What heavy element was the chernobyl nuclear accelarators supposed to make?
1 answer:
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(1)
Cheetah speed:
Its position at time t is given by
(1)
Gazelle speed:
the gazelle starts S0=96.8 m ahead, therefore its position at time t is given by
(2)
The cheetah reaches the gazelle when . Therefore, equalizing (1) and (2) and solving for t, we find the time the cheetah needs to catch the gazelle:
(2) To solve the problem, we have to calculate the distance that the two animals can cover in t=7.5 s.
Cheetah:
Gazelle:
So, the gazelle should be ahead of the cheetah of at least
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If west means the west of the axis x the velocity equal :
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