and inversely proportional to the channel width.
Hence, we can use the following steps to prove the above channel thickness equation: - Continuity equation: A1V1 = A2V2 - Where A is the cross-sectional area and V is the velocity of the fluid. - For a rectangular channel,
A = WD
where W is the channel width and D is the channel depth. - Rearranging the continuity equation for the ratio of channel depth to channel width,
we get: D1/W1 = D2/W2
Substitute D1/W1 = h1 and D2/W2 = h2 in the above equation. - We get the following expression: h1 = h2
The question is incomplete so this is general answer.
This proves that the channel thickness is constant along the flow and does not depend on the channel width or the velocity of the fluid.
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A wire of 2 mm² cross-sectional area and 2.5 cm long contains 2 ×1020 electrons. It has a 10 2 resistance. What is the drift velocity of the charges in the wire when 5 Volts battery is applied across it? A. 2x 10-4 m/s B. 7.8 x 10 m/s C. 1.6 x 10-3 m/s D. 3.9 x 10 m/s 0 Ibrahim,
The drift velocity of the charges in the wire when a 5 Volts battery is applied across it is approximately 7.8 × 10^3 m/s. The correct answer is option B. To find the drift velocity of charges in the wire, we can use the formula:
v_d = I / (n * A * q)
Where:
v_d is the drift velocity,
I is the current flowing through the wire,
n is the number of charge carriers per unit volume,
A is the cross-sectional area of the wire,
q is the charge of each carrier.
First, let's find the current I using Ohm's Law:
I = V / R
Where:
V is the voltage applied across the wire,
R is the resistance of the wire.
Given that the voltage is 5 Volts and the resistance is 10 Ω, we have:
I = 5 V / 10 Ω = 0.5 A
Next, we need to determine the number of charge carriers per unit volume. Given that the wire contains 2 × 10^20 electrons, we can assume that the number of charge carriers is the same, so:
n = 2 × 10^20 carriers/m^3
Now, we can calculate the drift velocity:
v_d = (0.5 A) / ((2 × 10^20 carriers/m^3) * (2 × 10^-6 m^2) * (1.6 × 10^-19 C))
Simplifying the expression:
v_d = (0.5 A) / (6.4 × 10^-5 carriers * m^-3 * C * m^2)
v_d = 7.8125 × 10^3 m/s
Therefore, the drift velocity of the charges in the wire when a 5 Volts battery is applied across it is approximately 7.8 × 10^3 m/s. The correct answer is option B.
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A motor run by 85 V battery has a 25 turn square coil with side of long 5.8 cm and total resistance 34 Ω When spinning the magnetic field fot by the wir in the cola 2.6 x 10⁻² T Part A What is the maximum torque on the motor? Express your answer using two significant figures r = ______________ m·N
A motor run by 85 V battery has a 25 turn square coil with side of long 5.8 cm and total resistance 34 Ω When spinning the magnetic field felt by the wire in the cola 2.6 x 10⁻² T. The maximum torque on the motor is approximately 0.021 N·m.
To find the maximum torque on the motor, we can use the formula for torque in a motor:
τ = B × A × N ×I
Where:
τ = torque
B = magnetic field strength
A = area of the coil
N = number of turns in the coil
I = current flowing through the coil
In this case, B = 2.6 x 10⁻² T, A = (5.8 cm)^2, N = 25 turns, and we need to find I.
First, let's convert the area to square meters:
A = (5.8 cm)^2 = (5.8 x 10⁻² m)^2 = 3.364 x 10⁻⁴ m²
Next, let's find the current flowing through the coil using Ohm's Law:
V = I × R
Where:
V = voltage (85 V)
R = resistance (34 Ω)
Rearranging the formula to solve for I:
I = V / R
I = 85 V / 34 Ω ≈ 2.5 A
Now, let's substitute the values into the torque formula:
τ = (2.6 x 10⁻² T) × (3.364 x 10⁻⁴ m²) × (25 turns) × (2.5 A)
Calculating:
τ ≈ 0.021 N·m
Therefore, the maximum torque on the motor is approximately 0.021 N·m.
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Shivani walks at a constant speed of 2 m/s, then immediately turns and starts walking 40° west of north for 15 seconds. She takes a break for 10 seconds as she figures which way she wants to walk. Finally, she walks 50° east of south for 30 seconds. (a) Draw a diagram on an xy-plane. (b) How far away is Shivani from where she started walking? (c) What is her distance travelled?
Shivani is 60.07 m away from where she started walking.
Shivani's distance travelled is 60 m.
Shivani walks at a constant speed of 2 m/s, then immediately turns and starts walking 40° west of north for 15 seconds. She takes a break for 10 seconds as she figures which way she wants to walk. Finally, she walks 50° east of south for 30 seconds. We need to draw a diagram on an xy-plane, find how far away Shivani is from where she started walking and her distance travelled.
a) To plot Shivani's movements in an xy-plane, follow the given directions. Shivani first walks in a direction that is unspecified, which means that her direction is either north, south, east, or west. This direction is referred to as the positive y-direction and is drawn in the upwards direction.Then Shivani walks 40° west of north for 15 seconds. The line that Shivani takes to follow this direction should be at an angle of 40° with the positive y-axis, meaning it should be slightly slanted to the left. Finally, Shivani walks 50° east of south for 30 seconds. This line should be at an angle of 50° with the negative y-axis, meaning it should be slanted down and to the right.
