PHYSICS COLLEGE

We know we have exerted of force even when we have done no work this is called _____

Answers

Answer 1

Answer:

Answer: The correct answer is zero work done.

Explanation:

Work is said to be done when the object moves through a distance when the force is applied to the object.

If the object does not move a distance even the force is exerted on the object then the work done is zero in this case.

Therefore, when the force is exerted even when no work is done then this is called zero work done.

Answer 2

Answer:

Final answer:

Force is experienced even when no work is done, such as when pushing against a wall. This is due to the fact that work in physics requires force to be applied over a distance. When no movement occurs, no work is done, yet a force was still exerted.

Explanation:

The concept you're referring to is known as force, a fundamental aspect in Newton's laws of motion. According to Newton's third law, every action has an equal and opposite reaction. So, when you push against a wall, it pushes back with an equal amount of force, even though no movement occurs, and therefore no work is done. This is due to the role distance plays in the calculation of work. In the physics sense, work is done when a force is applied over a certain distance.

This is also tied to the concept of potential energy. For example, when a force causes an object to deform, such as compressing a spring, the work done is stored as potential energy in the object until it is released. Yet, if the object does not move or deform, no work has been done, but a force was still exerted.

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0.67 m/s2
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54 m/s2

Answers

4.0 m/s2

it's 9 squared divided by 6

the answer is B 4.0 m/s2

An isotropic point source emits light at wavelength 500 nm, at the rate of 185 W. A light detector is positioned 380 m from the source. What is the maximum rate ∂B/∂t at which the magnetic component of the light changes with time at the detector's location?

Answers

Answer:

$(dB)/(dt) = 3.49 *10^(6) \ \ T/s$

Explanation:

Given that

An isotropic point source emits light at a wavelength $\lambda$ = 500 nm

Power = 185 W

Radius = 380 m

Let's first calculate the The intensity of the wave , which is = $(Power )/(Area)$

= $(Power)/(4 \pi r ^2)$

= $(185 \ W)/( 4 \pi (380)^2)$

= $1.0195*10^(-4) \ W/m^2$

Now;

The amplitude of the magnetic field is calculated afterwards by using poynting vector

i.e

$I = ((c)/(2 \mu_0 ))B_(max^2)$

$B_(max^2) = ((2 \mu_0 I)/( c))$

$B_(max^2) = ((2 *4 \pi *10^(-7)*1.0195*10^(-4))/( 3*10^8))$

$B_(max^2) = 8.5409*10^(-19)$

$B_(max) = \sqrt {8.5409*10^(-19)}$

$B_(max) = 9.242*10^(-10)$

The magnetic field wave equation can now be expressed as;

$B = B_(max) sin (kx - \omega t)$

Taking the differentiation

$(dB)/(dt)= - \omega B_(max) \ cos ( kx - \omega t)$

The maximum value ;

$(dB)/(dt) = \omega B _(max)$

where ;

$\omega = 2 \pi f\n\omega = (2 \pi c)/(\lambda)$

then

$(dB)/(dt) = (2 \pi c)/(\lambda) B _(max)$

$(dB)/(dt) = (2 \pi 3*10^8*9.242*10^(-10))/(500*10^(-9))$

$(dB)/(dt) = 3484751.917$

$(dB)/(dt) = 3.49 *10^(6) \ \ T/s$

The maximum rate $(∂B/∂t)$ at which the magnetic component of the light changes with time at the detector's location is approximately $6.8 x 10^9$ Tesla per second (T/s).

To find the maximum rate at which the magnetic component of the light changes with time at the detector's location, you can use the formula for the rate of change of magnetic field due to an electromagnetic wave. The formula is given by:

$∂B/∂t = (2π / λ) * E * c$

Where:

$∂B/∂t$ is the rate of change of the magnetic field.

λ is the wavelength of the light.

E is the electric field strength.

c is the speed of light in a vacuum, approximately $3 x 10^8 m/s.$

You have the wavelength (λ) as 500 nm, which is 500 x 10^-9 meters, and the electric field strength (E) can be calculated using the power (P) and the distance (r) from the source. The power emitted by the source is 185 W, and the distance from the source to the detector is 380 m.

