What is the impulse of a dodgeball hitting a person in the face if it's initial velocity was 10m/s and it's velocity after hitti
ng him in the face it is 5m/s? The ball weighs 3kg.
1 answer:
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Answer:
C. At the bottom of the circle.
Explanation:
Lets take
Radius of the circle = r
Mass = m
Tension = T
Angular speed = ω
The radial acceleration towards = a
a= ω² r
Weight due to gravity = mg
<h3>At the bottom condition</h3>T - m g = m a
T = m ω² r + m g
<h3>At the top condition</h3>T + m g = m a
T= m ω² r -m g
From above equation we can say that tension is grater when ball at bottom of the vertical circle.
Therefore the answer is C.
C. At the bottom of the circle.
Answer: ok, so total time is Tt = 6.65 seconds = time falling + time of sound traveling.
the time of the rock falling is given by: (1)
where h is the heigth of the hole, and t₀ is the time it took to reach the bottom.
the time of the sound traveling is given by 343*t₁ - h = 0 (2)
so t₁ + t₀ = 6.65s = total time
then you know that t₁ = 6.65s - t₀
so now you have two variables, t₀ and h, and we want to know the value of h, so we want to write t₀ as a function of h.
in the second equation we have now: 343*(6.65 - t₀) = h
so t₀ = (-h/343 + 6.65)s
replacing this on the first equation you have:
now you want to take h to the right side so
w
so if we replace g by 9.8 meters over seconds square we get
h =
where you will take the positive root
h = 61 meters