The lithosphere because it includes the outer region of the earth including the crust and outer mantle
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:
B) 100 J
Explanation:
Assuming the distance given is measured along the incline, the vertical change in height is (5 m)(sin 30°) = 2.5 m. Then the change in potential energy is ...
∆PE = mg(∆h) = (4 kg)(10 m/s^2)(2.5 m) = 100 J
Resistance = V / I
= 12 / 3 = 4
I think it’s the first option