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Gnoma [55]
3 years ago
13

A cue ball has a mass of 0.5 kg. During a game of pool, the cue ball is struck and now has a velocity of 3 . When it strikes a s

olid ball with a mass of 0.5 kg, the cue ball comes to a complete stop. What is the new velocity of the solid ball? Round your answer to the nearest whole number.
Physics
2 answers:
photoshop1234 [79]3 years ago
8 0

Answer: 3 m/s

Explanation:

We can solve the problem by using the law of conservation of momentum: during the collision between the two balls, the total momentum of the system before the collision and after the collision must be conserved:

p_i = p_f

The total momentum before the collision is given only by the cue ball, since the solid ball is initially at rest, therefore

p_i = m_c u_c = (0.5 kg)(3 m/s)=1.5 kg m/s

So, the final total momentum will also be

p_f = 1.5 kg m/s

And the total momentum after the collision is given only by the solid ball, since the cue ball is now at rest, therefore:

p_f = m_s v_s

from which we find the velocity of the solid ball

v_s = \frac{p_f}{m_s}=\frac{1.5 kg m/s}{0.5 kg}=3 m/s

Otrada [13]3 years ago
6 0

3, just did the assignment

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Explain the difference between the terms "rotation" and "revolution"
Novay_Z [31]

Answer:

Explanation:

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A car approaches you at a constant speed, sounding its horn, and you hear a frequency of 76 Hz. After the car goes by, you hear
Talja [164]

Answer:

70.07 Hz

Explanation:

Since the sound is moving away from the observer then

f_o = f_s\frac {(v+vs)}{v} and f_o = f_s\frac {(v-vs)}{v} when moving towards observer

With f_o of 76 then taking speed in air as 343 m/s we have

76 = f_s\times\frac {(343-vs)}{343}

f_s=\frac {343\times 76}{343-v_s}

Similarly, with f_o of 65 we have

65 = f_s\times\frac {(343+vs)}{343}\\f_s=\frac {343\times 65}{343+v_s}

Now

f_s=\frac {343\times 65}{343+v_s}=\frac {343\times 76}{343-v_s}

v_s=27.76 m/s

Substituting the above into  any of the first two equations then we obtain

f_s=70.07 Hz

4 0
3 years ago
An object is pushed from rest across a sheet of ice, accelerating at 8.0 m/s^2 [E] over a displacement of 1.05 m [E]. The object
tatiyna

Answer:

D_T=18.567m

Explanation:

From the question we are told that:

Acceleration a=8.0 m/s^2

Displacement d=1.05 m

Initial time t_1=6.0s

Final Time t_2=2.5s

Generally the equation for Velocity of 1.05 travel is mathematically given by

Using Newton's Law of Motion

 V^2=2as

 V=\sqrt{2*6*1.05}

 V=4.1m/s

Generally the equation for Distance traveled before stop is mathematically given by

 d_2=v*t_1

 d_2=3.098*4

 d_2=12.392

Generally the equation for Distance to stop is mathematically given by

Since For this Final section

Final velocity v_3=0 m/s

Initial velocity u_3=4.1 m/s

Therefore

Using Newton's Law of Motion

 -a_3=(4.1)/(2.5)

 -a_3=1.64m/s^2

Giving

 v_3^2=u^2-2ad_3

Therefore

 d_3=\frac{u_3^2}{2ad_3}

 d_3=\frac{4.1^2}{2*1.64}

 d_3=5.125m

Generally the Total Distance Traveled is mathematically given by

 D_T=d_1+d_2+d_3

 D_T=5.125m+12.392+1.05 m

 D_T=18.567m

6 0
2 years ago
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