The specific speed of the original pump (Ns1) is approximately 246.48, and the specific speed of the modified pump (Ns2) is approximately 245.52.
To predict the volume flow rate and net head of the modified pump, we can use the concept of pump affinity laws. These laws state that the flow rate and head developed by a pump are proportional to the speed of the pump.
Let's denote the subscript "1" for the original pump and "2" for the modified pump.
According to the pump affinity laws:
Flow Rate (Q2) = (N2 / N1) * (Q1)
Head (H2) = (N2 / N1)^2 * (H1)
where:
Q1 is the original flow rate (14.0 Lpm)
H1 is the original net head (1.5 m)
N1 is the original motor speed (1200 rpm)
N2 is the new motor speed (1800 rpm)
Let's calculate the values:
6.1. Predict the volume flow rate and net head of the modified pump:
Using the above formulas, we can calculate:
Q2 = (1800 rpm / 1200 rpm) * (14.0 Lpm)
= 21.0 Lpm
H2 = (1800 rpm / 1200 rpm)^2 * (1.5 m)
= 3.375 m
Therefore, the predicted volume flow rate of the modified pump is 21.0 Lpm, and the predicted net head is 3.375 m.
6.2. Calculate and compare the pump specific speed of both pumps:
The pump specific speed (Ns) is a dimensionless parameter that characterizes the geometry and performance of a pump. It is calculated using the following formula:
Ns = (N * Q^0.5) / H^0.75
where:
N is the pump speed (rpm)
Q is the flow rate (Lpm)
H is the head (m)
For the original pump:
Ns1 = (1200 rpm * (14.0 Lpm)^0.5) / (1.5 m)^0.75
For the modified pump:
Ns2 = (1800 rpm * (21.0 Lpm)^0.5) / (3.375 m)^0.75
Calculating the specific speeds:
Ns1 ≈ 246.48
Ns2 ≈ 245.52
Therefore, the specific speed of the original pump (Ns1) is approximately 246.48, and the specific speed of the modified pump (Ns2) is approximately 245.52.
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A negative charge of −0.510μC exerts an upward 0.600−N force Part A on an unknown charge that is located 0.400 m directly below the first charge, What is the value of the unknown charge (magnitude and sign)? Express your answer with the appropriate units.
The magnitude of the unknown charge is 1.85 × 10^-5 C, and it is positive.
We can use Coulomb's law to solve this problem. Coulomb's law states that the force between two charges is proportional to the product of the charges and inversely proportional to the square of the distance between them.
The force between the two charges is given by:
F = k * q1 * q2 / r^2
where:
* F is the force between the two charges (in N)
* k is the Coulomb constant (8.987551787 × 10^9 N m^2 / C^2)
* q1 is the magnitude of the first charge (in C)
* q2 is the magnitude of the second charge (in C)
* r is the distance between the two charges (in m)
In this problem, we know the following:
* F = 0.600 N
* q1 = −0.510μC = −5.10 × 10^-6 C
* r = 0.400 m
We need to find the value of q2.
Substituting the known values into the equation for Coulomb's law, we get:
F = k * q1 * q2 / r^2
0.600 N = 8.987551787 × 10^9 N m^2 / C^2 * (-5.10 × 10^-6 C) * q2 / (0.400 m)^2
Solving for q2, we get:
q2 = 1.85 × 10^-5 C
Since the force between the two charges is upward, and the first charge is negative, the second charge must be positive.
Therefore, the value of the unknown charge is 1.85 × 10^-5 C, and it is positive.
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During a lightning flash, there exists a potential difference of V
cloud
−V
ground
=2.2×10
9
V between a cloud and the ground. As a result, a charge of −28C is transferred from the ground to the cloud. (a) How much work W
ground -clowd
is done on the charge by the electric force? (b) If the work done by the electric force were used to accelerate a 1100-kg automobile from rest, what would be its final speed? (c) If the work done by the electric force were converted into heat, how many kilograms of water at 0
∘
C could be heated to 100
∘
C ? W
ground-cloud
= v
f
=∣ m=
(a) To calculate the work done (W) by the electric force, we can use the formula:
W = q * ΔV
where q is the charge transferred (given as -28 C) and ΔV is the potential difference (given as 2.2 × 10^9 V).
Plugging in the values, we have:
W = (-28 C) * (2.2 × 10^9 V)
= -61.6 × 10^9 J
Therefore, the work done on the charge by the electric force is -61.6 × 10^9 Joules.
(b) To find the final speed (v_f) of the 1100-kg automobile, we can use the work-energy theorem. The work done by the electric force is equal to the change in kinetic energy of the automobile.
W = (1/2) * m * v_f^2
Solving for v_f, we have:
v_f^2 = (2 * W) / m
v_f^2 = (2 * (-61.6 × 10^9 J)) / 1100 kg
v_f^2 = -112.0 × 10^6 m^2/s^2
Since velocity cannot be negative in this context, we take the positive square root:
v_f = √(-112.0 × 10^6 m^2/s^2) ≈ 10.6 × 10^3 m/s
Therefore, the final speed of the automobile would be approximately 10.6 × 10^3 m/s.
(c) To determine how many kilograms of water at 0 °C could be heated to 100 °C using the work done by the electric force, we need to consider the specific heat capacity of water.
The specific heat capacity of water is 4.186 J/g°C. We can calculate the amount of heat required to raise the temperature of the water from 0 °C to 100 °C using the formula:
Q = m * c * ΔT
where Q is the heat energy, m is the mass of the water, c is the specific heat capacity, and ΔT is the change in temperature.
Since the work done by the electric force is equal to the heat energy, we can equate the two:
W = Q
Substituting the values, we have:
-61.6 × 10^9 J = m * (4.186 J/g°C) * (100 °C - 0 °C)
Simplifying, we find:
m = (-61.6 × 10^9 J) / (4.186 J/g°C * 100 °C)
m ≈ -147.3 × 10^6 g
Therefore, the negative mass value implies that the work done by the electric force cannot be converted into heat to heat water.
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An infinite line charge of linear charge density +1.50μC/m lies on the z axis. Find the electric potential at distances from the line charge of (a) 2.00 m,(b)4.00 m, and (c) 12.0 m. Assume that we choose V=0 at a distance of 2.50 m from the line of charge. "SडM
The electric potential at distances of 2.00 m, 4.00 m, and 12.0 m from the line charge are approximately 1.35 * 10^6 V, 6.74 * 10^5 V, and 2.24 * 10^5 V, respectively, when referenced to a distance of 2.50 m from the line of charge where V = 0.To find the electric potential at different distances from the infinite line charge, we can use the formula for the electric potential due to a line charge:
V = kλ / r
where V is the electric potential, k is the electrostatic constant (8.99 * 10^9 N·m²/C²), λ is the linear charge density (in C/m), and r is the distance from the line charge.