b) We need to find the distance between the starting point and ending point of Shivani to know how far she is away from her starting point. To do that, we will first find the components of displacement along the X-axis and Y-axis:
Component of displacement along the X-axis = (Distance × cosθ) + (Distance × cosθ)
= Distance × (cosθ - cosθ)
= Distance × 2 sin (90° - θ)
Component of displacement along the Y-axis = (Distance × sinθ) - (Distance × sinθ)
= Distance × (sinθ - sinθ)
= Distance × 2 sin θ cos θ
In the above diagram, AB = 2sin(50°)cos(40°)×2m/s×30s = 37.07 m and CB = 2cos(50°)sin(40°)×2m/s×30s = 47.03 m
So, distance from the starting point = √(AB²+CB²) = √(37.07² + 47.03²) = 60.07 m
Thus, Shivani is 60.07 m away from where she started walking.
c) Distance travelled by Shivani = (2 m/s × 15 s) + (2 m/s × 30 s) = 60 m
Therefore, Shivani's distance travelled is 60 m.
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Determine the output voltage for the network of Figure 2 if V₁ = 2 mV and ra= 50 kn. (5 Marks) Marking Scheme: 1. Calculation using correct Formulae 2. Simulation using any available software 6.8 k V₂ S 91 MQ HF 15 MQ ww www www Figure 2 VGTH=3V k=0.4×10-3 3.3k2
The output voltage for the given network is 2.9 V.
In the given network if V₁ = 2 mV and ra= 50 kn, the output voltage can be determined . using Kirchoff's voltage law and Ohm's law. In Kirchoff's voltage law, the sum of the voltage drops in a closed loop equals the voltage rise in the same loop. In the network, a closed loop consists of a battery and the circuit's resistance.
Thus,Vin - Ira - Vds = 0 where Vin is the voltage drop across the battery, I is the current, ra is the resistance and Vds is the voltage drop across the resistor. Rearranging the equation, we getVout = Ira which is the voltage drop across the resistance. Using Ohm's law, I=Vds/ra. Substituting Vds=VGTH−Vout and simplifying,Vout=(VGTH-Vin)*ra=3V-2mV*50kΩ=3V-100V=2.9V.Vout = 2.9 V.
Simulation can be carried out using any available software.
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Answer the following question based on the lecture videos and the required readings. Give two examples of exceptions to the general rules of the patterns of motion in our solar system. Limit your answer to less than 100 words.
Two examples of exceptions to the general rules of the patterns of motion in our solar system are Retrograde motion and Irregular moons
Two examples of exceptions to the general rules of the patterns of motion in our solar system are retrograde motion and irregular moons.
1. Retrograde motion: Retrograde motion refers to the apparent backward or reverse motion of a planet in its orbit. Normally, planets move in a prograde or eastward direction around the Sun. However, due to the varying orbital speeds of planets, there are times when a planet appears to slow down, reverse its direction, and move westward relative to the background stars. This is known as retrograde motion. It occurs because of the differences in orbital periods and distances of planets from the Sun.
2. Irregular moons: Most moons in the solar system follow regular, predictable orbits around their parent planets. However, there are some moons, known as irregular moons, that have more eccentric and inclined orbits. These moons exhibit irregular patterns of motion compared to the regular, prograde motion of the larger moons. Their orbits may be highly elongated, inclined, or even retrograde. Examples of irregular moons include the moons of Jupiter, such as Ananke and Carme. These exceptions highlight the complexity and diversity of celestial motion within our solar system, demonstrating that not all celestial bodies follow the same predictable patterns of motion as the planets.
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Block A, with mass m A
, initially at rest on a horizontal floor. Block B, with mass m B
, is initially at rest on the horizontal top of A. The coefficient of static friction between the two blocks is μ s
. Block A is pulled with an increasing force. It begins to slide out from under B when its acceleration reaches:
The acceleration at which block A starts to slide out from under block B is [tex]a = (μs * mB * g) / mA[/tex].
When block A is pulled with an increasing force, it experiences a static friction force in the opposite direction. The maximum static friction force that can be exerted between the two blocks is given by the equation. [tex]a = (μs * mB * g) / mA[/tex]
Where μs is the coefficient of static friction, and N is the normal force. For block A to start sliding out from under block B, the maximum static friction force should equal the force pulling block A. Therefore, we have [tex]F_friction = μs * N = F_pull[/tex]
The normal force N is equal to the weight of block B acting downward, which is given by
[tex]N = mB * g[/tex]
Where mB is the mass of block B, and g is the acceleration due to gravity. Substituting N and F_pull into the equation, we get
[tex]μs * mB * g = F_pull[/tex]
Since the force pulling block A is equal to the product of its mass and acceleration ([tex](F_pull = mA * a)[/tex]), we have
[tex]μs * mB * g = mA * a.[/tex]
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A force, F, is applied to a 5.0 kg block of ice, initially at rest, on a smooth surface. What is the velocity of the block after 3.0 s?
When a force is applied to a 5.0 kg block of ice initially at rest on a smooth surface, we can determine the velocity of the block after 3.0 s using Newton's second law of motion.
Newton's second law states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. Mathematically, it can be expressed as:
F = m * a,
where F is the applied force, m is the mass of the block (5.0 kg), and a is the acceleration.
