Answer:
118.06 days
Explanation:
d = distance of the center of moon from surface of planet = 2.315 x 10⁵ km = 2.315 x 10⁸ m
R = radius of the planet = 4.15 x 10³ km = 4.15 x 10⁵ m
r = center to center distance between the planet and moon = R + d
M = mass of the planet = 7.15 x 10²² kg
T = Time period of revolution around the planet
Using Kepler's third law
[tex]T^{2}=\frac{4\pi ^{2}r^{3}}{GM}[/tex]
[tex]T^{2}=\frac{4\pi ^{2}(R + d)^{3}}{GM}[/tex]
[tex]T^{2}=\frac{4(3.14)^{2}((4.15\times 10^{5}) + (2.315\times 10^{8}))^{3}}{(6.67\times 10^{-11})(7.15\times 10^{22})}[/tex]
T = 1.02 x 10⁷ sec
we know that , 1 day = 24 h = 24 x 3600 sec = 86400 sec
T = [tex](1.02 \times 10^{7} sec)\frac{1 day}{86400 sec}[/tex]
T = 118.06 days
A 1400-kg car is traveling with a speed of 17.7 m/s. What is the magnitude of the horizontal net force that is required to bring the car to a halt in a distance of 33.1 m?
Answer:
The magnitude of the horizontal net force is 13244 N.
Explanation:
Given that,
Mass of car = 1400 kg
Speed = 17.7 m/s
Distance = 33.1 m
We need to calculate the acceleration
Using equation of motion
[tex]v^2-u^2=2as[/tex]
Where, u = initial velocity
v = final velocity
s = distance
Put the value in the equation
[tex]0-(17.7)^2=2a\times 33.1[/tex]
[tex]a=\dfrac{-(17.7)^2}{33.1}[/tex]
[tex]a=-9.46\ m/s^2[/tex]
Negative sign shows the deceleration.
We need to calculate the net force
Using newton's formula
[tex]F = ma[/tex]
[tex]F =1400\times(-9.46)[/tex]
[tex]F=-13244\ N[/tex]
Negative sign shows the force is opposite the direction of the motion.
The magnitude of the force is
[tex]|F| =13244\ N[/tex]
Hence, The magnitude of the horizontal net force is 13244 N.
A coil of 1000 turns of wire has a radius of 12 cm and carries a counterclockwise current of 15 A. If it is lying flat on the ground, and the Earth’s magnetic field points due north, has a magnitude of 5.8 x 10-5 T, and makes a downward angle of 25° with the vertical, what is the torque on the loop?
Answer:
0.01663 Nm
Explanation:
N = 1000, r = 12 cm = 0.12 m, i = 15 A, B = 5.8 x 10^-5 T, θ = 25
Torque = N i A B Sinθ
Torque = 1000 x 15 x 3.14 x 0.12 x 0.12 x 5.8 x 10^-5 x Sin 25
Torque = 0.01663 Nm
In case of a damped oscillation. (a) Amplitude increases (b) Remains constant (c) None (d) Amplitude reduces
Answer:
option (d)
Explanation:
There are two types of oscillations.
1. Damped oscillations: The oscillations in which the amplitude of a wave goes on decreasing continuously are called damped oscillations.
For example, the oscillations of a seismic wave.
2. Undamped oscillations: The oscillations in which the amplitude o a wave remains constant are called undamped oscillations.
for example, oscillations of light waves.
A fisherman notices that his boat is moving up and down periodically without any horizontal motion, owing to waves on the surface of the water. It takes a time of 2.80 s for the boat to travel from its highest point to its lowest, a total distance of 0.600 m. The fisherman sees that the wave crests are spaced a horizontal distance of 5.50 m apart.
Part A How fast are the waves traveling? Express the speed v in meters per second using three significant figures.
Answer:
Very fast!
Explanation:
The waves are traveling very fast. I know this because they're not traveling slow for sure.
If this is incorrect, they're probably traveling moderately fast.
A rock is thrown straight up. What is the net external force acting on the rock when it is at the top of its trajectory?
Answer:
Net external force acting on the rock when it is at the top of its trajectory is force due to gravity (mg).
Explanation:
The forces acting on a rock thrown up are force due to gravity and air resistance. Air resistance is directly proportional to velocity of rock, when velocity is zero air resistance is zero. When it is at the top of its trajectory its velocity is zero. So air resistance is also zero. Hence only gravitational force acts on the rock.
Net external force acting on the rock when it is at the top of its trajectory is force due to gravity (mg).
At the peak of its trajectory, the only significant force acting on a rock is the force of gravity. Thus, the net external force on the system is represented as -mg, where m is the mass of the rock and g is the acceleration due to gravity. At this point, the rock begins its descent due to the gravitational pull.
