No, it only does when entering an atmosphere
Answer:
A) K / K₀ = 4 b) v / v₀ = 4
Explanation:
A) For this exercise we can use the conservation of mechanical energy
in the problem it indicates that the displacement was doubled (x = 2xo)
starting point. At the position of maximum displacement
Em₀ = Ke = ½ k (2x₀)²
final point. In the equilibrium position
= K = ½ m v²
Em₀ = Em_{f}
½ k 4 x₀² = K
(½ K x₀²) = K₀
K = 4 K₀
K / K₀ = 4
B) the speed value
½ k 4 x₀² = ½ m v²
v = 4 (k / m) x₀
if we call
v₀ = k / m x₀
v = 4 v₀
v / v₀ = 4
Answer:
11
Explanation:
1. You are going to be rounding down.
2. change the metric ton to kg.
1.000 * 10^3 kg = 1000 kg
1000 / 87 = 11.49 = 11 people
Answer:
Explanation:
The processes are described on the image attached below. The isobaric process consists of an horizontal line, the adiabatic expansion is described by a polytropic curve:

Where:


Final pressure is:



Answer:
Explanation:
When the spring is compressed by .80 m , restoring force by spring on block
= 130 x .80
= 104 N , acting away from wall
External force = 82 N , acting towards wall
Force of friction acting towards wall = μmg
= .4 x 4 x 9.8
= 15.68 N
Net force away from wall
= 104 -15.68 - 82
= 6.32 N
Acceleration
= 6.32 / 4
= 1.58 m / s²
It will be away from wall
Energy released by compressed spring = 1/2 k x²
= .5 x 130 x .8²
= 41.6 J
Energy lost in friction
= μmg x .8
= .4 x 4 x 9.8 x .8
= 12.544 J
Energy available to block
= 41.6 - 12.544 J
= 29 J
Kinetic energy of block = 29
1/2 x 4 x v² = 29
v = 3.8 m / s
This will b speed of block as soon as spring relaxes. (x = 0 )