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fiasKO [112]
3 years ago
5

A person shooting at a target from a distance of 450 metres finds that the sound of the bullet hitting the target comes 1 / 2 se

conds after he fired. A person equidistant from the target and shooting point hears the bullet hit 3 seconds after he heard the gun. The speed of sound is ___.
Physics
1 answer:
White raven [17]3 years ago
3 0

Answer:

1350 m/s

Explanation:

Speed of bullet

Distance traveled by the bullet = 450 m

Distance traveled by sound = 450 m

Using S= V × t

==> t= S/V

So, time for bullet t1=450/vb

time for sound t2=450/vs

Because it takes 1/2 sec from firing to hearing of sound by the shooter

==>  t1+t2= 1/2 sec

==> 450/vb+450/vs=0.5sec  ---- (1)

Now a person at distance 'x' each from gun and target takes 3 sec to hear sound from firing to hitting the target.

Fire sound duration

Distance =x m

Speed of sound = V

time =T1

==> T1 = x/vs

Bullet hitting + target sound  duration

During this duration bullet travels a distance of 450 m and sound of hitting travels a distance 'x'

time taken by sound = 450/vb

time taken by sound to travel distance 'x'= x/vs

so let T2= 450/vb + x/vs

But all this happens in 3 sec i-e T = 3 sec

and because firing occurs before hitting the target so he hears strike sound in time T = T2-T1= 450/vb + x/vs -x/vs

But T= 3sec

==> 3= 450/vb

==> vb= 1350 m/s

From eq 1 and 2

again using same relation as above, i-e within 3 seconds he watches bullet traveling a distance of 450 m and hearing sound

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In billiards, the 0.165 kg cue ball is hit toward the 0.155 kg eight ball, which is stationary. The cue ball travels at 5.8 m/s
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Answer:

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Explanation:

given data

mass m1 = 0.165 kg

mass m2 = 0.155 kg

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to find out

after collision another ball velocity v4

solution

we consider here ball move in x axis and after collision 1st ball move upside of x axis with angle 35 degree and other ball move downside with x axis with angle θ

so from conservation of momentum we say

m1v1 = m1v3cos35 + m2v4cosθ   with x axis    .............1

m1v3sin35 = m2v4sinθ                   with y axis  .............2

so from 1 equation

0.165 × 5.8 = 0.165(3.2)cos35 + 0.155(v4)cosθ

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form 2 equation

0.165(3.2)sin35 = 0.155(v4)sinθ  

v4 sinθ = 1.95                                              ......................4

so magnitude of another ball velocity is square and adding equation 3 and 4

another ball velocity = √(3.39²+1.96²)

another ball velocity = 3.92 m/s

and direction is tanθ = 1.96/3.39

θ = 30° clockwise from initial direction

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