Ask question

Ask question

All categories

2 years ago

13

Imagine that you replace the block in the video with a happy or sad ball identical to the one used as a pendulum, so that the sa

d ball strikes a sad ball and the happy ball strikes a happy ball. The target balls are free to move, and all the balls have the same mass. In the collision between the sad balls, how much of the balls' kinetic energy is dissipated?
a) None of it
b) All of it
c) Half of it

1 answer:

gizmo_the_mogwai [7]2 years ago

6 0

Answer: a) The Answer to the question is option a) None of it.

Explanation:

The reason is because according to the law of conservation of energy Energy can neither be created nor destroyed but can be transformed from one form to the other. Therefore none of the kinetic energy was dissipated, rather it was transformed to another form of energy.

You might be interested in

Which of the following is NOT a good way to reduce fuel consumption?

Sidana [21]

Answer:

Explanation:

d

4 0

2 years ago

Read 2 more answers

A shift in one fringe in the Michelson-Morley experiment corresponds to a change in the round-trip travel time along one arm of

olya-2409 [2.1K]

Explanation:

When Michelson-Morley apparatus is turned through $90^{o}$ then position of two mirrors will be changed. The resultant path difference will be as follows.

$\frac{lv^{2}}{\lambda c^{2}} - (-\frac{lv^{2}}{\lambda c^{2}}) = \frac{2lv^{2}}{\lambda c^{2}}$

Formula for change in fringe shift is as follows.

n = $\frac{2lv^{2}}{\lambda c^{2}}$

$v^{2} = \frac{n \lambda c^{2}}{2l}$

v = $\sqrt{\frac{n \lambda c^{2}}{2l}}$

According to the given data change in fringe is n = 1. The data is Michelson and Morley experiment is as follows.

l = 11 m

$\lambda = 5.9 \times 10^{-7} m$

c = $3.0 \times 10^{8}$ m/s

Hence, putting the given values into the above formula as follows.

v = $\sqrt{\frac{n \lambda c^{2}}{2l}}$

= $\sqrt{\frac{1 \times (5.9 \times 10^{-7} m) \times (3.0 \times 10^{8})^{2}}{2 \times 11 m}}$

= $2.41363 \times 10^{9} m/s$

Thus, we can conclude that velocity deduced is $2.41363 \times 10^{9} m/s$ .

3 0

1 year ago

A wire with a length of 150 m and a radius of 0.15 mm carries a current with a uniform current density of 2.8 x 10^7A/m^2. The c

Mrac [35]

Answer:

The current is 2.0 A.

(A) is correct option.

Explanation:

Given that,

Length = 150 m

Radius = 0.15 mm

Current density $J=2.8\times10^{7}\ A/m^2$

We need to calculate the current

Using formula of current density

$J = \dfrac{I}{A}$

$I=J\timesA$

Where, J = current density

A = area

I = current

Put the value into the formula

$I=2.8\times10^{7}\times\pi\times(0.15\times10^{-3})^2$

$I=1.97=2.0\ A$

Hence, The current is 2.0 A.

7 0

1 year ago

A compact, dense object with a mass of 2.90 kg is attached to a spring and is able to oscillate horizontally with negligible fri

enot [183]

(a) 80 N/m

The spring constant can be found by using Hooke's law:

$F=kx$

where

F is the force on the spring

k is the spring constant

x is the displacement of the spring relative to the equilibrium position

At the beginning, we have

F = 16.0 N is the force applied

x = 0.200 m is the displacement from the equilibrium position

Solving the formula for k, we find

$k=\frac{F}{m}=\frac{16.0 N}{0.200 m}=80 N/m$

(b) 0.84 Hz

The frequency of oscillation of the system is given by

$f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}$

where

k = 80 N/m is the spring constant

m = 2.90 kg is the mass attached to the spring

Substituting the numbers into the formula, we find

$f=\frac{1}{2\pi}\sqrt{\frac{80 N/m}{2.90 kg}}=0.84 Hz$

(c) 1.05 m/s

The maximum speed of a spring-mass system is given by

$v=\omega A$

where

$\omega$ is the angular frequency

A is the amplitude of the motion

For this system, we have

$\omega=2\pi f=2\pi (0.84 Hz)=5.25 rad/s$

$A=0.200 m$ (the amplitude corresponds to the maximum displacement, so it is equal to the initial displacement)

