Answer:
181.48 N
Explanation:
Calculate the area :
Area = pi * r² ;
pi = 3.14 ; r1 = 90cm /100 = 0.9m ; r2 = 10/100 = 0.1m
Area 1, A1 = 3.14 * 0.1² = 0.0314 m²
Area 2, A2 = 3.14 * 0.9² = 2.5434 m²
Force, F = mass * acceleration due to gravity
F2 = 1500 * 9.8 = 14700 N
Force 1 / Area 1 = Force 2 / Area 2
Force 1 = (Force 2 / Area 2), * Area 1
Force 1 = (14700 / 2.5434) * 0.0314
Force = 5779.6650 * 0.0314
= 181.48 N
Missing question:
"Determine (a) the astronaut’s orbital speed v and (b) the period of the orbit"
Solution
part a) The center of the orbit of the third astronaut is located at the center of the moon. This means that the radius of the orbit is the sum of the Moon's radius r0 and the altitude (
) of the orbit:
This is a circular motion, where the centripetal acceleration is equal to the gravitational acceleration g at this altitude. The problem says that at this altitude,
. So we can write
where
is the centripetal acceleration and v is the speed of the astronaut. Re-arranging it we can find v:
part b) The orbit has a circumference of
, and the astronaut is covering it at a speed equal to v. Therefore, the period of the orbit is
So, the period of the orbit is 2.45 hours.
Energy is neither created nor destroyed
E in = E out
750 J in so 750 J out