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
so your saying the start is 0 N and when he/she hits the ball its inertia is 3 N. if that is so m*v=
.05*3=<u>.15</u>
NHTSA (National Highway Traffic Safety Administration) now recommends the technique known as 9 and 3. Place your left hand on the left portion of the steering wheel in a location approximate to where the nine would be if the wheel was a clock. Your right hand should be placed on the right portion of the wheel where the three would be located.
Answer:
Explanation:
The value of the frequency is not given. Lets assume it to be 14603 kHz. the question now becomes:
A radio technician measures the frequency of an AM radio transmitter. The frequency is 14603 kHz . What is the frequency in megahertz?
Answer:
kilo to mega = 10^-3
then
14603 kHz = 14603 *10^-3 MHz = 14.603 MHz