Since the block is initially at rest, its initial velocity is zero. We can use the kinematic equation to find the final velocity:
v = u + a * t,
where v is the final velocity, u is the initial velocity (zero in this case), a is the acceleration, and t is the time (3.0 s).
To find the acceleration, we rearrange Newton's second law:
a = F / m.
By plugging in the values, we can calculate the acceleration of the block:
a = F / m.
Once we have the acceleration, we can substitute it into the kinematic equation to find the final velocity:
v = 0 + (F / m) * t.
By applying the given force and the mass of the block, we can calculate the final velocity of the block after 3.0 s.
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A ball is launched with a horizontal velocity of 10.0 m/s from a 20.0−m cliff. How long will it be in the air? How far will it land from the base of the cliff?
The ball will land 20.2 m from the base of the cliff.
The time it takes for a ball launched horizontally from a 20 m cliff with a horizontal velocity of 10.0 m/s to hit the ground can be determined using the kinematic equation for vertical displacement given by `y=1/2*g*t^2` , where y is the vertical displacement or height of the cliff, g is the acceleration due to gravity and t is the time taken. The acceleration due to gravity is taken as -9.8 m/s^2 because it acts downwards.Using the formula,`y = 1/2*g*t^2 `=> t = √(2y/g) => t = √(2*20/9.8) => t = √4.08 => t = 2.02 sThe ball will take 2.02 seconds to reach the ground.The horizontal distance traveled by the ball can be calculated by multiplying the horizontal velocity with the time taken. Hence,Distance = velocity × time= 10.0 m/s × 2.02 s= 20.2 m. Therefore, the ball will land 20.2 m from the base of the cliff.
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The table lists the mass and charge of a proton and a neutron. A 3 column table with 2 rows. The first column is labeled particle with entries proton and neutron. The second column is labeled mass times 10 Superscript negative 27 baseline kg with entries 1.673, 1.675. The last column is labeled charge times 10 Superscript negative 19 baseline C with entries 1.61, 0. How do the gravitational and electrical forces between a proton and a neutron compare? The gravitational force is much smaller than the electrical force for any distance between the particles. The gravitational force is much larger than the electrical force for any distance between the particles. The gravitational force is much smaller than the electrical force for only very small distances between the particles. The gravitational force is much larger than the electrical force for only very small distances between the particles.
In comparing the gravitational and electrical forces between a proton and a neutron, we can conclude that the gravitational force is much smaller than the electrical force for any distance between the particles.
The gravitational and electrical forces between a proton and a neutron can be compared based on their respective masses and charges.
The mass of a proton is approximately 1.673 x 10^-27 kg, while the mass of a neutron is slightly higher at 1.675 x 10^-27 kg. Therefore, their masses are very similar.
However, when it comes to their charges, a proton has a charge of approximately 1.61 x 10^-19 C, while a neutron has no charge (0 C).
In terms of the gravitational force, which depends on the masses of the particles, the forces between a proton and a neutron would be similar since their masses are very close.
On the other hand, the electrical force, which depends on the charges of the particles, would be significantly different. The presence of a charge on the proton creates an electrical force, while the neutral neutron does not contribute to an electrical force.
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Answer: A
Explanation:
Find the net force on charge Q=5c due to other charges shown:
The net force on charge Q = 5C due to the other charges is 36N, directed to the left.
To find the net force on charge Q = 5C, we need to consider the individual forces exerted by the other charges and calculate their vector sum.
Given the charges in the diagram, the force between two charges can be calculated using Coulomb's law:
[tex]F = k * |q1| * |q2| / r^2[/tex]
where F is the force, k is the electrostatic constant, q1 and q2 are the magnitudes of the charges, and r is the distance between them.
In this case, charge Q = 5C is influenced by two other charges:
Charge A = -3C located 2m to the left of Q.
Charge B = +4C located 3m to the right of Q.
Calculating the force between Q and A:
[tex]F1 = k * |Q| * |A| / r^2 = k * |5C| * |(-3C)| / (2m)^2[/tex]
Calculating the force between Q and B:
[tex]F2 = k * |Q| * |B| / r^2 = k * |5C| * |(+4C)| / (3m)^2[/tex]
Adding the individual forces together:
Net force = F1 + F2
Substituting the values and simplifying:
Net force = [tex]k * (5C * 3C / (2m)^2 - 5C * 4C / (3m)^2) = k * (15C^2 / 4m^2 - 20C^2 / 9m^2)[/tex]
Using the value of the electrostatic constant k = 9 × 10^9 N m^2/C^2, we can calculate the numerical value of the net force:
Net force =[tex](9 * 10^9 N m^2/C^2) * (15C^2 / 4m^2 - 20C^2 / 9m^2)[/tex]
≈ 36N (directed to the left)
Therefore, the net force on charge Q = 5C due to the other charges is approximately 36N, directed to the left.
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The complete question is:
Find the net force on charge Q=5c due to other charges shown,
A capacitor has a capacitance of 3.7 x 10-6 F. In the charging process, electrons are removed from one plate and placed on the other plate. When the potential difference between the plates is 610 V, how many electrons have been transferred?
Approximately 1.4106 x 10^16 electrons have been transferred during the charging process of the capacitor.
To find the number of electrons transferred during the charging process of a capacitor, we can use the equation:
Q = CV
Where:
Q is the charge transferred (in Coulombs),
C is the capacitance of the capacitor (in Farads),
V is the potential difference across the capacitor (in Volts).
