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
Reflection from such a rough surface is called diffuse reflection and appears matte
Answer:
2. Earth layers compare in egg layer it is hard on the outside
Explanation:
Is the component perpendicular to the surface on contact of the contact force <span />
The time when the two players will collide is 0.96 s.
The equation of motion of the two players is given as;
x1 = 0.1 m + (–3.9 m/s )t
x2 = –6.3 m + (2.8 m/s )t
The time when the two players collide, their displacement is equal or the difference in their position will be zero.
Thus, the time when the two players will collide is 0.96 s.
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