Given that the linear charge density is +1.50 μC/m (1.50 * 10^-6 C/m), we can calculate the electric potential at the given distances:
a) At a distance of 2.00 m:
V₁ = (8.99 * 10^9 N·m²/C²) * (1.50 * 10^-6 C/m) / 2.00 m
b) At a distance of 4.00 m:
V₂ = (8.99 * 10^9 N·m²/C²) * (1.50 * 10^-6 C/m) / 4.00 m
c) At a distance of 12.0 m:
V₃ = (8.99 * 10^9 N·m²/C²) * (1.50 * 10^-6 C/m) / 12.0 m
Now, to calculate the potential with respect to a reference point at a distance of 2.50 m from the line of charge, we subtract the potential at that reference point from each of the calculated potentials:
V₁' = V₁ - V(2.50 m)
V₂' = V₂ - V(2.50 m)
V₃' = V₃ - V(2.50 m)
Given that V(2.50 m) = 0 (as chosen in the problem), the equations simplify to:
V₁' = V₁
V₂' = V₂
V₃' = V₃
Now, we can substitute the known values and calculate the electric potential at each distance:
a) At a distance of 2.00 m:
V₁' = (8.99 * 10^9 N·m²/C²) * (1.50 * 10^-6 C/m) / 2.00 m
b) At a distance of 4.00 m:
V₂' = (8.99 * 10^9 N·m²/C²) * (1.50 * 10^-6 C/m) / 4.00 m
c) At a distance of 12.0 m:
V₃' = (8.99 * 10^9 N·m²/C²) * (1.50 * 10^-6 C/m) / 12.0 m
Calculating the results:
a) V₁' ≈ 1.35 * 10^6 V
b) V₂' ≈ 6.74 * 10^5 V
c) V₃' ≈ 2.24 * 10^5 V
Therefore, the electric potential at distances of 2.00 m, 4.00 m, and 12.0 m from the line charge are approximately 1.35 * 10^6 V, 6.74 * 10^5 V, and 2.24 * 10^5 V, respectively, when referenced to a distance of 2.50 m from the line of charge where V = 0.
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A rectangular loop of wire is situated so that one end (herght h ) is between the plates of a parallel-Plate capacitor (fig. 7.9), oriented paraliel to the fied E. The other end is way outside, where the field is essentrally zero. What is the emf in this loop? if the dotal resistance is R, What corrent flows? Explain.
The emf in the loop depends on the potential difference inside the capacitor, and the current is determined by the loop's resistance. The external electric field has minimal effect.
In the given scenario, the rectangular loop of wire is situated between the plates of a parallel-plate capacitor. The loop is oriented parallel to the electric field (E) created by the capacitor. One end of the loop is inside the capacitor where the electric field is present, while the other end is outside where the field is essentially zero.
The emf (electromotive force) in the loop can be determined by considering the change in electric potential across the loop. Since the electric field is parallel to the loop, there is no change in electric potential along the sides of the loop that are parallel to the field. However, there is a change in electric potential across the end of the loop that is inside the capacitor.
The emf (ε) in the loop can be calculated using the formula:
ε = ΔV
Where ΔV is the change in electric potential across the end of the loop inside the capacitor.The current flowing through the loop can be determined using Ohm's law:
I = ε / R
Where R is the total resistance of the loop.Since the electric field is essentially zero outside the capacitor, the change in electric potential across the end of the loop outside the capacitor is negligible, and thus the emf in the loop is primarily determined by the change in electric potential across the end inside the capacitor.
Therefore, the emf in the loop is determined by the change in electric potential across the end inside the capacitor, and the current flowing through the loop is given by Ohm's law, with the total resistance of the loop.
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A nonconducting sphere (radius =5.0 cm ) which has a charge of +9nC distributed uniformly throughout its volume. What is the magnitude of the electric field (in N/C) at a point that is 3 cm from the center of the sphere? Your Answer: Answer The answer should be in integer form (No decimals).
The magnitude of the electric field at a point that is 3 cm from the center of the sphere is 18 N/C.The answer to this problem is:The magnitude of the electric field (in N/C) at a point that is 3 cm from the center of the sphere is 18 N/C.
The formula to calculate the electric field due to a uniformly charged sphere at any point is given as:E = k * (Q/R³) * r Where,E = Electric field K = Coulomb's constant = 9 × [tex]10^9[/tex] N.m²/C²Q = Total charge of the sphere R = Radius of the sphere at which the charge is distributed.r = Distance of the point from the center of the sphere
Put the given values in the above formula:R = 5.0 cm = 0.05 mQ = + 9 nC = + 9 × [tex]10^-9[/tex] C (here, charge is given in nano-coulombs, so converting it into coulombs) r = 3 cm = 0.03 m
Therefore,E = k * (Q/R³) * rE = (9 × [tex]10^9[/tex]) * {9 × [tex]10^-9[/tex]/(0.05)³} * (0.03)E = 18 N/C
Therefore, the magnitude of the electric field at a point that is 3 cm from the center of the sphere is 18 N/C.
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How do you know if an object is receiving more heat than it is passing on, receiving less heat than it is passing on, or receiving and passing on the same amount?
To determine if an object is receiving more heat than it is passing on, receiving less heat than it is passing on, or receiving and passing on the same amount, we need to consider the concept of thermal equilibrium.
In thermal equilibrium, two objects that are in contact with each other or part of the same system reach a point where there is no net heat transfer between them. This means that the rate of heat transfer from one object to the other is equal to the rate of heat transfer from the second object back to the first.
If an object is receiving more heat than it is passing on, its temperature will increase over time. This indicates that it is gaining more thermal energy than it is losing, and there is a net heat transfer into the object.
If an object is receiving less heat than it is passing on, its temperature will decrease over time. This indicates that it is losing more thermal energy than it is gaining, and there is a net heat transfer out of the object.
If an object is receiving and passing on the same amount of heat, it will reach a steady state where its temperature remains constant. This indicates that the object is in thermal equilibrium, with the rate of heat transfer into the object equal to the rate of heat transfer out of the object.