Given:
C = 3.7 x 10^(-6) F
V = 610 V
Substituting these values into the equation, we have:
Q = (3.7 x 10^(-6) F)(610 V)
Q = 2.257 x 10^(-3) C
Now, we know that the charge of one electron is approximately 1.6 x 10^(-19) C. To find the number of electrons transferred, we can divide the total charge by the charge of one electron:
Number of electrons = Q / (1.6 x 10^(-19) C)
Number of electrons = (2.257 x 10^(-3) C) / (1.6 x 10^(-19) C)
Performing the calculation, we get:
Number of electrons = 1.4106 x 10^(16)
Therefore, approximately 1.4106 x 10^16 electrons have been transferred during the charging process of the capacitor.
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A satellite of mass 658.5 kg is moving in a stable circular orbit about the Earth at a height of 11RE, where RE = 6400km = 6.400 x 106 m = 6.400 Mega-meters is Earth’s radius. The gravitational force (in newtons) on the satellite while in orbit is:
A satellite of mass 658.5 kg is moving in a stable circular orbit about the Earth at a height of 11RE, where RE = 6400km = 6.400 x 106 m = 6.400 Mega-meters is Earth’s radius. the gravitational force acting on the satellite while in orbit is approximately [tex]2.443 * 10^4 Newtons.[/tex] Newtons.
The gravitational force acting on the satellite while in orbit can be calculated using the equation for gravitational force:
Force = (Gravitational constant * Mass of satellite * Mass of Earth) / (Distance from satellite to center of Earth)^2
The gravitational constant is denoted by G and is approximately [tex]6.674 * 10^-11 N(m/kg)^2[/tex] The mass of the Earth is approximately [tex]5.972 * 10^{24} kg.[/tex]
The distance from the satellite to the center of the Earth is the sum of the Earth's radius (RE) and the height of the satellite (11RE). Substituting the given values into the equation, we have:
Force =[tex](6.674 * 10^-11 N(m/kg)^2 * 658.5 kg * 5.972 * 10^{24} kg) / ((11 * 6.400 * 10^6 m)^2)[/tex]
Simplifying the expression:
Force ≈ [tex]2.443 * 10^4 Newtons.[/tex]
Therefore, the gravitational force acting on the satellite while in orbit is approximately[tex]2.443 * 10^4 Newtons.[/tex]
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A point charge is 10 µc. Find the field and potential at a distance of 30 cm?
The electric field at a distance of 30 cm from a point charge of 10 µC is 3.33 × 10^4 N/C directed radially outward from the charge. The electric potential at that distance is 9 × 10^4 V.
The electric field at a distance of 30 cm from a point charge can be calculated using Coulomb's law: Electric field (E) = k * (Q / r^2),
E = (9 × 10^9 N m^2/C^2) * (10 × 10^-6 C) / (0.3 m)^2 = 3.33 × 10^4 N/C.
Therefore, the electric field at a distance of 30 cm from the point charge is 3.33 × 10^4 N/C
The potential at a distance from a point charge can be calculated using the equation: Potential (V) = k * (Q / r),
V = (9 × 10^9 N m^2/C^2) * (10 × 10^-6 C) / (0.3 m) = 9x 10^4 V.
Therefore, the potential at a distance of 30 cm from the point charge is 9x 10^4 V.
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Find the self inductance for the following
inductors.
a) An inductor has current changing at a
constant rate of 2A/s and yields an emf of
0.5V
b) A solenoid with 20 turns/cm has a
magnetic field which changes at a rate of
0.5T/s. The resulting EMF is 1.7V
c) A current given by I(t) = 10e~(-at) induces an emf of 20V after 2.0 s. I0 = 1.5A and a 3.5s^-1
The self inductance for each scenario is: (a) -0.25 H, (b) -3.4 H and (c) 2 H. To find the self inductance for each of the given inductors, we can use the formula for self-induced emf:
ε = -L (dI/dt)
where ε is the induced emf, L is the self inductance, and (dI/dt) is the rate of change of current. Rearranging the formula, we have:
L = -ε / (dI/dt)
Let's calculate the self inductance for each scenario:
a) An inductor has current changing at a constant rate of 2A/s and yields an emf of 0.5V.
Here, the rate of change of current (dI/dt) = 2A/s, and the induced emf ε = 0.5V. Plugging these values into the formula:
L = -0.5V / 2A/s
L = -0.25 H (henries)
b) A solenoid with 20 turns/cm has a magnetic field which changes at a rate of 0.5T/s. The resulting EMF is 1.7V.
In this case, we need to convert the turns per centimeter to turns per meter.
Since there are 100 cm in a meter, the solenoid has 20 turns/100 cm = 0.2 turns/meter.
