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:
n1 sin θ1 = n2 sin θ2 Snell's Law (θ1 is the angle of incidence)
sin θ2 = n1 / n2 * sin θ1
sin θ2 = 2.4 / 1.33 * sin θ1
sin θ2 = 1.80 * .407 = .734
θ2 = 47.2 deg
Answer:
It is because it cannot be used time again and again.