In summary:
- If an object's temperature is increasing, it is receiving more heat than it is passing on.
- If an object's temperature is decreasing, it is receiving less heat than it is passing on.
- If an object's temperature remains constant, it is receiving and passing on the same amount of heat, and it is in thermal equilibrium.
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cannonball was shot with an initial velocity of 35 m/s at a 50 degree angle from the ground. what is the maximum height achieved by the ball above the ground in meters
The maximum height achieved by the cannonball above the ground is approximately 24.14 meters.
To find the maximum height achieved by the cannonball, we can analyze its motion using the principles of projectile motion.
Given:
Initial velocity (v₀) = 35 m/s
Launch angle (θ) = 50 degrees
Acceleration due to gravity (g) = 9.8 m/s²
The motion of the cannonball can be divided into horizontal and vertical components.
Vertical motion:
The initial vertical velocity (v₀y) can be calculated using the launch angle:
v₀y = v₀ * sin(θ)
The time taken to reach the maximum height (t_max) can be determined using the equation:
t_max = v₀y / g
Using this time, we can find the maximum height (h_max) reached by the cannonball above the ground using the equation:
h_max = (v₀y)² / (2 * g)
Substituting the given values:
v₀y = 35 m/s * sin(50°)
t_max = (35 m/s * sin(50°)) / 9.8 m/s²
h_max = [(35 m/s * sin(50°))^2] / (2 * 9.8 m/s²)
Calculating these expressions gives us:
v₀y ≈ 27.01 m/s
t_max ≈ 1.46 s
h_max ≈ 24.14 m
Therefore, the maximum height achieved by the cannonball above the ground is approximately 24.14 meters.
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A \( 195 \mathrm{~g} \) block in pressed against a spring of force constant \( 1.50 \mathrm{kN} / \mathrm{m} \) until the block compretistes the spring \( 10.0 \mathrm{~cm} \). The spring rests at the
The spring rests at a position where it exerts a force of 0.15 kN.
When a block is pressed against a spring, it compresses the spring due to the applied force. In this case, a 195 g block is pressed against a spring with a force constant of 1.50 kN/m. The block compresses the spring by 10.0 cm.
To find the position where the spring rests, we can use Hooke's Law, which states that the force exerted by a spring is proportional to its displacement. The formula for Hooke's Law is:
F = k × x
Where F is the force, k is the force constant, and x is the displacement.
Given that the force constant is 1.50 kN/m and the displacement is 10.0 cm, we can convert the displacement to meters:
x = 10.0 cm = 0.10 m
Now we can calculate the force exerted by the spring:
F = (1.50 kN/m) * (0.10 m) = 0.15 kN
Therefore, the spring rests at a position where it exerts a force of 0.15 kN.
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Two long straight wires are parallel and 7.8 cm apart. They are to carry equal currents such that the magnetic field at a point halfway between them has magnitude 230μT. (a) Should the currents be in the same or opposite directions? (b) How much current is needed?
Two long straight wires are parallel and 7.8 cm apart. The currents should be in the same direction. Around 0.0186 A current is needed for the two wires to produce a magnetic field of 230 μT at the midpoint.
(a) To determine whether the currents should be in the same or opposite directions, we can apply the right-hand rule for the magnetic field between parallel wires.
The right-hand rule states that if we point the thumb of our right hand in the direction of the first wire's current, and the fingers in the direction of the second wire's current, the magnetic field at the point between the wires will be in the direction indicated by the extended palm.
Since the magnitude of the magnetic field is given to be non-zero at the midpoint, the currents must be in the same direction. Therefore, the currents should be in the same direction.
(b) To calculate the amount of current needed, we can use the formula for the magnetic field between parallel wires:
B = (μ₀ * I) / (2π * r)
Where B is the magnetic field, μ₀ is the permeability of free space (4π × [tex]10^-^7[/tex] T·m/A), I is the current, and r is the distance between the wires.
Given:
Distance between the wires, r = 7.8 cm = 0.078 m
Magnetic field, B = 230 μT = 230 × [tex]10^-^6\\[/tex] T
Rearranging the formula, we can solve for I:
I = (B * 2π * r) / μ₀
Substituting the given values:
I = (230 × [tex]10^-^6\\[/tex] T * 2π * 0.078 m) / (4π × [tex]10^-^7[/tex] T·m/A)
Simplifying:
I = 0.0186 A
Therefore, a current of approximately 0.0186 A (or 18.6 mA) is needed for the two wires to produce a magnetic field of 230 μT at the midpoint.
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A heart defibrillator passes 11.2 A through a patient's torso for 5.00 ms in an attempt to restore normal beating. (a) How much charge passed? c (b) What voltage was applied if 506 J of energy was dissipated? kV (c) What was the path's resistance? ks (d) Find the temperature increase caused in the 8.00 kg of affected tissue. The specific heat of tissue is 3500 J/(kg⋅
∘
C).
∘
C
The answer can be calculated by the following relations:
Q =I *t (Q= charge, I=current, t=time)
V =I*R (V=voltage, I=current, R=resistance)
E = VIt (E=energy, V=voltage, I=current, t=time)
E = mCΔT (E=energy, m=mass, C= specific heat, ΔT= temperature change)
Current, I = 11.2 A Time, t = 5.00 ms = 5 × 10⁻³ s, Energy dissipated, E = 506 J, Specific heat of tissue, c = 3500 J/(kg.°C), Mass of affected tissue, m = 8.00 kg
(a) Charge, Q passed = I × t = 11.2 A × 5.00 × 10⁻³ s = 0.056 C
(b) Voltage, V applied = E/It = E/Q = 506 J / 0.056 C = 9028.57 V ≈ 9.03 kV
Therefore, the voltage applied is 9.03 kV.
(c) Resistance, R = V/I = (9028.57 V) / (11.2 A) = 806.25 Ω ≈ 806 Ω
Therefore, the path's resistance is 806 Ω.
(d) The formula for the temperature increase caused is given by:ΔT = E / (mc) where ΔT is the temperature increase, c is the specific heat, m is the mass of the affected tissue, and E is the energy released or gained.
Substituting the given values, we get:ΔT = 506 J / (8.00 kg × 3500 J/(kg.°C))
ΔT = 0.0182 °C
Therefore, the temperature increase caused in the 8.00 kg of affected tissue is 0.0182°C.