First, calculate the electric field strength (E):

$E = sqrt(P / (2π * r^2))$

$E = sqrt(185 W / (2π * (380 m)^2))$

$E = sqrt(185 W / (2π * 144400 m^2))$

$E ≈ 6.325 x 10^-5 N/C$

Now, you can calculate the rate of change of the magnetic field:

$∂B/∂t = (2π / λ) * E * c$

$∂B/∂t = (2π / (500 x 10^-9 m)) * (6.325 x 10^-5 N/C) * (3 x 10^8 m/s)$

$∂B/∂t ≈ (3.77 x 10^15 Hz) * (6.325 x 10^-5 N/C) * (3 x 10^8 m/s)$

$∂B/∂t ≈ 6.8 x 10^9 T/s$

So, the maximum rate at which the magnetic component of the light changes with time at the detector's location is approximately $6.8 x 10^9$ Tesla per second (T/s).

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A 220 kg crate hangs from the end of a rope of length L = 14.0 m. You push horizontally on the crate with a varying force F to move it 4.00 m to the side.(a) What is the magnitude of F when the crate is in this final position? During the crate's displacement, what are
(b) the total work done on it,
(c) the work done by the gravitational force on the crate, and
(d) the work done by the pull on the crate from the rope?
(e) Knowing that the crate is motionless before and after its displacement, use the answers to (b), (c), and (d) to find the work your force F does on the crate.
(f) Why is the work of your force not equal to the product of the horizontal displacement and the answer to (a)?

Answers

Answer:

(a) magnitude of F = 797 N

(b)the total work done W = 0

(c)work done by the gravitational force = -1.55 kJ

(d)the work done by the pull = 0

(e) work your force F does on the crate = 1.55 kJ

Explanation:

Given:

Mass of the crate, m = 220 kg

Length of the rope, L = 14.0m

Distance, d = 4.00m

(a) What is the magnitude of F when the crate is in this final position

Let us first determine vertical angle as follows

=> $Sin \theta = (d )/(L)$

=> $\theta = Sin^(-1) (d)/(L)$ =

Now substituting thje values

=> $\theta = Sin^(-1) (4)/(12)$ =

=> $\theta = Sin^(-1) (1)/(3)$

=> $\theta = Sin^(-1)(0.333)$

=> $\theta = 19.5^(\circ)$

Now the tension in the string resolve into components

The vertical component supports the weight

=> $Tcos\theta = mg$

=> $T = (mg)/(cos\theta)$

=> $T = (230 * 9.8 )/(cos(19.5))$

=> $T = (2254 )/(cos(19.5))$

=> $T = (2254 )/(0.9426)$

=>T =2391N

Therefore the horizontal force

$F = TSin(19.5)$

F = 797 N

b) The total work done on it

As there is no change in Kinetic energy

The total work done W = 0

c) The work done by the gravitational force on the crate

The work done by gravity

Wg = Fs.d = - mgh

Wg = - mgL ( 1 - Cosθ )

Substituting the values

= $-230 * 9.8* 12 ( 1 - cos(19.5) )$

= $-230 * 9.8* 12 ( 1 - 0.9426) )$

= $-230 * 9.8* 12 (0.0574)$

= $-230 * 9.8* 0.6888$

= $-230 * 6.750$

= -1552.55 J

The work done by gravity = -1.55 kJ

d) the work done by the pull on the crate from the rope

Since the pull is perpendicular to the direction of motion,

The work done = 0

e)Find the work your force F does on the crate.

Work done by the Force on the crate

WF = - Wg

WF = -(-1.55)

WF = 1.55 kJ

(f) Why is the work of your force not equal to the product of the horizontal displacement and the answer to (a)

Here the work done by force is not equal to F*d

and it is equal to product of the cos angle and F*d

So, it is not equal to the product of the horizontal displacement and the answer to (a)

A grandfather clock keeps time using a pendulum consisting of a light rod connected to a small heavy mass. With a rod of length L, the period of oscillation is 2.00 s. What should the length of the rod be for the period of the oscillations to be 1.00 s?