The rate of change of magnetic field (dI/dt) = 0.5 T/s, and the induced emf ε = 1.7V. Plugging these values into the formula:
L = -1.7V / (0.5 T/s)
L = -3.4 H (henries)
c) A current given by I(t) = 10 [tex]e^{-at}[/tex] induces an emf of 20V after 2.0s. I0 = 1.5A and a = 3.5[tex]s^{-1}.[/tex]
To find the self inductance in this case, we need to find the rate of change of current (dI/dt) at t = 2.0s. Differentiating the current equation:
dI/dt = -10a * [tex]e^{-at}[/tex]
At t = 2.0s, the current is I(t) = [tex]10e^{-a*2}[/tex]= 10[tex]e^{-2a}[/tex]. Given I0 = 1.5A, we can solve for a:
1.5A = 10[tex]e^{-2a}[/tex]
[tex]e^{-2a}[/tex] = 1.5/10
-2a = ln(1.5/10)
a = -(ln(1.5/10))/2
Now, we can substitute the values into the formula:
L = -20V / (-10a * [tex]e^{-2a}[/tex])
L = 2 H (henries)
Therefore, the self inductance for each scenario is:
a) -0.25 H (henries)
b) -3.4 H (henries)
c) 2 H (henries)
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A 1000 kg motor vehicle starts from an initial velocity 1 m/s and after traveling at a distance of 113 m on a straight-line path, its speed is found to be 28 m/s. What is the magnitude of the average net acceleration of the car during the travel on this straight-line path? No need to write the unit. Please write the answer in one decimal place. (eg 1.234 should be written as 1.2).
Answer:
The acceleration is 3.5.
Explanation:
According to the question, the initial velocity is given as 1 m/s, the distance travelled is given as 113 m and the final velocity is given as 28 m/s.
Observe equation 1, [tex]v^{2} = u^{2} +2a s[/tex] where v is the final velocity, u is the initial velocity, a is the acceleration and s is the distance. Rearranging for acceleration gives
[tex]a = \frac{v^{2} -u^{2} }{2s}[/tex]
Thus, [tex]a = \frac{28^{2}-1^{2} }{226}[/tex]
Therefore, acceleration is 3.4646 which is 3.5 to 1 decimal place.
Water is being transported via a pipe at 1.2m/s, with a pipe being raised higher at the outlet than the inlet. At the inlet, the pressure of the water is measured to be 26000 Pa and 10000 Pa at the outlet. Assuming that the process is isothermal, calculate how much higher the outlet of the pipe is than the inlet (which has a height of 0). Answer in m.
The height difference between the outlet and inlet of the pipe is approximately 2.1 meters. The height difference between the outlet and inlet of the pipe, we can use Bernoulli's equation, which relates the pressure, velocity, and elevation of a fluid flowing in a pipe.
Bernoulli's equation states:
P₁ + (1/2)ρv₁² + ρgh₁ = P₂ + (1/2)ρv₂² + ρgh₂,
where P₁ and P₂ are the pressures at the inlet and outlet, respectively, ρ is the density of the fluid, v₁ and v₂ are the velocities at the inlet and outlet, h₁ and h₂ are the elevations at the inlet and outlet, and g is the acceleration due to gravity.
In this case, since the process is isothermal, there is no change in the fluid's internal energy. Therefore, the term (1/2)ρv₁² + ρgh₁ = (1/2)ρv₂² + ρgh₂ can be simplified as:
(1/2)ρv₁² + ρgh₁ = (1/2)ρv₂² + ρgh₂.
Since the height at the inlet is given as 0 (h₁ = 0), the equation becomes:
(1/2)ρv₁² = (1/2)ρv₂² + ρgh₂.
We can rearrange the equation to solve for the height difference (h₂ - h₁ = Δh):
Δh = (v₁² - v₂²) / (2g).
Given that the velocity at the inlet (v₁) is 1.2 m/s and the pressures at the inlet and outlet are 26000 Pa and 10000 Pa, respectively, we can use Bernoulli's equation to determine the velocity at the outlet (v₂) using the pressure difference:
P₁ + (1/2)ρv₁² = P₂ + (1/2)ρv₂².
Substituting the given values:
26000 + (1/2)ρ(1.2)² = 10000 + (1/2)ρv₂².
Simplifying and rearranging:
(1/2)ρv₂² = 26000 - 10000 + (1/2)ρ(1.2)².
Substituting the density of water (ρ = 1000 kg/m³):
(1/2)(1000)v₂² = 16000 + (1/2)(1000)(1.2)².
Simplifying and solving for v₂:
v₂ = √((16000 + 600) / 1000) ≈ 4.3 m/s.
Now we can substitute the values of v₁ = 1.2 m/s, v₂ = 4.3 m/s, and g = 9.8 m/s² into the equation for the height difference:
Δh = (1.2² - 4.3²) / (2 * 9.8) ≈ -2.1 m.
The negative sign indicates that the outlet of the pipe is 2.1 meters lower than the inlet.
Therefore, the height difference between the outlet and inlet of the pipe is approximately 2.1 meters.
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Given x1(t) = cos (t), x2(t) = sin (πt) and x3(t) = xi(t) + x2(t). a. Determine the fundamentals period of TI and T2 b. Determine if T3 is periodic or nonperiodic and shows the evident c. Determine the powers P1, P2 and P3 of each signal
The fundamental period (TI) for x1(t) is 2π, (T2) for x2(t) is 2 and x3(t) is nonperiodic. Powers P1 and P2 values are 1/2 while the power of P3 cant be determined since x3(t) is nonperiodic.