Explanation:The question is asking about the net external force acting on a rock when it is at the top of its trajectory. At the peak of its trajectory, the only significant force acting on the rock is the force of gravity (weight of the rock), assuming air resistance is negligible. Thus, the net external force on the system, if air resistance is neglected, is represented as -mg, where m is the mass of the rock and g is the acceleration due to gravity.
For example, if a rock is thrown straight upwards with an initial velocity, as it reaches the topmost point of its trajectory, the vertical velocity momentarily becomes zero before the rock starts descending again. At this point, the only major force acting on the rock is the gravitational pull exerted by the Earth, also known as the rock's weight.
It is important to note that this net external force (-mg) is what determines the acceleration of the object at the top of its trajectory. Simply put, even at the top of its path, the rock is still experiencing the force of gravity which pulls it back towards Earth, leading to its eventual fall back to the ground.
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While finding the spring constant, if X1 = 12 cm, X2 = 15 cm, and hanging mass = 22 grams, the value of spring constant K would be:________ (write your answer in newtons/meter)
Answer:
If x₁=12 cm then k=1.7985 N/m
If x₂=15 cm then k=1.4388 N/m
Explanation:
Hanging mass= 22 g=0.022 kg
Acceleration due to gravity g=9.81 m/s²
If x₁=displacement= 12 cm=0.12 m
k= spring constant
[tex]F=ma\\\Rightarrow F=0.022\times 9.81\\\Rightarrow F=0.21582\ N[/tex]
[tex]\text {For spring}\\F=kx\\\Rightarrow 0.21582=k\times 0.012\\\Rightarrow k=1.7985\ N/m\\[/tex]
∴k = 1.7985 N/m
If x₂=15 cm=0.15 m
Force of the hanging mass is same however the spring constant will change
[tex]\text {For spring}\\F=kx\\\Rightarrow 0.21582=k\times 0.015\\\Rightarrow k=1.4388\ N/m\\[/tex]
∴k = 1.4388 N/m
As the mass is not changing the spring constant has to change. That means that here there are two spring one with k=1.7985 N/m and the other with k= 1.4388 N/m
The waste products of combustion leave the internal combustion engine through the: A. crankshaft, B. exhaust valve. C. cylinder. D. intake valve.
Answer:
B.Exhaust Value
Explanation:
The waste products of combustion leave the internal combustion engine through the Exhuast Value.
Answer:
Exhaust Value
Explanation:
A 52.5-turn circular coil of radius 5.35 cm can be oriented in any direction in a uniform magnetic field having a magnitude of 0.455 T. If the coil carries a current of 25.3 mA, find the magnitude of the maximum possible torque exerted on the coil.
Answer:
5.43 x 10^-3 Nm
Explanation:
N = 52.5, radius, r = 5.35 cm = 0.0535 m, B = 0.455 T, I = 25.3 mA = 0.0253 A
Torque = N I A B Sin theta
Here, theta = 90 degree
Torque = 52.5 x 0.0253 x 3.14 x 0.0535 x 0.0535 x 0.455
Torque = 5.43 x 10^-3 Nm
Gas is confined in a tank at a pressure of 9.8 atm and a temperature of 29.0°C. If two-thirds of the gas is withdrawn and the temperature is raised to 83.0°C, what is the pressure of the gas remaining in the tank? 9.1e-21 Incorrect: Your answer is incorrect.
Answer:
3.9 atm
Explanation:
Ideal gas law states:
PV = nRT
where P is absolute pressure,
V is volume,
n is number of moles,
R is universal gas constant,
and T is absolute temperature.
The volume of the tank is constant, so we can say:
V₁ = V₂
Using ideal gas law to write in terms of P, n, R, and T:
n₁ R T₁ / P₁ = n₂ R T₂ / P₂
n₁ T₁ / P₁ = n₂ T₂ / P₂
Initially:
P₁ = 9.8 atm
T₁ = 29.0°C = 302.15 K
n₁ = n
Afterwards:
P₂ = P
T₂ = 83.0°C = 356.15 K
n₂ = n/3
Substituting:
n (302.15 K) / (9.8 atm) = (n/3) (356.15 K) / P
(302.15 K) / (9.8 atm) = (1/3) (356.15 K) / P
P = 3.85 atm
Rounding to 2 significant figures, P = 3.9 atm.
When a 0.15 kg baseball is hit, its velocity changes from +17 m/s to -17 m/s.
(a) What is the magnitude of the impulse delivered by the bat to the ball?
N·s
(b) If the baseball is in contact with the bat for 1.5 ms, what is the magnitude of the average force exerted by the bat on the ball?
