Answer:
P₂ = 2 P₁
we conclude that in the second time the power used is double that in the first rise
Explanation:
In this exercise we are asked the power to climb the stairs, if we assume that we go up with constant speed, we use an energy equal to the potential energy due to the difference in height of the stairs, as this height is constant the potential energy does not change and therefore therefore the energy used by us does not change either.
Now we can analyze the required power,
P = W / t
From the analysis of the previous paragraph the work is equal to the energy used, according to the work energy theorem,
therefore the first time the power is
P₁ = E / 10
P₁ = 0.1 E
for the second time the power is
P₂ = E / 5
P₂ = 0.2 E
we see that the power in the second case is
P₂ = 2 P₁
Therefore, we conclude that in the second time the power used is double that in the first rise.
(a) The stone moves by uniform accelerated motion, with constant acceleration

directed downwards, and its initial vertical position at time t=0 is 750 m. So, the vertical position (in meters) at any time t can be written as

(b) The time the stone takes to reach the ground is the time at which the vertical position of the stone becomes zero: y(t)=0. So, we can write

from which we find the time t after which the stone reaches the ground:

(c) The velocity of the stone at time t can be written as

because it is an accelerated motion with initial speed zero. Substituting t=12.37 s, we find the final velocity of the stone:

(d) if the stone has an initial velocity of

, then its law of motion would be

and we can find the time it needs to reach the ground by requiring again y(t)=0:

which has two solutions: one is negative so we neglect it, while the second one is t=11.78 s, so this is the time after which the stone reaches the ground.
Explanation:
power=f×v. recall= distances/ time
= f× d/t
= 30 × 2/5
=12watt
Answer:
D
Explanation:
the formula is F=mgh
so now you can write it like
m= 4×50=200
200×1.5×10=30000