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You are running in the woods, and you notice a bear running towards you 20m away from you. You know you are close to your car, which is in the opposite direction. Your speed is 4m/s and you know bears can run as fast as 6m/s. If you are 50m away from your car, will you make it safely? Assume you are both running at a constant speed.
a. What is the maximum possible distance you could be from the car to make it safely?
A bear is running towards you 20 m away from you. If you are 50 m away from your car, you would not be able to make it safely to your car. The maximum possible distance you could be from the is less than 20 m.
To determine whether you will make it safely to your car, let's analyze the situation:
Given:
- Your speed: 4 m/s
- Bear's speed: 6 m/s
- Distance between you and the bear: 20 m
- Distance between you and your car: 50 m
To calculate the maximum distance you could be from the car to make it safely, we need to consider the relative speeds between you and the bear. Since you are running in the opposite direction of the bear, your effective speed relative to the bear would be the difference between your speed and the bear's speed.
Relative speed = Your speed - Bear's speed
Relative speed = 4 m/s - 6 m/s
Relative speed = -2 m/s
The negative sign indicates that you and the bear are moving in opposite directions.
Now, let's calculate the time it takes for the bear to reach you:
Time = Distance / Relative speed
Time = 20 m / (-2 m/s)
Time = -10 seconds
The negative time value indicates that the bear would reach you in 10 seconds. However, since time cannot be negative, this implies that the bear has already reached you.
Therefore, if the bear is 20 meters away from you, you would not be able to make it safely to your car, which is 50 meters away. The maximum possible distance you could be from the car to make it safely is less than 20 meters.
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A proton and a bare helium nucleus (which has two protons and two neutrons) are at rest a distance d apart. Find the point P on the line connecting these particles where their combined electric field E is zero, and express P ′s distance (in terms of d ) from the proton
The electric field is given by the formula; E = kQ/r² where k is Coulomb's constant, Q is the magnitude of the charge, and r is the distance from the charge. The force due to an electric field on a particle is given by the formula; F = QE. If the electric field is zero at a point, then the force acting on a particle of charge Q at that point will also be zero. Therefore, the proton will be at equilibrium and won't move.
The equilibrium point on the line connecting the proton and the bare helium nucleus will be at a distance of d/3 from the proton. This point is called the center of mass of the system.The electric field at a distance x from the proton due to the bare helium nucleus is given by;E1 = kQ/r² --- equation (1)where Q is the magnitude of the charge, k is Coulomb's constant, and r is the distance from the charge.The electric field at a distance (d - x) from the proton due to the proton is given by;E2 = kQ/(d - r)² --- equation (2)where Q is the magnitude of the charge, k is Coulomb's constant, and (d - r) is the distance from the charge.
The electric field at point P due to the proton and the bare helium nucleus is given by;E = E1 + E2 --- equation (3)Substituting equations (1) and (2) into equation (3), we get;E = kQ/r² + kQ/(d - r)² --- equation (4)To find the distance x, where the electric field is zero, we will differentiate equation (4) with respect to r and equate the result to zero. This is because a zero electric field at a point implies that the force due to the electric field is also zero, which means that the net force acting on a particle at that point is zero.
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A ball thrown horizontally at 25 m/s travels a horizontal distance of 55 m before hitting the ground From what height was the ball thrown? You may want to review Express your answer using two significant figures with the appropriate units.
The ball was thrown from a height of approximately 15 meters.
To determine the initial height of the ball, we can use the horizontal distance traveled (55 m) and the horizontal velocity (25 m/s). Using the equation h = (1/2) * g * t^2, where g is the acceleration due to gravity (approximately 9.8 m/s^2) and t is the time taken for the ball to hit the ground, we find that the time is approximately 2.2 seconds. Plugging this value into the equation, we calculate the initial height to be approximately 15 meters.
Thus, the ball was thrown from a height of approximately 15 meters.
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What change in entropy occurs when a 0.12 kg steam at 121 deg C is transformed into ice at -28 deg C?
Use the 2nd law of thermodynamics to explain calculated result.
The change in entropy that occurs when a 0.12 kg steam at 121°C is converted into ice at -28°C is calculated to be -0.145 J/K. The 2nd law of thermodynamics explanation is as provided below.
Entropy is a measure of the disorder of a system, and the second law of thermodynamics establishes that the entropy of an isolated system increases over time.
The entropy of a system can be calculated using the equation ΔS = Q/T, where ΔS is the change in entropy, Q is the amount of heat transferred, and T is the absolute temperature of the system.
When a 0.12 kg steam at 121°C is transformed into ice at -28°C, there is a significant change in entropy. The steam has a high level of entropy due to its high temperature and disordered molecular arrangement, while the ice has a low level of entropy due to its low temperature and highly ordered molecular arrangement.
The process of transforming the steam into ice involves a decrease in temperature, which means that heat energy is released from the system.
This results in a negative value for Q in the equation ΔS = Q/T. Since the absolute temperature of the system is decreasing, the value of T is also decreasing, which means that the change in entropy is negative.
To calculate the change in entropy, we need to know the amount of heat transferred and the absolute temperatures of the steam and ice.
The specific heat capacity of water is 4.18 J/g°C, so the amount of heat transferred can be calculated using the equation Q = mcΔT, where m is the mass of the steam, c is the specific heat capacity of water, and ΔT is the change in temperature.
Using this equation, we can calculate that the amount of heat transferred is
Q = (0.12 kg)(4.18 J/g°C)(121°C - (-28°C)) = 22,380 J.
The absolute temperatures of the steam and ice are 394 K and 245 K, respectively, so we can calculate the change in entropy using the equation
ΔS = Q/T = (22,380 J)/(394 K) - (245 K)) = -0.145 J/K.
The negative value for the change in entropy indicates that the process of transforming steam into ice results in a decrease in disorder and an increase in order.
This is in accordance with the second law of thermodynamics, which states that the entropy of an isolated system increases over time.
The transformation of steam into ice represents a decrease in entropy, but the overall entropy of the universe still increases due to the dissipation of heat energy into the surroundings.
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A piece of Nichrome wire has a radius of 6.0×10
−4
m. It is used in a laboratory to make a heater that dissipates 4.00×10
2
W of power when connected to a voltage source of 110 V. Ignoring the effect of temperature on resistance, estimate the necessary length of wire.
The necessary length of the Nichrome wire is approximately 8.68 meters.
To estimate the necessary length of the Nichrome wire, we can use the formula for power dissipation in a resistor:
P = (V^2) / R
where P is the power dissipated, V is the voltage across the resistor, and R is the resistance of the wire.