Final answer:
The magnitude of the impulse delivered by the bat to the ball is 5.1 kg·m/s. The magnitude of the average force exerted by the bat on the ball is 3.4 × 10^3 N.
Explanation:
To find the magnitude of the impulse delivered by the bat to the ball, we need to use the formula for impulse:
Impulse = change in momentum
Momentum is given by the product of mass and velocity (momentum = mass × velocity). The change in momentum is the final momentum minus the initial momentum (change in momentum = final momentum - initial momentum).
Given that the mass of the baseball is 0.15 kg, the initial velocity is +17 m/s, and the final velocity is -17 m/s, we can calculate the magnitude of the impulse:
Impulse = (0.15 kg × -17 m/s) - (0.15 kg × 17 m/s) = -5.1 kg·m/s
So, the magnitude of the impulse delivered by the bat to the ball is 5.1 kg·m/s.
To find the magnitude of the average force exerted by the bat on the ball, we can use the formula for average force:
Average force = impulse / time
Given that the time the ball is in contact with the bat is 1.5 ms (1.5 × 10^-3 s) and the magnitude of the impulse is 5.1 kg·m/s, we can calculate the magnitude of the average force:
Average force = 5.1 kg·m/s / (1.5 × 10^-3 s) = 3.4 × 10^3 N
Therefore, the magnitude of the average force exerted by the bat on the ball is 3.4 × 10^3 N.
A circular coil that has 100 turns and a radius of 10.0 cm lies in a magnetic field that has a magnitude of 0.0650 T directed perpendicular to the coil. (a) What is the magnetic flux through the coil? (b) The magnetic field through the coil is increased steadily to 0.100 T over a time interval of 0.500 s. What is the magnitude of the emf induced in the coil during the time interval?
Answer:
(a) 0.204 Weber
(b) 0.22 Volt
Explanation:
N = 100, radius, r = 10 cm = 0.1 m, B = 0.0650 T, angle is 90 degree with the plane of coil, so theta = 0 degree with the normal of coil.
(a) Magnetic flux, Ф = N x B x A
Ф = 100 x 0.0650 x 3.14 x 0.1 0.1
Ф = 0.204 Weber
(b) B1 = 0.0650 T, B2 = 0.1 T, dt = 0.5 s
dB / dt = (B2 - B1) / dt = (0.1 - 0.0650) / 0.5 = 0.07 T / s
induced emf, e = N dФ/dt
e = N x A x dB/dt
e = 100 x 3.14 x 0.1 x 0.1 x 0.07 = 0.22 V
(a) The magnetic flux through a coil 0.204 Weber
(b) The emf induced in the coil during the time interval 0.22 Volt
What will be the magnetic flux and emf induced in the coil?It is given that
Number of the turns N = 100,
Radius of the coil r = 10 cm = 0.1 m,
The magnetic field B = 0.0650 T
Since the angle is 90 degrees with the plane of the coil, so [tex]\Theta[/tex]= 0 degrees with the normal coil.
(a) The Magnetic flux, will be given as
[tex]\rm\phi =N\times B\times A[/tex]
By putting the values
[tex]\phi =100\times0.0650\times3.14\times 0.1\times0.1[/tex]
[tex]\phi=0.204 \rm \ weber[/tex]
(b) The emf induced will to be given as
[tex]B_1=0.0650T\ \ B_2=0.1T\ \ dt=0.5s[/tex]
[tex]\dfrac{dB}{dt} =\dfrac{B_2-B_1}{dt} =\dfrac{0.1-0.650}{0.5} =0.07\dfrac{T}{s}[/tex]
induced emf,
[tex]e=N\dfrac{d\phi}{dt}[/tex]
[tex]e=N\times A \times \dfrac{dB}{dt}[/tex]
[tex]e=100\times 3.14\times0.1\times0.1\times0.07=0.22V[/tex]
Thus
(a) The magnetic flux through a coil 0.204 Weber
(b) The emf induced in the coil during the time interval 0.22 Volt
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A piece of wood with density of 824 kg/m^3 is tied to the bottom of a pool and the wood does not move. The volume of the wood is 1.2 m^3. What is the tension is in the rope?
Answer:
2069.76 N
Explanation:
density of wood = 824 kg/m^3
density of water = 1000 kg/m^3
Volume of wood = 1.2 m^3
True weight of wood = volume of wood x density of wood x gravity
True weight of wood = 1.2 x 824 x 9.8 = 9690.24 N
Buoyant force acting on wood = volume of wood x density of water x gravity
Buoyant force acting on wood = 1.2 x 1000 x 9.8 = 11760 N
Tension in the rope = Buoyant force - True weight
tension in rope = 11760 - 9690.24 = 2069.76 N
Photons with wavelength 1 pm are incident on electrons. What is the frequency of the Compton-scattered photons at an angle of 60°?