Answers

Answer:

The length of the rod should be

$(L)/(4) \n$

Explanation:

Period of simple pendulum is given by

$T=2\pi\sqrt{(l)/(g)} \n$

We have

$(T_1^2)/(T_2^2)=(l_1)/(l_2)\n\n(2^2)/(1^2)=(L)/(l_2)\n\nl_2=(L)/(4) \n$

The length of the rod should be

$(L)/(4) \n$

A rigid tank initially contains 3kg of carbon dioxide (CO2) at a pressure of 3bar.The tank is connected by a valve to a frictionless piston-cylinder assembly located vertically above, initially containing 0.5 m^3 of CO2. The piston area is 0.1 m^2. Initially the pressure of the CO2 in the piston-cylinder assembly is 2 bar. The ambient pressure and temperature are 1 bar and 290 K. Although the valveis closed, a small leak allows CO2 to flow slowly into the cylinder from the tank. Owing to heat transfer, the temperature of the CO2 throughout the tank and the piston-cylinder assembly stays constant at 290K. You can assume ideal gas behavior for CO2.Determine the following:

a. The total amount of energy transfer by work (kJ)
b. The total amount of energy transfer by heat (kJ)

Answers

Answer:

Part a: The total amount of energy transfer by the work done is 54.81 kJ.

Part b: The total amount of energy transfer by the heat is 54.81 kJ

Explanation:

Mass of Carbon Dioxide is given as m1=3 kg

Pressure is given as P1=3 bar =300 kPA

Volume is given as V1=0.5 m^3

Pressure in tank 2 is given as P2=2 bar=200 kPa

T=290 K

Now the Molecular weight of $CO_2$ is given as

M=44 kg/kmol

the gas constant is given as

$R=\frac{\bar{R}}{M}\nR=(8.314)/(44)\nR=0.189 kJ/kg.K$

Volume of the tank is given as

$V=(mRT)/(P_1)\nV=(3 * 0.189 * 290)/(300 )\nV=0.5481 m^3$

Final mass is given as

$m_2=(P_2V)/(RT)\nm_2=(200* 0.5481)/(0.189* 290)\nm_2=2 kg$

Mass of the CO2 moved to the cylinder

$m=m_1-m_3\nm=3-2=1 kg$

The initial mass in the cylinder is given as

$m_((cyl)_1)=(P_((cyl)_1)V_1)/(RT)\nm_((cyl)_1)=(200* 0.5)/(0.189 * 290)\nm_((cyl)_1)=1.82 kg$

The mass after the process is

$m_((cyl)_2)=m_((cyl)_1)+m\nm_((cyl)_2)=1.82+1\nm_((cyl)_2)=2.82\n$

Now the volume 2 of the cylinder is given as

$V_((cyl)_2)=(m_((cyl)_2)RT)/(P_2)\nm_((cyl)_2)=(2.82* 0.189* 290)/(200)\nm_((cyl)_1)=0.774 m^3$

Part a:

So the Work done is given as

$W=P(V_2-V_1)\nW=200(0.774-0.5)\nW=54.81 kJ$

The total amount of energy transfer by the work done is 54.81 kJ.

Part b:

The total energy transfer by heat is given as

$Q=\Delta U+W\nQ=0+W\nQ=54.81 kJ$

As the temperature is constant thus change in internal energy is 0.

The total amount of energy transfer by the heat is 54.81 kJ

Select all the statements regarding electric field line drawings that are correct. Group of answer choices:
1. Electric field lines are the same thing as electric field vectors.
2. Electric field line drawings allow you to determine the approximate direction of the electric field at a point in space.
3. The number of electric field lines that start or end at a charged particle is proportional to the amount of charge on the particle.
4. The electric field is strongest where the electric field lines are close together.

Answers

Answer:

All statement are correct.

Explanation:

1. Electric field lines are the same thing as electric field vectors, electric field are mathematically vectors quantity. These vectors point in the direction in which a positive test charge would move.

2. Electric field line drawings allow you to determine the approximate direction of the electric field at a point in space. Yes it is correct tangent drawn at any point on these lines gives the direction of electric filed at that point.

3. The number of electric field lines that start or end at a charged particle is proportional to the magnitude of charge on the particle, is a correct statement.

4.The electric field is strongest where the electric field lines are close together, again a correct statement as relative closeness of field lines indicate a stronger strength of electric field.

Hence we can say that all the statement are correct.

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When water freezes, its volume increases by 9.05% (that is, ΔV / V0 = 9.05 × 10-2). What force per unit area is water capable of exerting on a container when it freezes? (It is acceptable to use the bulk modulus of water, B = 2.2 × 109 N/m2, for this problem.) Give your answer in N/cm2.

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Impulse is the______of the force and time of contact

A 6.0-cm-diameter horizontal pipe gradually narrows to 4.0 cm. When water flows through this pipe at a certain rate, the gauge pressure in these two sections is 32.0 kPa and 24.0 kPa, respectively. What is the volume rate of flow?