Power dissipation (P) = 4.00 x 10^2 W
Voltage (V) = 110 V
Radius of the wire (r) = 6.0 x 10^-4 m
First, we need to calculate the resistance of the wire using the formula:
R = (ρ * L) / A
where ρ is the resistivity of the wire, L is the length of the wire, and A is the cross-sectional area of the wire.
The resistivity of Nichrome wire is typically around 1.10 x 10^-6 Ω·m.
Now, we can rearrange the formula for resistance to solve for the length of the wire:
L = (R * A) / ρ
To find the cross-sectional area (A) of the wire, we use the formula for the area of a circle:
A = π * r^2
Substituting the values:
A = π * (6.0 x 10^-4 m)^2
A ≈ 3.14 x 10^-10 m²
Now we can calculate the resistance (R) using the power dissipation and voltage:
R = (V^2) / P
R = (110 V)^2 / 4.00 x 10^2 W
R ≈ 30.25 Ω
Finally, we can calculate the length (L) of the wire:
L = (R * A) / ρ
L = (30.25 Ω * 3.14 x 10^-10 m²) / (1.10 x 10^-6 Ω·m)
L ≈ 8.68 m
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A rock with a mass of 518 g in air is found to have an apparent mass of 325 g when submerged in water. a. What mass of water is displaced? m _w = g b. What is the volume of the rock? V_r =cm ^{3}
c. What is its average density? rho_r =g/cm ^{3}
a. The mass of water displaced is calculated as follows: The mass of water displaced is 193 g.b. The volume of the rock is 518 cm³.c. The average density of the rock is 1 g/cm³.
Apparent loss in mass = Mass in air - Mass in water
Apparent loss in mass = 518 - 325
= 193 g
The mass of water displaced is equal to the mass of the water that has been pushed aside by the rock. As a result, the mass of water displaced is equal to the mass that has been lost by the rock
.Apparent loss in mass = Mass of displaced water
Mass of displaced water = Apparent loss in mass
Mass of displaced water = 193 g
Therefore, the mass of water displaced is 193 g.b. The volume of the rock is calculated using the following formula
:Volume of rock = (Mass of rock) / (Density of water)
The mass of the rock is 518 g. The density of water is 1 g/cm³
.Volume of rock = (518 g) / (1 g/cm³)
Volume of rock = 518 cm³
Therefore, the volume of the rock is 518 cm³.c. The average density of the rock is calculated using the following formula:
Average density of rock = (Mass of rock) / (Volume of rock)
The mass of the rock is 518 g. The volume of the rock is 518 cm³
.Average density of rock = (518 g) / (518 cm³)
Average density of rock = 1 g/cm³
Therefore, the average density of the rock is 1 g/cm³.
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Suppose you have a hardware device that operates an FFT with a window size of 512 samples at 44100 samples per second. You want to use it to completely \& accurately analyze a bass drum with frequencies that range from 50 Hz to 30 Hz; can the device serve this purpose as stated? Yes. No. Yes but only on Friday. Maybe.
Yes, the hardware device can serve the purpose of accurately analyzing a bass drum with frequencies ranging from 50 Hz to 30 Hz. Here's why:
1. The window size of 512 samples refers to the number of samples the device can analyze at once. In this case, it can process 512 samples.
2. The sample rate of 44100 samples per second means that the device can capture 44100 samples in one second.
3. To accurately analyze a signal, we need to ensure that the highest frequency of interest is properly represented in the samples. According to the Nyquist-Shannon sampling theorem, we need to sample at a rate that is at least twice the maximum frequency. In this case, the maximum frequency is 50 Hz.
4. Since the sample rate of 44100 samples per second is more than twice the maximum frequency of 50 Hz, the device can accurately capture the bass drum frequencies ranging from 50 Hz to 30 Hz.
5. The device's window size of 512 samples allows for detailed analysis of the bass drum signal within the specified frequency range. The FFT algorithm can provide information about the different frequencies present in the signal, including the bass drum frequencies.
Therefore, the hardware device can serve the purpose of accurately analyzing a bass drum with frequencies ranging from 50 Hz to 30 Hz.
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please help im not sure what to do coefficient: 0.20 i. The stick man has a mass of \( 90 \mathrm{~kg} \). Assume that the co-efficient of friction is the same 2 between the box and the ground as it is between the stick man's shoe and the ground. Use Ne
The coefficient of friction is a measure of the frictional force between two surfaces in contact. In this case, if the coefficient of friction is the same between the box and the ground as it is between the stick man's shoe and the ground, we can use that value to calculate the frictional force acting on the stick man.
The frictional force can be calculated using the equation[tex]\( f_{\text{friction}} = \mu \cdot N \), where \( f_{\text{friction}} \)[/tex] is the frictional force,[tex]\( \mu \)[/tex] is the coefficient of friction, and[tex]\( N \)[/tex]is the normal force.
To find the normal force, we can use the equation [tex]\( N = m \cdot g \), where \( m \)[/tex] is the mass of the stick man and \( g \) is the acceleration due to gravity.
Substituting the given values, we have[tex]\( N = 90 \, \text{kg} \times 9.8 \, \text{m/s}^2 \).[/tex]
Next, we can calculate the frictional force by multiplying the coefficient of friction by the normal force. Using the given coefficient of friction of 0.20, we have [tex]\( f_{\text{friction}} = 0.20 \times N \).[/tex]
Finally, substituting the value of [tex]\( N \)[/tex] into the equation, we can calculate the frictional force acting on the stick man.
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8. Two long bar magnets are aligned so that north poles face each other. The magnets are separated by 1 cm, and a repulsive force between the north poles is 0.10 N. When the separation is increased to 2 cm the force will be A. 0.10 N. B. 0.050 N. C. 0.025 N. D. 0.20 N. E. 0.40 N.
The force will be D. 0.20 N. Given that two long bar magnets are aligned so that north poles face each other. The magnets are separated by 1 cm, and a repulsive force between the north poles is 0.10 N. We have to find the force between two long bar magnets when the separation is increased to 2 cm.