Answer:
f = 1.354*10^{20} Hz
Explanation:
By conservation of linear momentum, wavelength shift due to collision of photon to electron is given by following formula
[tex]\lambda ^{'}-\lambda =\frac{h}{m_{o}c}(1-cos\theta )[/tex]
where h is plank constant = 6.626*10^{-34}
c = speed of light = 3*10^{8} m/s
scattered angle = 60 degree
m = rest mass of electron = 9*10^{-31}
[tex]\lambda ^{'}=10^{-12} +\frac{6.626*10^{-34}}{9*10^{-31}*3*10^{8}}(1-cos60^{o} )[/tex]
[tex]\lambda ^{'}= 2.215 pm[/tex]
we know that 1 pm = 10^{-12}m
[tex]f = \frac{c}{\lambda ^{'}}[/tex]
[tex]f = \frac{3*10^{8}}{2.215 *10^{-12}} = 1.354*10^{20} Hz[/tex]
f = 1.354*10^{20} Hz
A coil of self- induction 0.5 H, its ohmic resistance is 10 ⦠carrying a dc current of intensity 2 A, The voltage at its terminals isâ¦â¦ A. Zero
B. 20
C. 5
D. 0.2
Answer:
Option (B)
Explanation:
L = 0.5 H, I = 2 A, R = 10 Ohm
V = I R
In case of dc the inductor cannot works.
V = 2 x 10 = 20 V
Can the dielectric permittivity of an insulator be negative?
Max (mass = 15 kg) is hanging from one end of a 13-m long bungee cord that has its other end fixed to a bridge above. The bungee cord has a circular cross section with a diameter of 2.5 cm, and a Young’s modulus of 17 MPa. What is the stress in the bungee cord due to Max’s weight?
The stress in the bungee cord due to the Max's weight is [tex]\fbox{\begin \\ 299511\text{ N}/\text{m}^{2}\end{minispace}}[/tex] or [tex]\fbox{\begin\\ 2.99\times10^{5}\text{ N}/\text{m}^{2}\end{minispace}}[/tex] or [tex]\fbox{\begin\\3.0\times10^{5}\text{ N}/\text{m}^{2}\end{minispace}}[/tex].
Further Explanation:
The stress developed in the bungee cord is the amount of tensile force developed inside the bungee cord due to the Max's weight.
Given:
The diameter d of the bungee cord is [tex]2.5\text{ cm}[/tex].
The mass m of Max is [tex]15\text{ kg}[/tex].
Concept:
The stress developed in the bungee cord is given by:
[tex]\fbox{\begin \sigma =\dfrac{F}{A}\end{minispace}}[/tex] ... (1)
Here, [tex]\sigma[/tex] is the stress developed in the bungee cord, [tex]F[/tex]is the force due to Max's weight and [tex]A[/tex] is the area of cross-section of the bungee cord.
The radius of the bungee cord will be the half of its diameter.
[tex]r=\dfrac{d}{2}[/tex]
Substitute [tex]2.5\text{ cm}[/tex] for [tex]d[/tex].
[tex]r= \dfrac{2.5\text{ cm}}{2}\\r=1.25 \text{ cm}[/tex]
Convert the radius of the bungee cord in meter.
[tex]r=\dfrac{1.25}{100} \text{ m}\\r=0.0125\text{ m}[/tex]
The area of cross-section of the bungee cord is given by:
[tex]A=\pi r^{2}[/tex]
Substitute [tex]0.0125\text{ m}[/tex] for [tex]r[/tex].
[tex]A=\pi (0.0125\text{ m})^{2}\\A=4.908\times10^{-4} m^{2}[/tex]
The force on the bungee cord due to Max's mass is the gravitational force acting on Max's body. The gravitational force acting on Max's body is given by:
[tex]F=mg[/tex]
Here, m is the mass of Max's body and g is acceleration due to gravity.
Consider the value of acceleration due to gravity on Earth be [tex]9.80\text{ m}/\text{s}^{2}[/tex].
Substitute [tex]15\text{ kg}[/tex] for m and [tex]9.80\text{ m}/\text{s}^{2}[/tex] for g in above expression.
[tex]F=(15\text{ kg})\times(9.80\text{ m}/\text{s}^{2})\\F=147\text{ N}[/tex]
Substitute [tex]147\text{ N}[/tex] for F and [tex]4.980\times10^{-4}\text{ m}^{2}[/tex] for A in equation (1).
