Answer:
F₁ / F₂ = 10
therefore the first out is 10 times greater than the second barrier
Explanation:
For this exercise let's use the relationship between momentum and momentum.
I = F t = Δp
in this case the final velocity is zero
F t = 0 -m v₀
F = m v₀ / t
in order to answer the question we must assume that the two vehicles have the same mass and speed
concrete barrier
F₁ = -p₀ / 0.1
F₁ = - 10 p₀
barrier collapses
F₂ = -p₀ / 1
let's look for the relationship of the forces
F₁ / F₂ = 10
therefore the first out is 10 times greater than the second barrier
Answer:
<h3>62.5N</h3>
Explanation:
The pressure at one end of the piston is equal to the pressure on the second piston.
Pressure = Force/Area
F1/A1 = F2/A2
Given
F1 = 250N
A1 = 2.0m²
A2 = 0.5m²
F2 = ?
Substituting the given values in the formula;
250/2 = F2/0.5
cross multiply
250*0.5 = 2F2
125 = 2F2
F2 = 125/2
F2 = 62.5N
Hence the force needed to lift this piston if the area of the second piston is 0.5 m^2 is 62.5N
Answer:
A simple pulley is a wheel with a rope that allows you to pull one end and have it lift whatever is on the other end. A modern, common example of this is a crane, often used in construction.
Explanation:
<h2>Thus the force of friction is 235 N</h2>
Explanation:
When the bear was at the height of 14 m . Its potential energy = m g h
here m is the mass of bear , g is acceleration due to gravity and h is the height .
Thus P.E = 27 x 10 x 14 = 3780 J
The K.E of the bear just before hitting = 
m v²
= x 27 x ( 6.1 )² = 490 J
The force of friction f = P.E - K.E = 3290 J
Because the work done = Force x Distance
Thus frictional force = 
= 235 N