According to Coulomb's law, the force between two magnetic poles is given by:F = k * (m₁ * m₂)/r²Where,F = Force between two magnetic poles
m₁, m₂ = Strength of poles
r = Distance between two poles
k = Coulomb's constant
For two north poles, the force is repulsive, and k = 10⁻⁷ Nm²/C²We know that, the force between two magnets is given as 0.10 N when the distance between the magnets is 1 cm, i.e. r₁ = 1 cm = 0.01 m
k = 10⁻⁷ Nm²/C²Using above values, we can find the strength of poles as:
m₁ * m₂ = F * r₁²/km₁ * m₂ = 0.10 * (0.01)² / 10⁻⁷m₁ * m₂ = 10⁻⁵ Nm²/C²Now, we need to find the force when the distance is 2 cm, i.e. r₂ = 0.02 mF₂ = k * (m₁ * m₂)/r₂²F₂ = 10⁻⁷ * (10⁻⁵) / 0.02²F₂ = 0.125 N
Approximate answer to two decimal places is 0.13 N.
Therefore, the correct option is D. 0.20 N.
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#2. In a certain cyclotron, a proton moves perpendicular to a magnetic field with a radius of 0.25 m at a frequency of 3.2×10 6Hz. (a) Find the magnitude of the magnetic field.
The magnitude of the magnetic field in the cyclotron is approximately 2.082 Tesla.
In a cyclotron, the centripetal force required for a charged particle to move in a circular path is provided by the magnetic field. The centripetal force is given by:
F = (m*v^2) / r
where F is the centripetal force, m is the mass of the proton, v is its velocity, and r is the radius of the circular path.
In this case, the proton moves perpendicular to the magnetic field, so the magnetic force is:
F = q * v * B
where q is the charge of the proton and B is the magnetic field.
Setting the centripetal force equal to the magnetic force:
(m*v^2) / r = q * v * B
Simplifying and solving for B:
B = (mv) / (qr)
Given that the radius is 0.25 m, the frequency is 3.2 × 10^6 Hz, and the charge of a proton is q = 1.6 × 10^-19 C, we can calculate the magnetic field B.
The velocity v of the proton can be found using the relationship between frequency and velocity in a cyclotron:
v = 2π * r * f
Substituting the values:
v = 2π * (0.25 m) * (3.2 × 10^6 Hz)
Calculating the velocity:
v ≈ 5.024 × 10^6 m/s
Now we can calculate the magnetic field B:
B = (m * v) / (q * r)
= [(1.67 × 10^-27 kg) * (5.024 × 10^6 m/s)] / [(1.6 × 10^-19 C) * (0.25 m)]
≈ 2.082 T
Therefore, The magnitude of the magnetic field in the cyclotron is approximately 2.082 Tesla.
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A long, nonconducting, solid cylinder of radius 5.4 cm has a nonuniform volume charge density rho that is a function of radial distance r from the cylinder axis: rho=Ar
2
. For A=2.1μC/m
5
, what is the magnitude of the electric field at (a)r=1.6 cm and (b)r=11 cm.
To find the magnitude of the electric field at two different radial distances, we can use Gauss's Law.
Which states that the electric field at a point outside a uniformly charged cylinder is directly proportional to the linear charge density.
Given:6
Radius of the cylinder, r = 5.4 cm = 0.054 m
Volume charge density, ρ = Ar^2
A = 2.1 μC/m^5 = 2.1 × 10^-6 C/m^5
(a) At r = 1.6 cm = 0.016 m:
The electric field can be calculated using Gauss's Law as follows:
Let's consider a Gaussian surface in the form of a cylinder of radius r and length L, coaxial with the given solid cylinder.
The charge enclosed within the Gaussian surface is equal to the total charge within the solid cylinder.
Charge enclosed, Q_enclosed = ρ × V
V = πr^2L
Substituting the given values:l
Q_enclosed = (2.1 × 10^-6 C/m^5) × (π × (0.054 m)^2 × L)
= (2.1 × 10^-6) × (π × 0.002916 m^3 × L)
= 6.1092 × 10^-9 L C
Applying Gauss's Law, we know that the electric field outside the cylinder is given by:
E × 2πrL = Q_enclosed / ε₀
E × 2πrL = (6.1092 × 10^-9 L C) / ε₀
Simplifying the equation:
E = (6.1092 × 10^-9 L C) / (2πrL × ε₀)
E = 3.0546 × 10^-9 C / (πrε₀)
Now, we need to calculate the value of ε₀, the permittivity of free space:
ε₀ ≈ 8.854 × 10^-12 C^2/(N·m^2)
Substituting the value of ε₀:
E = 3.0546 × 10^-9 C / (π × 0.016 m × 8.854 × 10^-12 C^2/(N·m^2))
E ≈ 5.4787 × 10^9 N/C
Therefore, the magnitude of the electric field at r = 1.6 cm is approximately 5.4787 × 10^9 N/C.
(b) At r = 11 cm = 0.11 m:
Using the same approach as above, we can calculate the magnitude of the electric field at this distance:
E = 3.0546 × 10^-9 C / (π × 0.11 m × 8.854 × 10^-12 C^2/(N·m^2))
E ≈ 1.3294 × 10^8 N/C
Therefore, the magnitude of the electric field at r = 11 cm is approximately 1.3294 × 10^8 N/C.
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Suppose a point charge creates a 12500 N/C electric field at a distance of 0.45 m. A 50% Part (a) What is the magnitude of the point charge in coulombs? ∣Q∣= Hints: deduction per hint. Hints remaining: Feedback: 0% deduction per feedback.
According to the question the magnitude of the point charge is approximately 0.01125 C.
We can use the equation for the electric field [tex](\(E\))[/tex] created by a point charge [tex](\(Q\))[/tex] at a distance [tex](\(r\))[/tex] from the charge:
[tex]\[E = \frac{kQ}{r^2}\][/tex]
where [tex]\(k\)[/tex] is the Coulomb's constant.
In this case, we are given that the electric field [tex](\(E\))[/tex] is 12500 N/C and the distance [tex](\(r\))[/tex] is 0.45 m. We need to solve for the magnitude of the point charge [tex](\(|Q|\))[/tex].
Rearranging the equation, we have:
[tex]\[|Q| = \frac{Er^2}{k}\][/tex]
Substituting the given values and the value of the Coulomb's constant [tex](\(k = 9 \times 10^9 \, \text{N}\cdot\text{m}^2/\text{C}^2\))[/tex], we can calculate the magnitude of the point charge:
[tex]\[|Q| = \frac{(12500 \, \text{N/C})(0.45 \, \text{m})^2}{9 \times 10^9 \, \text{N}\cdot\text{m}^2/\text{C}^2}\][/tex]
Simplifying the expression:
[tex]\[|Q| = 0.01125 \, \text{C}\][/tex]
Therefore, the magnitude of the point charge is approximately 0.01125 C.