[tex]\sigma=\dfrac{147\text{ N}}{4.908\times10^{-4}\text{ m}^{2}}\\\sigma=299511\text{ N}/\text{m}^{2}[/tex]
Thus, the stress developed in the bungee cord is [tex]\fbox{\begin \\ 299511\text{ N}/\text{m}^{2}\end{minispace}}[/tex] or [tex]\fbox{\begin\\ 2.99\times10^{5}\text{ N}/\text{m}^{2}\end{minispace}}[/tex] or [tex]\fbox{\begin\\3.0\times10^{5}\text{ N}/\text{m}^{2}\end{minispace}}[/tex].
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Answer Details:
Grade: Senior school
Subject: Physics
Chapter: Stress and Strain
Keywords:
Max, Max's weight, force, stress, area, 299511 N/m2, 299511 N/m^2, 2.99x10^5 N/m2, gravitational, radius, bungee, cord, weight, developed, mass, 3x10^5 N/m^2.
The stress in the bungee cord due to Max’s weight is 294,000 N/m².
The given parameters;
mass of Max, m = 15 kgdiameter of the cord, d = 2.5 cmradius of the cord, r = = 1.25 cm = 0.0125 mYoung's modulus of the rod, E = 17 MPaThe area of the bungee cord is calculated as follows;
[tex]A = \pi r^2\\\\A = \pi \times (0.0125)^2\\\\ A= 0.0005 \ m^2[/tex]
The weight of Max is calculated as follows;
[tex]F = mg\\\\F = 15 \times 9.8\\\\F = 147 \ N[/tex]
The stress in the bungee cord due to Max’s weight is calculated as;
[tex]\sigma = \frac{F}{A} \\\\\sigma = \frac{147}{0.0005} \\\\\sigma = 294,000 \ N/m^2[/tex]
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A 2.07-kg fish is attached to the lower end of an unstretched vertical spring and released. The fish drops 0.131 m before momentarily coming to rest. (a) What is the spring constant of the spring? (b) What is the period of the oscillations of the fish? ?
Answer:
part a)
k = 310 N/m
part b)
T = 0.51 s
Explanation:
Part A)
As per work energy theorem we have
Work done by gravity + work done by spring = change in kinetic energy
[tex]mgx - \frac{1}{2}kx^2 = 0[/tex]
[tex](2.07)(9.8)(0.131) - \frac{1}{2}k(0.131)^2 = 0[/tex]
now we will have
[tex] k = 310 N/m[/tex]
Part B)
Time period of oscillation is given as
[tex]T = 2\pi\sqrt{\frac{m}{k}}[/tex]
[tex]T = 2\pi\sqrt{\frac{2.07}{310}}[/tex]
[tex]T = 0.51 s[/tex]
A box rests on the back of a truck. The coefficient of friction between the box and the surface is 0.32. (3 marks) a. When the truck accelerates forward, what force accelerates the box? b. Find the maximum acceleration the truck can have before the box slides.
Answer:
Part a)
here friction force will accelerate the box in forward direction
Part b)
a = 3.14 m/s/s
Explanation:
Part a)
When truck accelerates forward direction then the box placed on the truck will also move with the truck in same direction
Here if they both moves together with same acceleration then the force on the box is due to friction force between the box and the surface of the truck
So here friction force will accelerate the box in forward direction
Part b)
The maximum value of friction force on the box is known as limiting friction
it is given by the formula
[tex]F = \mu mg[/tex]
so we have
[tex]F = ma = \mu mg[/tex]
now the acceleration is given as
[tex]a = \mu g[/tex]
[tex]a = (0.32)(9.8) = 3.14 m/s^2[/tex]
Final answer:
The force that accelerates the box when the truck moves forward is the force of static friction. The maximum acceleration the truck can have before the box slides can be found using the equation: max acceleration = coefficient of friction x acceleration due to gravity.
Explanation:
When the truck accelerates forward, the force that accelerates the box is the force of static friction between the box and the surface of the truck bed. This force opposes the motion of the box and allows it to stay in place on the truck.
The maximum acceleration the truck can have before the box slides can be found using the equation:
max acceleration = coefficient of friction x acceleration due to gravity
Using the given coefficient of friction between the box and the surface, you can substitute the value and solve for the maximum acceleration.
A driver parked his car on a steep hill and forgot to set the emergency brake. As a result, the car rolled down the hill and crashed into a parked truck. If the car was moving at 10 mph (4.5 m/s) when it hit the truck 7.0 seconds after it began to move, what was the car’s average acceleration while it rolled down the hill? Define the positive x direction to be down the hill.