Substituting into the formula, we find the maximum speed:

$v=(5.25 rad/s)(0.200 m)=1.05 m/s$

(d) x = 0

The maximum speed in a simple harmonic motion occurs at the equilibrium position. In fact, the total mechanical energy of the system is equal to the sum of the elastic potential energy (U) and the kinetic energy (K):

$E=U+K=\frac{1}{2}kx^2+\frac{1}{2}mv^2$

where

k is the spring constant

x is the displacement

m is the mass

v is the speed

The mechanical energy E is constant: this means that when U increases, K decreases, and viceversa. Therefore, the maximum kinetic energy (and so the maximum speed) will occur when the elastic potential energy is minimum (zero), and this occurs when x=0.

(e) 5.51 m/s^2

In a simple harmonic motion, the maximum acceleration is given by

$a=\omega^2 A$

Using the numbers we calculated in part c):

$\omega=2\pi f=2\pi (0.84 Hz)=5.25 rad/s$

$A=0.200 m$

we find immediately the maximum acceleration:

$a=(5.25 rad/s)^2(0.200 m)=5.51 m/s^2$

(f) At the position of maximum displacement: $x=\pm 0.200 m$

According to Newton's second law, the acceleration is directly proportional to the force on the mass:

$a=\frac{F}{m}$

this means that the acceleration will be maximum when the force is maximum.

However, the force is given by Hooke's law:

$F=kx$

so, the force is maximum when the displacement x is maximum: so, the maximum acceleration occurs at the position of maximum displacement.

(g) 1.60 J

The total mechanical energy of the system can be found by calculating the kinetic energy of the system at the equilibrium position, where x=0 and so the elastic potential energy U is zero. So we have

$E=K=\frac{1}{2}mv_{max}^2$

where

m = 2.90 kg is the mass

$v_{max}=1.05 m/s$ is the maximum speed

Solving for E, we find

$E=\frac{1}{2}(2.90 kg)(1.05 m/s)^2=1.60 J$

(h) 0.99 m/s

When the position is equal to 1/3 of the maximum displacement, we have

$x=\frac{1}{3}(0.200 m)=0.0667 m$

so the elastic potential energy is

$U=\frac{1}{2}kx^2=\frac{1}{2}(80 N/m)(0.0667 m)^2=0.18 J$

and since the total energy E = 1.60 J is conserved, the kinetic energy is

$K=E-U=1.60 J-0.18 J=1.42 J$

And from the relationship between kinetic energy and speed, we can find the speed of the system:

$v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(1.42 J)}{2.90 kg}}=0.99 m/s$

(i) 1.84 m/s^2

When the position is equal to 1/3 of the maximum displacement, we have

$x=\frac{1}{3}(0.200 m)=0.0667 m$

So the restoring force exerted by the spring on the mass is

$F=kx=(80 N/m)(0.0667 m)=5.34 N$

And so, we can calculate the acceleration by using Newton's second law:

$a=\frac{F}{m}=\frac{5.34 N}{2.90 kg}=1.84 m/s^2$

8 0

1 year ago

A photon is scattered from an initially stationary electron within a metal. How does the frequency of the photon change upon sca

sammy [17]

Answer:

The frequency of the photon decreases upon scattering

Explanation:

Here we note that when a photon is scattered by a charged particle, it is referred to as Compton scattering.

Compton scattering results in a reduction of the energy of the photon and hence an increase in the wavelength (from λ to λ') of the photon known as Compton effect.

Therefore, since the wavelength increases, we have from

λf = λ'f' = c

f = c/λ

Where:

f and f' = The frequency of the motion of the photon before and after the scattering

c = Speed of light (constant)

We have that the frequency, f, is inversely proportional to the wavelength, λ as follows;

f = c/λ

As λ = increases, and c is constant, f decreases, therefore, the frequency of the photon decreases upon scattering.

5 0

2 years ago