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geologist is surveying a cave. She follows a path 170 m straight West, then 250 m in the direction 45° East of South and then 280m 30° East of North. Find the magnitude and direction of her total displacement using the three methods: (1) Parallelogram method (2) Polygon method (3) Component method
The geologist's total displacement using three methods is as follows:
1) Parallelogram method:
Magnitude: Approximately 158.5 mDirection: Approximately 60.4° North of East2) Polygon method:
Magnitude: Approximately 252.5 mDirection: Approximately -8.4° South of East (or 171.6° East of South)3) Component method:
Magnitude: Approximately 252.5 mDirection: Approximately -8.4° South of East (or 171.6° East of South)Thus, regardless of the method used, the geologist's total displacement is approximately 252.5 m, and the direction is approximately -8.4° South of East (or 171.6° East of South).
To find the magnitude and direction of the geologist's total displacement using the three methods, let's calculate them step by step:
(1) Parallelogram Method:
In the parallelogram method, we create a parallelogram using the two longest displacement vectors (170 m straight West and 280 m 30° East of North). The diagonal of the parallelogram represents the resultant displacement.
First, let's resolve the displacement vectors into their horizontal (x) and vertical (y) components:
170 m West: -170 m in the x-direction
280 m 30° East of North:
280 * cos(30°) ≈ 242.51 m in the x-direction, 2
80 * sin(30°) ≈ 140 m in the y-direction
Next, we add the x and y components separately:
Total x-component = -170 m + 242.51 m
≈ 72.51 m
Total y-component = 140 m
Using the Pythagorean theorem, we can find the magnitude of the resultant displacement:
Magnitude = √(Total x-component^2 + Total y-component^2)
Magnitude = √((72.51)^2 + (140)^2)
Magnitude ≈ 158.5 m
To find the direction of the resultant displacement, we can use trigonometry:
Direction = tan^(-1)(Total y-component / Total x-component)
Direction = tan^(-1)(140 / 72.51)
Direction ≈ 60.4° North of East
Therefore, using the parallelogram method, the magnitude of the geologist's total displacement is approximately 158.5 m, and the direction is approximately 60.4° North of East.
(2) Polygon Method:
In the polygon method, we create a polygon by connecting the displacement vectors in sequential order and draw the resultant vector from the starting point to the end point.
To calculate the magnitude and direction using the polygon method, we can add the x and y components of the displacement vectors:
170 m West: -170 m in the x-direction
250 m 45° East of South:
250 * cos(45°) ≈ 176.78 m in the x-direction, -
250 * sin(45°) ≈ -176.78 m in the y-direction (opposite direction)
280 m 30° East of North:
280 * cos(30°) ≈ 242.51 m in the x-direction,
280 * sin(30°) ≈ 140 m in the y-direction
Total x-component = -170 m + 176.78 m + 242.51 m
≈ 249.29 m
Total y-component = -176.78 m + 140 m = -36.78 m
Magnitude = √(Total x-component^2 + Total y-component^2)
Magnitude = √((249.29)^2 + (-36.78)^2)
Magnitude ≈ 252.5 m
Direction = tan^(-1)(Total y-component / Total x-component)
Direction = tan^(-1)(-36.78 / 249.29)
Direction ≈ -8.4° South of East (or 171.6° East of South)
Therefore, using the polygon method, the magnitude of the geologist's total displacement is approximately 252.5 m, and the direction is approximately -8.4° South of East (or 171.6° East of South).
(3) Component Method:
In the component method, we find the horizontal and vertical components of each displacement vector and then add them to find the resultant displacement.
Resolve each displacement vector into its horizontal (x) and vertical (y) components:
170 m West: -170 m in the x-direction, 0 m in the y-direction
250 m 45° East of South:
250 * cos(45°) ≈ 176.78 m in the x-direction, -
250 * sin(45°) ≈ -176.78 m in the y-direction (opposite direction)
280 m 30° East of North:
280 * cos(30°) ≈ 242.51 m in the x-direction,
280 * sin(30°) ≈ 140 m in the y-direction
Total x-component = -170 m + 176.78 m + 242.51 m ≈ 249.29 m
Total y-component = 0 m - 176.78 m + 140 m = -36.78 m
Magnitude = √(Total x-component^2 + Total y-component^2)
Magnitude = √((249.29)^2 + (-36.78)^2)
Magnitude ≈ 252.5 m
Direction = tan^(-1)(Total y-component / Total x-component)
Direction = tan^(-1)(-36.78 / 249.29)
Direction ≈ -8.4° South of East (or 171.6° East of South)
Therefore, using the component method, the magnitude of the geologist's total displacement is approximately 252.5 m, and the direction is approximately -8.4° South of East (or 171.6° East of South).
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A 8.9-kg child sits in a 3.7-kg high chair. Find the normal force exerted by the floor on the chair.
The normal force exerted by the floor on the chair is approximately 123.48 Newtons.
To find the normal force exerted by the floor on the chair, we need to consider the gravitational forces acting on the child and the chair.
The gravitational force acting on the child can be calculated using the formula:
F_child = m_child × g
where
m_child = mass of the child = 8.9 kg
g = acceleration due to gravity = 9.8 m/s²
F_child = 8.9 kg × 9.8 m/s²
F_child = 87.22 N
Similarly, the gravitational force acting on the chair can be calculated using the same formula:
F_chair = m_chair × g
where
m_chair = mass of the chair = 3.7 kg
g = acceleration due to gravity = 9.8 m/s²
F_chair = 3.7 kg × 9.8 m/s²
F_chair = 36.26 N
Since the child is sitting in the chair, the normal force exerted by the floor on the chair is equal to the sum of the child's weight and the chair's weight:
Normal force = F_child + F_chair
Normal force = 87.22 N + 36.26 N
Normal force = 123.48 N
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If the depth of a swimming pool is 3 m and the density of water in the swimming pool is 10^3kg/m^3, the gage pressure measured at the bottom of the swimming pool is: Multiple Choice a) 29.4kPa b) 23.5kPa c) 14.4kPa d) 3.0kPa
The depth of a swimming pool is 3 m and the density of water in the swimming pool is 10^3 kg/m³. Therefore, we need to determine the gauge pressure measured at the bottom of the swimming pool.