Answer:0.642
Explanation:
Given
initially car is at rest i.e u=0
When car hits the truck it's velocity is 4.5 m/s
it hits the truck after 7 sec
Now average acceleration of car is =[tex]\frac{change\ in\ velocity}{time\ taken}[/tex]
average acceleration of car =[tex]\frac{\left ( final velocity\right )-\left ( initial velocity\right )}{time}[/tex]
average acceleration of car=[tex]\frac{\left ( 4.5\right )-\left ( 0\right )}{7}[/tex]
average acceleration of car=[tex]0.642 m/s^{2}[/tex]
Taking Down hill as positive x axis
average acceleration of car=[tex]0.642\hat{i}[/tex]
A 66 g66 g ball is thrown from a point 1.05 m1.05 m above the ground with a speed of 15.1 ms/15.1 ms . When it has reached a height of 1.59 m1.59 m , its speed is 10.9 ms/10.9 ms . What was the change in the mechanical energy of the ball-Earth system because of air drag
Answer:
[tex]\Delta E = 8.20 - 4.95 = 3.25 J[/tex]
Explanation:
Initial total mechanical energy is given as
[tex]ME = U + KE[/tex]
here we will have
[tex]U = mgh[/tex]
[tex]U = (0.066)(9.81)(1.05) = 0.68 J[/tex]
also we have
[tex]KE = \frac{1}{2}mv^2[/tex]
[tex]KE = \frac{1}{2}(0.066)(15.1)^2[/tex]
[tex]KE = 7.52 J[/tex]
[tex]ME_i = 0.68 + 7.52 = 8.2 J[/tex]
Now similarly final mechanical energy is given as
[tex]U = mgh[/tex]
[tex]U = (0.066)(9.81)(1.59) = 1.03 J[/tex]
also we have
[tex]KE = \frac{1}{2}mv^2[/tex]
[tex]KE = \frac{1}{2}(0.066)(10.9)^2[/tex]
[tex]KE = 3.92 J[/tex]
[tex]ME_f = 1.03 + 3.92 = 4.95 J[/tex]
Now change in mechanical energy is given as
[tex]\Delta E = ME_i - ME_f[/tex]
[tex]\Delta E = 8.20 - 4.95 = 3.25 J[/tex]
Which of the following is created by fans? a) Air flow b) Water flow c) Wastewater flow d) Vacuum flow
Starting from rest, the boy runs outward in the radial direction from the center of the platform with a constant acceleration of 0.5 m/s2 . The platform rotates at a constant rate of 0.2 rad/s. Determine his velocity (magnitude) when t = 3 sec
Answer:
His velocity when t= 3 sec is V= 1.56 m/s
Explanation:
a= 0.5 m/s²
ω= 0.2 rad/s
t= 3 sec
Vr= a*t
Vr= 1.5 m/s
r= a*t²/2
r= 2.25m
Vt= w*r
Vt= 0.45 m/s
V= √(Vr²+Vt²)
V= 1.56 m/s
The boy's tangential velocity, after accelerating radially for 3 seconds at 0.5 m/s^2 while the platform rotates at 0.2 rad/s, is 1.5 m/s.
Explanation:To determine the boy's velocity (magnitude) at t = 3 sec while he runs outward radially from the center of a rotating platform, we must consider both his radial (tangential) velocity due to his acceleration and the rotational movement of the platform.
The boy starts from rest with a constant radial acceleration of 0.5 m/s2. Using the kinematic equation v = u + at (where u is the initial velocity, a is the acceleration, and t is the time), we can calculate his radial velocity after 3 seconds as:
vradial = 0 + (0.5 m/s2)(3 s) = 1.5 m/s
The platform's constant rotation rate is given as 0.2 rad/s. This rotational movement does not change the boy's radial (tangential) velocity directly, as the question only asks for his velocity in the radial direction. Therefore, the total magnitude of the boy's velocity after 3 seconds is 1.5 m/s tangentially.
An elevator car, with a mass of 450 kg is suspended by a single cable. At time = 0s, the elevator car is raised upward. The tension on the cable is constant at 5000N during the first 3 seconds of operation. Determine the magnitude of the velocity of the elevator at time = 3 seconds.
Answer:
33.33 m/s
Explanation:
m = 450 kg. T = 5000 N, t = 3 seconds,
let the net acceleration is a.
T = m a
a = 5000 / 450 = 11.11 m/s^2
u = 0 , v = ?
Let v be the velocity after 3 seconds.
Use first equation of motion
v = u + a t
v = 0 + 11.11 x 3 = 33.33 m/s
A capacitor is connected to a 9 V battery and acquires a charge Q. What is the charge on the capacitor if it is connected instead to an 18 V battery? A capacitor is connected to a 9 V battery and acquires a charge Q. What is the charge on the capacitor if it is connected instead to an 18 V battery? Q Q/2 2Q 4Q
Answer:
Option C is the correct answer.
