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3 years ago

5

The video shows an animated billiards experiment in which a cue ball strikes a glued-in-place eight-ball. which of the following

explains why the momentum of the eight-ball is conserved?

1 answer:

riadik2000 [5.3K]3 years ago

8 0

Suppose object A<span> is a cue ball and object </span>B<span> is an eight ball on a pool table. If the cue ball strikes the eight ball, the cue ball exerts a force on the eight ball that sends it rolling toward the pocket. At the same time, the eight ball exerts an equal and opposite force on the cue ball that brings it to a stop. Note that both the cue ball and the eight ball each experience a change in momentum. However, the sum of the momentum of the cue ball and the momentum of the eight ball remains constant throughout.</span>

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"A uniform cylinder of mass M and radius R is rolling without slipping. The velocity of its center of mass is v. What is the cyl

Goshia [24]

Answer:

Explanation:

Given

mass of cylinder is M

radius R

velocity of center of mass is v

As there is no slipping therefore cylinder will rotate as well as translate

Moment of inertia of cylinder $I=\frac{MR^2}{2}$

Kinetic Energy of cylinder $K.E.=\frac{1}{2}Mv^2$

Rotational energy $R.E.=\frac{1}{2}I\omega ^2$

for rolling

$v=\omega \times r$

where $\omega =angular\ velocity\ of\ cylinder$

$R.E.=\frac{1}{2}\times \frac{MR^2}{2}\times (\frac{v}{R})^2=\frac{1}{4}Mv^2$

Total kinetic Energy $=\frac{1}{2}Mv^2+\frac{1}{4}Mv^2=\frac{3}{4}Mv^2$

6 0

3 years ago

In 2007, michael carter (u.s.) set a world record in the shot put with a throw of 24.77 m. what was the initial speed of the sho

Serga [27]

The initial speed of the shot is 15.02 m/s.

The Shot put is released at a height y<em> </em>from the ground with a speed u. It is released at an angle θ to the horizontal. In a time t, the shot put travels a distance <em>R</em> horizontally.

Pl refer to the attached diagram.

Resolve the velocity u into horizontal and vertical components, u ₓ=ucosθ and uy=u sinθ. The horizontal component remains constant in the absence of air resistance, while the vertical component varies due to the action of the gravitational force.

Write an expression for R.

$R=u_xt=(ucos \theta)t$

Therefore,

$t=\frac{R}{ucos\theta} .......(1)$

In the time t, the net displacement of the shotput is y in the downward direction.

Use the equation of motion,

$y=u_yt-\frac{1}{2}gt^2=(usin\theta) t-\frac{1}{2}gt^2$

Substitute the value of t from equation (1).

$y=(ucos\theta)(\frac{R}{ucos\theta} )-\frac{1}{2} g(\frac{R}{ucos\theta} )^2\\ =Rtan\theta-(\frac{gR^2}{2u^2cos^2\theta} )$

Substitute -2.10 m for y, 24.77 m for R and 38.0° for θ and solve for u.

$y=Rtan\theta-(\frac{gR^2}{2u^2cos^2\theta} )\\ (-2.10m)=(24.77 m)(tan38.0^o)-\frac{(9.8 m/s^2)(24.77m)^2}{2u^2(cos38.0^o)^2} \\ u^2=225.71(m/s)^2\\ u=15.02m/s$

The shot put was thrown with a speed 15.02 m/s.

7 0

3 years ago

The low areas created as a sound wave propagates are called rarefactions

qwelly [4]

The answer is; pressure

The sound is a longitudinal wave meaning the particles vibrate parallel to the direction of the wave. Sound waves, therefore, produce compression (akin to the crest in a transverse wave) and rarefaction regions (akin to a trough in a transverse wave) as its energy is propagated in the medium.

8 0

3 years ago

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How does work affect energy between objects so it can cause a change in the form of energy?

mihalych1998 [28]

Answer:

Doing work' is a way of transferring energy from one object to another, energy is transferred when a force moves through a distance.

Explanation: So more energy, more work done bc u transferred more energy to move the object and doing the work. and if you only use a little of energy, the work done also only a little.

5 0

2 years ago

Read 2 more answers

g a small smetal sphere, carrying a net charge is held stationarry. what is the speed are 0.4 m apart

weeeeeb [17]

Complete Question

A small metal sphere, carrying a net charge q1=−2μC, is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of q2= -8μC and mass 1.50g, is projected toward q1. When the two spheres are 0.80m apart, q2 is moving toward q1 with speed 20ms−1. Assume that the two spheres can be treated as point charges. You can ignore the force of gravity.The speed of q2 when the spheres are 0.400m apart is.

Answer:

The value $v_2 = 4 \sqrt{10} \ m/s$

Explanation:

From the question we are told that

The charge on the first sphere is $q_1 = 2\mu C = 2*10^{-6} \ C$

The charge on the second sphere is $q_2 = 8 \mu C = 8*10^{-6} \ C$

The mass of the second charge is $m = 1.50 \ g = 1.50 *10^{-3} \ kg$

The distance apart is $d = 0.4 \ m$

The speed of the second sphere is $v_1 = 20 \ ms^{-1}$

Generally the total energy possessed by when $q_2$ and $q_1$ are separated by $0.8 \ m$ is mathematically represented

$Q = KE + U$

Here KE is the kinetic energy which is mathematically represented as

$KE = \frac{1 }{2} m (v_1)^2$

substituting value

$KE = \frac{1 }{2} * ( 1.50 *10^{-3}) (20 )^2$

$KE = 0.3 \ J$

And U is the potential energy which is mathematically represented as

$U = \frac{k * q_1 * q_2 }{d }$

substituting values

$U = \frac{9*10^9 * 2*10^{-6} * 8*10^{-6} }{0.8 }$

$U = 0.18 \ J$

So

$Q = 0.3 + 0.18$

$Q = 0.48 \ J$

Generally the total energy possessed by when $q_2$ and $q_1$ are separated by $0.4 \ m$ is mathematically represented

$Q_f = KE_f + U_f$

Here $KE_f$ is the kinetic energy which is mathematically represented as

$KE_f = \frac{1 }{2} m (v_2^2$

substituting value

$KE_f = \frac{1 }{2} * ( 1.50 *10^{-3}) (v_2 )^2$

$KE_f = 7.50 *10^{ -4} (v_2 )^2$

And $U_f$ is the potential energy which is mathematically represented as

$U_f = \frac{k * q_1 * q_2 }{d }$

substituting values

$U_f = \frac{9*10^9 * 2*10^{-6} * 8*10^{-6} }{0.4 }$

$U_f = 0.36 \ J$

From the law of energy conservation

$Q = Q_f$

So

$0.48 = 0.36 +(7.50 *10^{-4} v_2^2)$

$v_2 = 4 \sqrt{10} \ m/s$

6 0

3 years ago