An archer shoots an arrow into a target using a bow. If the "action" force is the force of the bowstring against the arrow, what
1 answer:
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Answer:
a) v = √(v₀² + 2g h), b) Δt = 2 v₀ / g
Explanation:
For this exercise we will use the mathematical expressions, where the directional towards at is considered positive.
The velocity of each ball is
ball 1. thrown upwards vo is positive
v² = v₀² - 2 g (y-y₀)
in this case the height y is zero and the height i = h
v = √(v₀² + 2g h)
ball 2 thrown down, in this case vo is negative
v = √(v₀² + 2g h)
The times to get to the ground
ball 1
v = v₀ - g t₁
t₁ =
ball 2
v = -v₀ - g t₂
t₂ = - \frac{v_{o} + v }{ g}
From the previous part, we saw that the speeds of the two balls are the same when reaching the ground, so the time difference is
Δt = t₂ -t₁
Δt =
Δt = 2 v₀ / g
Answer:
Explanation:
A general wave function is given by:
A: amplitude of the wave = 0.075m
k: wave number
w: angular frequency
a) You use the following expressions for the calculation of k, w, T and λ:
b) Hence, the wave function is:
c) for x=3m you have:
d) the speed of the medium:
you can see the velocity of the medium for example for x = 0:
Explanation:
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