Explanation:
We have charge stored in a capacitor
Q = CV
C is the capacitance and V is the voltage.
A capacitor is connected to a 9 V battery and acquires a charge Q.
That is
Q = C x 9 = 9C
What is the charge on the capacitor if it is connected instead to an 18 V battery
That is
Q' = C x 18 = 9C x 2 = 2Q
Option C is the correct answer.
When the voltage on a capacitor is doubled, the charge on the capacitor doubles as well. Hence, if a capacitor initially has a charge Q at 9V, it will have a charge of 2Q at 18V.
Explanation:The subject of the question is the behavior of a capacitor when it is connected to various voltage sources. The charge Q on a capacitor is given by the equation Q = CV, where C is the capacitance of the capacitor and V is the voltage across it. So if a capacitor is charged by a 9V battery (Q = C*9V), and then connected to an 18V battery, its new charge (lets call it Q1) would be C*18V. Therefore, Q1 = 2Q, since the voltage is doubled, hence, the charge on the capacitor also doubles.
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A blacksmith heats a 1.5 kg iron horseshoe to 500 C, containing 20 kg of water at 18 C then plunges it into a bucket What is the equilbrlum temperature? Express your answer using two significant figures.
Answer:
21.85 C
Explanation:
mass of iron = 1.5 kg, initial temperature of iron, T1 = 500 C
mass of water = 20 kg, initial temperature of water, T2 = 18 C
let T be the equilibrium temperature.
Specific heat of iron = 449 J/kg C
specific heat of water = 4186 J/kg C
Use the principle of caloriemetry
heat lost by the hot body = heat gained by the cold body
mass of iron x specific heat of iron x decrease in temperature = mass of water x specific heat of water x increase in temperature
1.5 x 449 x (500 - T) = 20 x 4186 x (T - 18)
336750 - 673.5 T = 83720 T - 1506960
1843710 = 84393.5 T
T = 21.85 C
A Nichrome wire is used as a heating element in a toaster. From the moment the toaster is first turned on to the moment the wire reaches it maximum temperature, the current in the wire drops by 20% from its initial value. What is the temperature change in the wire? The temperature coefficient for Nichrome is α = 0.0004 (°C)-1.
Answer:
Temperature change in the wire is 625°C
Explanation:
Let the initial current in the wire be I₀
Thus,
the final current in the wire will be, I = (1 - 0.20) × I₀ = 0.80I₀
Given α = 0.0004 1/°C
Now,
Resistance = (potential difference (E))/current (I)
also
R = R₀ [1 + αΔT]
Where ΔT is the increase in temperature , thus
E/I = E/I₀ [1 + α × ΔT)
on substituting the values in the above equation, we get
1/(0.80I₀) = 1/(I₀) × [1 + 0.0004 × ΔT]
or
1/(0.80) = [1 + 0.0004 × ΔT)]
or
ΔT = 625°C
The temperature change in the Nichrome wire, based on a 20% drop in current, is calculated to be approximately 625°C.
A Nichrome wire's resistance increases with temperature, primarily determined by its temperature coefficient.
Step-by-Step Solution :
Let I₀ be the initial current and I be the current after heating.E/I = E/I₀ [1 + α × ΔT)
on substituting the values in the above equation, we get
= 1/(0.80I₀) = 1/(I₀) × [1 + 0.0004 × ΔT]
= 1/(0.80) = [1 + 0.0004 × ΔT)]
= ΔT = 625°C
The temperature change in the Nichrome wire is approximately 625°C.
A generator is built with 500 turns.The area of each turn is 0.02 m^2 , and the external magnetic field through the coils is 0.30 T. If a turbine rotates the coil at 960 rpm, what is the induced rms voltage?
Answer:
213.18 volt
Explanation:
N = 500, A = 0.02 m^2, B = 0.30 T, f = 960 rpm = 960 / 60 = 16 rps
The maximum value of induced emf is given by
e0 = N x B x A x ω
e0 = N x B x A x 2 x π x f
e0 = 500 x 0.3 x 0.02 x 2 x 3.14 x 16
e0 = 301.44 Volt
The rms value of induced emf is given by
[tex]e_{rms} = \frac{e_{0}}{\sqrt{2}}[/tex]
[tex]e_{rms} = \frac{301.44}{\sqrt{2}}[/tex]
e rms = 213.18 volt
The mean time between collisions for electrons in room temperature copper is 2.5 x 10-14 s. What is the electron current in a 2 mm diameter copper wire where the internal electric field strength is 0.01 V/m?
