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:
C)evaporating of water
Explanation:that is because waters physical property is lost after evaporating
Answer:
The rate of change of distance between the two ships is 18.63 km/h
Explanation:
Given;
distance between the two ships, d = 140 km
speed of ship A = 30 km/h
speed of ship B = 25 km/h
between noon (12 pm) to 4 pm = 4 hours
The displacement of ship A at 4pm = 140 km - (30 km/h x 4h) =
140 km - 120 km = 20 km
(the subtraction is because A is moving away from the initial position and the distance between the two ships is decreasing)
The displacement of ship B at 4pm = 25 km/h x 4h = 100 km
Using Pythagoras theorem, the resultant displacement of the two ships at 4pm is calculated as;
r² = a² + b²
r² = 20² + 100²
r = √10,400
r = 101.98 km
The rate of change of this distance is calculated as;
r² = a² + b²
r = 101.98 km, a = 20 km, b = 100 km

Some one already asked this question and you can copy paste and google it but I believe it is c you may want to double check