A student throws a baseball at a large gong 52 m away and hears the sound of the gong 1.73333 s later. The speed of sound in air
1 answer:
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Answer:
v = 15.8 m/s
Explanation:
Let's analyze the situation a little, we have a compressed spring so it has an elastic energy that will become part kinetic energy and a potential part for the man to get out of the barrel, in addition there is a friction force that they perform work against the movement. So the variation of mechanical energy is equal to the work of the fictional force
= ΔEm =
-Em₀
Let's write the mechanical energy at each point
Initial
Em₀ = Ke = ½ k x²
Final
= K + U = ½ m v² + mg y
Let's use Hooke's law to find compression
F = - k x
x = -F / k
x = 4400/1100
x = - 4 m
Let's write the energy equation
fr d = ½ m v² + mgy - ½ k x²
Let's clear the speed
v² = (fr d + ½ kx² - mg y) 2 / m
v² = (40 4.00 + ½ 1100 4² - 60.0 9.8 2.50) 2/60.0
v² = (160 + 8800 - 1470) / 30
v = √ (229.66)
v = 15.8 m/s
Answer:
Acceleration of the bullet will be 1778835.6
Explanation:
We have given length of the barrel refile s= 0.855 m
When the bullet leaves the muzzle its velocity is 553 m/sec
So final velocity v = 553 m/sec
Initial velocity will be 0 that is u = 0 m/sec
According to third equation of motion