Given signals are;x1(t) = cos(t) x2(t) = sin(πt) x3(t) = x1(t) + x2(t)a) To find the fundamental period of T1;The fundamental period of a signal x(t) is denoted by T0, and it is defined as the smallest value of T such that x(t) = x(t+T) for all values of t. Therefore, x1(t) = x1(t+T1), whereT1= 2π/ω1= 2π/1= 2π. Thus, the fundamental period of x1(t) is T1= 2π.b) To find the fundamental period of T2;x2(t) = x2(t+T2), whereT2 = 2π/ω2= 2π/π= 2Thus, the fundamental period of x2(t) is T2 = 2.c) To determine if T3 is periodic or non-periodic and show the evident;x3(t) = x1(t) + x2(t) Therefore,x3(t) = cos(t) + sin(πt)If we assume T3 exists, then we can say thatx3(t) = x3(t + T3)cos(t) + sin(πt) = cos(t + T3) + sin(π(t + T3))
Therefore, the function will be periodic if the following conditions are satisfied: cos(t + T3) = cos(t)sin(π(t + T3)) = sin(πt)Expanding the above expression, cos(t + T3) = cos(t)sin(πt)cos(T3) + cos(πt)sin(πt)sin(T3) = sin(πt). Simplifying, cos(T3) = 1Therefore, T3 is a multiple of 2π. Also, sin(T3) = 0.If T3 exists, it must be a multiple of T1 and T2.LCM(T1, T2) = LCM(2π, 2) = 2πThe multiple of 2π is 2π itself. Therefore, T3 = 2πd, where d is a constant. But since sin(T3) = 0, d must be an even integer.T3 is periodic with a fundamental period of 2πd. Thus, T3 = 4π.d) To determine the power P1, P2 and P3 of each signal; Power is defined as the average value of the energy carried by the signal over the given time.T1 = 2π, ω1 = 1; P1 = (1/T1)∫(T1/2)^(T1/2)x1^2(t) dt= (1/2π) ∫π^(-π) cos^2(t) dt= 1/2.T2 = 2, ω2 = π; P2 = (1/T2)∫(T2/2)^0x2^2(t) dt= (1/4) ∫2^0 sin^2(πt) dt= 1/4.T3 = 4π; P3 = (1/T3)∫(T3/2)^(-T3/2)x3^2(t) dt= (1/8π) ∫2π^(-2π) (cos(t) + sin(πt))^2 dt= (1/8π) [π + 2] = (π + 2)/8π.
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Task 1
Describe what happens at a p-n junction. Your description must
include reference to electrons, holes, depletion regions and
forward and reverse biasing.
At a p-n junction, the diffusion and recombination of charge carriers form a depletion region, and when forward biased, it allows current flow, while reverse bias inhibits current flow.
What is a p-n junction?A P-N junction is an interface or a boundary between two semiconductor material types, namely the p-type and the n-type, inside a semiconductor.
In a p-type semiconductor, the majority carriers are holes, which are essentially positively charged vacancies in the valence band.
In contrast, an n-type semiconductor has excess electrons as the majority carriers. At a p-n junction, the diffusion and recombination of charge carriers lead to the formation of a depletion region.
Forward bias reduces the potential barrier, allowing current flow, while reverse bias increases the barrier, inhibiting current flow.
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A GM counter is a gas-filled detector. Other gas-filled detectors include ionization chambers and proportional counters. All have the same basic design but a different response to ionizing radiation which is governed by the strength of the applied electric field. Draw a schematic diagram of applied voltage vs the number of ion pairs produced and label the following regions:
(a) Recombination region
(b) Ionization region
(c) Proportional region
(d) Limited proportionality region
(e) GM region
(f) Continuous discharge region
If a beam of light is incident from water (n = 1.33) to crown glass (n1.52) with an incident angle of 40.0 degrees, what is the angle of the refracted beam of light?
The refracted angle of the light beam, when it moves from water (n = 1.33) into a crown glass (n = 1.52) at an incident angle of 40.0 degrees, is approximately 30.7 degrees.
This value is calculated using Snell's law of refraction, which relates the ratio of the sine of the angles of incidence and refraction to the inverse ratio of the indices of refraction of the two mediums. Snell's law, or the law of refraction, states that the ratio of the sine of the angle of incidence to the sine of the angle of refraction is equivalent to the reciprocal of the ratio of the indices of refraction. In mathematical terms, n1*sin(θ1) = n2*sin(θ2). Here, n1 and n2 are the refractive indices of the first and second medium respectively, and θ1 and θ2 are the angles of incidence and refraction. Given the refractive indices of water (n1 = 1.33) and crown glass (n2 = 1.52), and the angle of incidence (θ1 = 40.0 degrees), we can calculate the angle of refraction (θ2) using this law. This calculation yields an angle of approximately 30.7 degrees.
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A1 to bintang ball that is mading at 2.90 m* tres her pool table and bounces straight back * 2.2 ts original soced). The colorata 700 (tume that the same as me pestive direction Calculate the weagufurca { act on the body the burre te direction at the spot worrower ) ( How much kinetic roergy in joules is het during the contre magte (what percent of the origin?
When a ball of mass 2.90 kg strikes a pool table and bounces straight back with a speed of 2.2 m/s, the change in momentum can be calculated by subtracting the initial momentum from the final momentum.
The weight force acting on the ball can be determined by multiplying the mass of the ball by the acceleration due to gravity. The kinetic energy lost during the collision can be calculated as the difference between the initial kinetic energy and the final kinetic energy. The percentage of the original kinetic energy lost can be found by dividing the lost kinetic energy by the initial kinetic energy and multiplying by 100.