Answer:
1.87 A
Explanation:
τ = mean time between collisions for electrons = 2.5 x 10⁻¹⁴ s
d = diameter of copper wire = 2 mm = 2 x 10⁻³ m
Area of cross-section of copper wire is given as
A = (0.25) πd²
A = (0.25) (3.14) (2 x 10⁻³)²
A = 3.14 x 10⁻⁶ m²
E = magnitude of electric field = 0.01 V/m
e = magnitude of charge on electron = 1.6 x 10⁻¹⁹ C
m = mass of electron = 9.1 x 10⁻³¹ kg
n = number density of free electrons in copper = 8.47 x 10²² cm⁻³ = 8.47 x 10²⁸ m⁻³
[tex]i[/tex] = magnitude of current
magnitude of current is given as
[tex]i = \frac{Ane^{2}\tau E}{m}[/tex]
[tex]i = \frac{(3.14\times 10^{-6})(8.47\times 10^{28})(1.6\times 10^{-19})^{2}(2.5\times 10^{-14}) (0.01)}{(9.1\times 10^{-31})}[/tex]
[tex]i[/tex] = 1.87 A
Determine the force between two long parallel wires, that are separated by a dis- tance of 0.1m the wire on the left has a current of 10A and the one on the right BT has a current of 15A, both going up. 31十131
Answer:
3 x 10^-4 N/m
Attractive
Explanation:
r = 0.1 m, i1 = 10 A, i2 = 15 A
The force per unit length between two parallel wires is given by
[tex]F = \frac{\mu _{0}}{4\pi }\times \frac{2i_{1}i_{2}}{r}[/tex]
[tex]F = \frac{10^{-7}\times 2\times 10\times 15}{0.1}[/tex]
F = 3 x 10^-4 N/m
Thus, the force per unit length between two wires is 3 x 10^-4 N/m, the force is attractive in nature because the direction of flow of current in both the wires is same.
Three balls are tossed with same initial speed from a fourth floor dormitory window. Ball A is launched 45 degrees above the horizontal, Ball B is launched 45 degrees below the horizontal, and Ball C is launched horizontally. Which ball hits the ground with the greatest speed
Answer:
All the balls hit the ground with same velocity.
Explanation:
Let the speed of throwing be u and height of building be h.
Ball A is launched 45 degrees above the horizontal
Initial vertical speed = usin45
Vertical acceleration = -g
Height = -h
Substituting in v² = u² +2as
v² = (usin45)² -2 x g x (-h)
v² = 0.5u² +2 x g x h
Final vertical speed² = 0.5u² +2 x g x h
Initial horizontal speed = final horizontal speed = ucos45
Final horizontal speed² = 0.5u²
Magnitude of final velocity
[tex]v=\sqrt{0.5u^2+2gh+0.5u^2}=\sqrt{u^2+2gh}[/tex]
Ball C is launched horizontally
Initial vertical speed = 0
Vertical acceleration = -g
Height = -h
Substituting in v² = 0² +2as
v² = 0² -2 x g x (-h)
v² = 2 x g x h
Final vertical speed² = 2 x g x h
Initial horizontal speed = final horizontal speed = u
Final horizontal speed² =u²
Magnitude of final velocity
[tex]v=\sqrt{2gh+u^2}=\sqrt{u^2+2gh}[/tex]
Ball B is launched 45 degrees below the horizontal
Initial vertical speed = usin45
Vertical acceleration = g
Height = h
Substituting in v² = u² +2as
v² = (usin45)² +2 x g x h
v² = 0.5u² +2 x g x h
Final vertical speed² = 0.5u² +2 x g x h
Initial horizontal speed = final horizontal speed = ucos45
Final horizontal speed² = 0.5u²
Magnitude of final velocity
[tex]v=\sqrt{0.5u^2+2gh+0.5u^2}=\sqrt{u^2+2gh}[/tex]
All the magnitudes are same so all the balls hit the ground with same velocity.
The ball C which was launched horizontally will hit the ground fastest.
Time of motion of the balls
With respect to angle of projection above the horizontal, the time of motion each ball is calculated as follows;
[tex]T = \frac{u sin\theta }{g}[/tex]
where;
θ is the angle of projection above the horizontalu is the initial velocityT is the time to hit the groundWhen θ is 45 degrees above the horizontal;
[tex]T = \frac{u \times sin(45)}{9.8}\\\\ T = 0.072u[/tex]
When θ is 45 degrees below the horizontal = 135 degrees above the horizontal
[tex]T = \frac{ u \times sin(135)}{9.8} \\\\T = 0.072 \ s[/tex]
When launched horizontally, θ = 90 degrees above the horizontal
[tex]T = \frac{u \times sin(90)}{9.8} \\\\T = 0.1 u[/tex]
Thus, the ball C which was launched horizontally will hit the ground fastest.
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