We can begin the solution to this problem by first using the formula for gauge pressure, which is given as: P gauge = ρghwhere P gauge is the gauge pressure, ρ is the density of the fluid, g is the acceleration due to gravity, and h is the depth of the fluid from the surface.
Let's use this formula to determine the gauge pressure measured at the bottom of the swimming pool:
P gauge = ρghP gauge = (10³ kg/m³)(9.81 m/s²)
(3 m)P gauge = 29,430 Pa We have converted 29,430
Pa to kilopascals (kPa) by dividing by 1000.
The gauge pressure measured at the bottom of the swimming pool is 29.4 kPa.Option (a) is the correct answer.
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A hot-air balloon is rising upward with a constant speed of 3.95 m/s. When the balloon is 7.29 m above the ground, the balloonist accidentally drops a compass over the side of the balloon. How much time elapses before the compass hits the ground? Number Units
To find the time it takes for the compass to hit the ground, we can use the kinematic equation:
\[ h = \frac{1}{2}gt^2 \]
Where:
- \( h \) is the height (7.29 m)
- \( g \) is the acceleration due to gravity (9.8 m/s\(^2\))
- \( t \) is the time we're trying to find
Rearranging the equation, we have:
\[ t = \sqrt{\frac{2h}{g}} \]
Substituting the given values, we can calculate the time:
\[ t = \sqrt{\frac{2 \times 7.29}{9.8}} \]
\[ t \approx 1.13 \text{ seconds} \]
Therefore, it takes approximately 1.13 seconds for the compass to hit the ground after it is dropped from the hot-air balloon.
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the velocity components for a two-dimensional flow are
The velocity components for a two-dimensional flow are Vx = V cos θVy = V sin θ, where V = 150/√(tan²θ + 1)
The velocity components for a two-dimensional flow are as follows: Let V be the velocity and θ be the angle that the velocity vector makes with the x-axis.
Vx = V cos θ
Vy = V sin θ
Given,150 = √(Vx²+Vy²).....(1)
Using the above expression (1), we can find V.
Now, Vx/Vy = tan θ ⇒ Vx = V tan θ
Putting the value of Vx in terms of V in the expression (1),
we have 150 = √(V²tan²θ + V²)150 = V√(tan²θ + 1)
Dividing by √(tan²θ + 1), we get150/√(tan²θ + 1) = V
Therefore, the velocity components for a two-dimensional flow are Vx = V cos θVy = V sin θ, where V = 150/√(tan²θ + 1)
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Talk about the Thermal Requests in Machining Processes contemplating in your answer: (1) definition; (2) causes (associate with the chip root regions); (3) heat dissipation pathways; (4) consequences of high heat; (5) the factors that can influence the amount of heat generated and how they influence it.
Thermal requests in machining processes refer to the excess heat that is generated during the process, which can lead to detrimental effects to both the tool and workpiece. In this answer, we will explore thermal requests in machining processes in more detail, including the definition, causes, heat dissipation pathways, consequences of high heat, and factors that can influence the amount of heat generated and how they influence it.
Thermal requests in machining processes are the excess heat generated during the process that can cause damage to the tool and workpiece. The amount of heat generated is influenced by several factors, including the cutting speed, feed rate, and depth of cut, as well as the material properties of the tool and workpiece.
The causes of thermal requests are primarily due to the chip root regions where the temperature is higher than other regions. The cutting speed, feed rate, and depth of cut can cause localized heating in the chip root regions, leading to an increase in the temperature of the tool and workpiece.
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76 kg runner comes to a stop at the end of the race, slowing from 6 m/s to rest in the space of 4 seconds, moving a distance of 12 m in the process. How much work did the ground do on the runner in this process? Question 2 1 pts What is the most reasonable estimate for the power output of a person who is greatly exerting themselves? 100,000 W 1,000,000 W 100 W 10 W 1 W 0.1 W 10,000 W 0.01 W 1.000 W
To calculate the work done by the ground on the runner, we need to use the work-energy principle. The work done is equal to the change in kinetic energy.
Given:
Mass of the runner (m) = 76 kg
Initial velocity (u) = 6 m/s
Final velocity (v) = 0 m/s
Time taken (t) = 4 seconds
Distance traveled (s) = 12 m
First, we can calculate the change in kinetic energy:
Change in kinetic energy = (1/2) * m * (v^2 - u^2)
Substituting the values:
Change in kinetic energy = (1/2) * 76 * (0^2 - 6^2) = -1368 J
Since the runner comes to a stop, the work done by the ground is equal to the negative change in kinetic energy:
Work done by the ground = -(-1368) J = 1368 J
Therefore, the ground did 1368 Joules of work on the runner.
For Question 2:
The most reasonable estimate for the power output of a person who is greatly exerting themselves can vary depending on the context and individual. However, a reasonable range for power output during intense physical activity is typically between 100 W to 1000 W. Therefore, the options 100 W, 1000 W, and 10,000 W would be the most reasonable estimates for the power output of a person who is greatly exerting themselves.
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A small point charge with q=20mC and mass 1 gram is being held in place 0.5 cm from the negatively charged plate of a capacitor as shown in the figure below - which also shows the total capacitor gap distance of 0.6 cm and voltage across the plates of 20Volts. The charged particle is released. What will its velocity be when it is 0.25 cm from the negatively charged plate of the capacitor?
The velocity of the charged particle, when it is 0.25 cm from the negatively charged plate of the capacitor, is approximately 89.44 m/s.
To determine the velocity of the charged particle when it is 0.25 cm from the negatively charged plate of the capacitor, we can use the principle of conservation of energy.
The initial potential energy of the charged particle, when it is held in place, is given by:
U_initial = q *
where q is the charge of the particle (20 mC) and V is the voltage across the capacitor plates (20 V).
The final kinetic energy of the particle, when it is 0.25 cm from the plate, can be calculated using:
K_final = (1/2) * m * v^2
where m is the mass of the particle (1 gram = 0.001 kg) and v is its velocity.
Since there is no non-conservative work done on the particle, the initial potential energy is converted into kinetic energy at the final position:
U_initial = K_final
Substituting the values and solving for v:
(20 mC * 20 V) = (0.5 * 10^(-3) kg) * (v^2)
v^2 = (20 mC * 20 V) / (0.5 * 10^(-3) kg)
v^2 = 8000 m^2/s^2
v ≈ 89.44 m/s
Therefore, the velocity of the charged particle when it is 0.25 cm from the negatively charged plate of the capacitor is approximately 89.44 m/s.
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