To determine the change in momentum of the ball, we subtract the final momentum from the initial momentum. The initial momentum is given by the product of the mass and the initial velocity, which is 2.90 kg * 0 m/s since the ball is at rest. The final momentum is given by the product of the mass and the final velocity, which is 2.90 kg * (-2.2 m/s) since the ball bounces back in the opposite direction.
The weight force acting on the ball can be calculated by multiplying the mass of the ball (2.90 kg) by the acceleration due to gravity (approximately 9.8 m/s^2). This will give us the weight force in Newtons.
To calculate the kinetic energy lost during the collision, we subtract the final kinetic energy from the initial kinetic energy. The initial kinetic energy is given by (1/2) * mass * (initial velocity)^2, and the final kinetic energy is given by (1/2) * mass * (final velocity)^2.
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Consider a makeup mirror that produces a magnification of 1.35 when a person's face is 11.5 cm away. What is the focal length of the makeup mirror in meters?
f = ______
The focal length of the makeup mirror in meters, f = 0.0122 m
Magnification formula is given by,
Magnification (m) = height of image (h′) / height of object (h)
If f is the focal length of the mirror, the distance from the object to the mirror is given by d = f and the distance from the image to the mirror is also d = f.
The magnification of the makeup mirror is given as 1.35.
Distance of the object from the mirror, d = 11.5 cm = 0.115 m
Magnification, m = 1.35So,
using the formula of magnification we have,
h′ / h = 1.35
Since
h = height of object and h′ = height of image, we can say that,
h′ = 1.35h
Using mirror formula we have,
1/f = 1/d + 1/d'
1/f = 1/d + 1/dh′ / h = d′ / d
d′ = 1.35h × d
Now, using similar triangles, we can say that,
d′ / d = h′ / h
d = d′h / h′
Now substituting the value of d in mirror formula we get,
1/f = 1/d + 1/d'
1/f = 1/d + h′ / dh
1/f = 1/d + 1.35h / (d × h′)
Putting the values, we have
1/f = 1/0.115 + 1.35 / (0.115 × h′)
1/f = 8.7 + 1.35 / (0.115 × h′)
1/f = (11.9 / h′)
m = h′ / h = 1.35
h′ = 1.35h
Substituting this value in above equation we have,
1/f = (11.9 / 1.35h)
f = (1.35h / 11.9) = (1.35 / 11.9) × h
f = (1.35 / 11.9) × 0.115 m
Therefore, the focal length of the makeup mirror in meters is 0.0122 m
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An object is being dragged across a flat level surface using a rope that is applying a constant 43.9 lb force to the side of the object at an angle that is 27.0 degrees above the horizontal. If this force is used to drag the object through a displacement of 26.8 ft, then how much work was done by this force in ftlb?
The work done by the force of the rope dragging the object is approximately 1049.84 ft-lb.
The work done by a force is given by the product of the magnitude of the force, the displacement, and the cosine of the angle between the force and the direction of displacement. In this case, the force applied by the rope is 43.9 lb and the displacement is 26.8 ft.
Using the given angle of 27.0 degrees, we can calculate the work done as follows:
W = 43.9 lb * 26.8 ft * cos(27.0°).
To evaluate the cosine function, the angle needs to be in radians. Converting 27.0 degrees to radians gives 0.471 radians.
Substituting the values into the equation, we get:
W = 43.9 lb * 26.8 ft * cos(0.471).
Evaluating the cosine function, we find cos(0.471) ≈ 0.920.
Finally, we can calculate the work done:
W = 43.9 lb * 26.8 ft * 0.920 ≈ 1049.84 ft-lb.
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The electric potential in a certain region is V = ax² + bx +c where a = 11 V/m², where b = −10 V/m², and c = 63 V. Determine the position where the electric field is zero.
The electric potential in a certain region is V = ax² + bx +c where a = 11 V/m², b = −10 V/m², and c = 63 V. We are supposed to find the position where the electric field is zero. Electric field is the negative of the gradient of potential, i.e.,
`E= -grad(V)`
Hence, to find where electric field is zero, we have to find the position where the gradient of potential is zero and then check whether that point is a point of minimum or maximum.
So, `E= -grad(V) = -(∂V/∂x) î`
For the given potential, `V = ax² + bx + c = 11x² - 10x + 63`
So, `E= -grad(V) = -(∂V/∂x) î = (-22x + 10) î`
Hence, electric field is zero when, `(-22x + 10) î = 0 => x = 5/11 m`
Therefore, the position where the electric field is zero is 5/11 m.
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why aeroplanes and boat having bird like structure
People have looked up at birds for years and they have inspired us to fly. Airplanes have wings, just like birds. They also have a light skeleton (or framework) to decrease their weight, and they have a streamlined shape to decrease drag.
When observing a galaxy the calcium absorption line, which has a rest wavelength of 3933 A is observed redshifted to 3936.5397 A. a)Using the Doppler shift formula calculate the cosmological recession velocity Vr, (c = 300 000km/s). b)Evaluate the Hubble constant H (in units of km/s/Mpc), assuming that the Hubble law Vr = Hd holds for this galaxy. The distance to the galaxy is measured to be 4 Mpc.
