<h3>Answer</h3>
1104 km/hour
<h3>Explanation</h3>
Distance between Dallas Texas to New York = 2760 km
Time the plane took from Dallas to New York = 2 hours
Time the plane took from New York back to Dallas = 2.5 hours
Formula to use
<h3>distance = speed x time </h3>
Speed the plane took from Dallas to New York
2760 = 2 x speed
speed = 2760 / 2
= 1380 km/hour
Speed the plane took from New York to Dallas (ROUND TRIP)
2760 = 2.5 x speed
speed = 2760 / 2.5
= 1104 km/hour
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:
Both of them reach the lake at the same time.
Explanation:
We have equation of motion s = ut + 0.5at²
Vertical motion of James : -
Initial velocity, u = 0 m/s
Acceleration, a = g
Displacement, s = h
Substituting,
s = ut + 0.5 at²
h = 0 x t + 0.5 x g x t²

Vertical motion of John : -
Initial velocity, u = 0 m/s
Acceleration, a = g
Displacement, s = h
Substituting,
s = ut + 0.5 at²
h = 0 x t + 0.5 x g x t²

So both times are same.
Both of them reach the lake at the same time.
They will hit the ground at the same time, as mass is negligible when calculating the acceleration of gravity when there is no air resistance