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
F - False.
The law of conservation of momentum states that the total momentum is conserved.
Answer: Technician A is correct
Explanation:
The intake manifold is the compactment that all fuel and air supply to the cylinders. It's connected to the engine so it has to be disconnected while the exhaust manifold receives all the exhaust gases from the cylinders and releases the gas through a single or double exhaust gases outlet.
Answer:
The kinetic energy of the proton at the end of the motion is 1.425 x 10⁻¹⁶ J.
Explanation:
Given;
initial velocity of proton, 
= 3 x 10⁵ m/s
distance moved by the proton, d = 3.5 m
electric field strength, E = 120 N/C
The kinetic energy of the proton at the end of the motion is calculated as follows.
Consider work-energy theorem;
W = ΔK.E
where;
K.Ef is the final kinetic energy
W is work done in moving the proton = F x d = (EQ) x d = EQd
Therefore, the kinetic energy of the proton at the end of the motion is 1.425 x 10⁻¹⁶ J.
Answer:
0.280 s
Explanation:
I set it up as 5.22=(55)(0.0266)/x and then solved for x to be 2.80.