The cosmological recession velocity (Vr) is approximately 272.2272 km/s.the Hubble constant (H) is approximately 2.21 * 10^(-18) km^(-1) s^(-1).
a) To calculate the cosmological recession velocity (Vr) using the Doppler shift formula, we can use the following equation:
Vr = (λ - λ₀) / λ₀ * c
Where:
λ is the observed wavelength
λ₀ is the rest wavelength
c is the speed of light (300,000 km/s)
Given:
λ = 3936.5397 Å
λ₀ = 3933 Å
c = 300,000 km/s
Let's calculate Vr:
Vr = (3936.5397 - 3933) / 3933 * 300,000
= 0.000907424 * 300,000
= 272.2272 km/s
Therefore, the cosmological recession velocity (Vr) is approximately 272.2272 km/s.
b) The Hubble constant (H) can be evaluated using the Hubble law equation:
Vr = Hd
Where:
Vr is the cosmological recession velocity
H is the Hubble constant
d is the distance to the galaxy
Given:
Vr = 272.2272 km/s
d = 4 Mpc = 4 million parsecs = 4 * 3.09 * 10^19 km
Let's calculate H:
H = Vr / d
= 272.2272 / [tex](4 * 3.09 * 10^{19})[/tex]
≈ 2.21 * [tex]10^{(-18)} km^{(-1)} s^{(-1)}[/tex]
Therefore, the Hubble constant (H) is approximately 2.21 * [tex]10^{(-18)} km^{(-1)} s^{(-1)}[/tex].
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The ground state of an electron has an energy E1=−15eV while its excited state has an energy E2=−10eV. The electron can absorb a photon with an energy of 2.4×10 ∧
−18 J None of the options 8×10 ∧
−19 J 1.6×10 ∧
−18 J
The electron can absorb a photon with an energy of 1.6x10^-18 J to transition from its ground state to its excited state.
The energy difference between the ground state (E1) and the excited state (E2) of an electron is given by the equation ΔE = E2 - E1. Substituting the given values, we have:
ΔE = (-10 eV) - (-15 eV)
= 5 eV
To convert this energy difference to joules, we use the conversion factor: 1 eV = 1.6x10^-19 J. Thus, ΔE in joules is:
ΔE = 5 eV * (1.6x10^-19 J/eV)
= 8x10^-19 J
Comparing this value to the photon energy of 2.4x10^-18 J, we see that it is smaller.
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The Starship Enterprise returns from warp drive to ordinary space with a forward speed of 60 km/s. To the crew's great surprise, a Klingon ship is 150 km directly ahead, traveling in the same direction at a mere 22 km/s. Without evasive action, the Enterprise will overtake and collide with the Klingons in just about 3.9 s . The Enterprise's computers react instantly to brake the ship. 6 of 6 Review | Constants Part A What magnitude acceleration does the Enterprise need to just barely avoid a collision with the Klingon ship? Assume the acceleration is constant.
Hint: Draw a position-versus-time graph showing the motions of both the Enterprise and the Klingon ship. Let zo = 0km be the location of the Enterprise as it returns from warp drive. How do you show graphically the situation in which the collision is "barely avoided"? Once you decide what it looks like graphically, express that situation mathematically.
The Enterprise needs to come to a stop just as it reaches position of Klingon ship. Therefore position-versus-time graph for Enterprise would be a straight line with a positive slope initially, representing its initial velocity of 60 km/s.
At the moment of collision avoidance, the Enterprise's position should match that of the Klingon ship. This means the two lines on the graph should intersect at the same point.
Mathematically, this can be expressed by setting the equations for the positions of the Enterprise and the Klingon ship equal to each other:
60t = 22t + 150
By rearranging the equation, we have: 60t - 22t = 150
38t = 150
t ≈ 3.95 seconds
Therefore, to just barely avoid a collision with the Klingon ship, the Enterprise needs to achieve an acceleration that brings it to a stop within approximately 3.95 seconds.
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A block of mass 4.0 kg and a block of mass 6.0 kg are linked by a spring balance of negligible mass. The blocks are placed on a frictionless horizontal surface. A force of 18.0 N is applied to the 6.0 kg block as shown. What is the reading on the spring balance?
The reading on the spring balance is 0.4 N.
When a force of 18.0 N is applied to the 6.0 kg block and there is no friction between the blocks and the horizontal surface. A spring balance is connected between two blocks. We are required to find the reading on the spring balance.
For that, we can use the formula of force that acts between the blocks connected by a spring balance. The formula is given as below:
F = kx where F is the force that acts between two blocks, k is the spring constant, and x is the displacement of the spring.The force that acts on the blocks is equal to the force applied on the heavier block. i.e., 18.0 N
The mass of the two blocks is M = 4.0 + 6.0 = 10.0 kg
The acceleration of the two blocks is given as follows:
For the heavier block 6.0 kg:
F = m₁a where m₁ is mass of the block
F = 18 N, m₁ = 6.0 kg
So, a = 18.0/6.0 = 3.0 m/s²
For the lighter block 4.0 kg:F = m₂a where m₂ is mass of the block m₂ = 4.0 kg
So, a = 3.0 m/s²
Using the force formula F = kxk = F/x = 18.0/0.4 = 45.0 N/m
The force on the spring is given as:F = kx
So, x = F/k = 18.0/45.0 = 0.4 m
Therefore, the reading on the spring balance is 0.4 m or 0.4 N (because 1 N/m = 1 N/m)
Answer: The reading on the spring balance is 0.4 N.
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