I assume you're talking about a pilot. If the ejection seat has an acceleration of 8<em>g</em>, then it would exert a normal force of 8<em>g</em> (70 kg) ≈ 5600 N.
(This is assuming the pilot is flying horizontally at a constant speed, and the seat is ejected vertically upward.)
To reiterate, this is *only* the force exerted by the seat on the pilot. Contrast this with the <em>net</em> force on the pilot, which would be the normal force minus the pilot's weight, 5600 N - (70 kg)<em>g</em> ≈ 4900 N.
If instead the seat ejects the pilot directly downward, the force exerted by the seat would have the same magnitude of 5600 N, but its direction would be reversed to point downward, making it negative. But the <em>net</em> force would change to -5600 N - (70 kg)<em>g</em> ≈ -6300 N
Answer:
The scale will read less than 700 N
Explanation:
Given;
normal weight of the person, W = 700 N
The upward acceleration of the elevator is given by Newton's second law of motion;
F = ma
Also, the weight of the person, W = mg
Net force on the person when the elevator accelerates upward is given as;
R = ma + mg
When the elevator slows down at a steady rate, then net force on the person is given as;
R = mg - ma
R = m(g - a), this net force on the person is also the reading of the scale.
Thus, the scale will read less than 700 N.
0.23 Hz
Period being the reciprocal of frequency
Answer:
v = 26.52 m/s
Explanation:
Here we know that as she passes the flag she will gain some speed
so here we can say that
Work done by all forces = change in kinetic energy of the system
so we will have
here we know that
h = 39.78 m
d = 27.81 m
so we have
