Answer:
b) True. the force of air drag on him is equal to his weight.
Explanation:
Let us propose the solution of the problem in order to analyze the given statements.
The problem must be solved with Newton's second law.
When he jumps off the plane
fr - w = ma
Where the friction force has some form of type.
fr = G v + H v²
Let's replace
(G v + H v²) - mg = m dv / dt
We can see that the friction force increases as the speed increases
At the equilibrium point
fr - w = 0
fr = mg
(G v + H v2) = mg
For low speeds the quadratic depended is not important, so we can reduce the equation to
G v = mg
v = mg / G
This is the terminal speed.
Now let's analyze the claims
a) False is g between the friction force constant
b) True.
c) False. It is equal to the weight
d) False. In the terminal speed the acceleration is zero
e) False. The friction force is equal to the weight
Answer:
Explanation:
Block A sits on block B and force is applied on block A . Block A will experience two forces 1) force P and 2 ) friction force in opposite direction of motion . Block B will experience one force that is force of friction in the direction of motion .
Let force on block A be P . friction force on it will be equal to kinetic friction, that is μ mg , where μ is coefficient of friction and m is mass of block A
friction force = .4 x 2.5 x 9.8
= 9.8 N
net force on block A = P - 9.8
acceleration = ( P - 9.8 ) / 2.5
force on block B = 9.8
acceleration = force / mass
= 9.8 / 6
for common acceleration
( P - 9.8 ) / 2.5 = 9.8 / 6
( P - 9.8 ) / 2.5 = 1.63333
P = 13.88 N .
If you only know its speed, that's not enough information to catch it. You could even chase it at DOUBLE that speed, and you'd never catch it if you were chasing in the wrong direction.
You also have to know the DIRECTION the runaway car is going, so that you can chase in the same direction.
Now that you know its speed AND direction, you know its velocity. You need that information to have any chance of catching it.
