Answer:
P = 2 pi R / v period of space station
F / m = v^2 / R centripetal force per unit of mass
So F / m = 4 pi^2 R^2 / (P^2 * R) = 4 pi^2 R / P^2
Also, F / m = 9.8 m/s^2 earth's gravitational attraction
So 9.8 = 4 pi^2 R / P^2 or R = 9.8 P^2 / 4 * pi^2) = 195 m
Or D = 2 R = 390 m the diameter required
Answer:
Explanation:
Gravitational law states that, the force of attraction or repulsion between two masses is directly proportional to the product of the two masses and inversely proportional to the square of their distance apart.
So,
Let the masses be M1 and M2,
F ∝ M1 × M2
Let the distance apart be R
F ∝ 1 / R²
Combining the two equation
F ∝ M1•M2 / R²
G is the constant of proportional and it is called gravitational constant
F = G•M1•M2 / R²
So, to increase the gravitational force, the masses to the object must be increased and the distance apart must be reduced.
So, option c is correct
C. Both objects have large masses and are close together.
My answer i believe is simply 250 Hz, because sounds or vibrations travel in 1 cycle/second, meaning the number of cycles, in your case 250, divided by the time,1 second, will ultimately be 250 Hertz. For every Cycle/second it will equal 1 Hz, so 250/1 = 250Hz
final velocity = initial
velocity + (acceleration x time) <span>
3.9 m/s = 0 m/s + (acceleration x 0.11 s)
3.9 m/s / 0.11 s = acceleration
30.45 m/s^2 = acceleration
distance = (initial velocity x time) +
1/2(acceleration)(time^2)
distance (0 m/s x 0.11 s) + 1/2(30.45 m/s^2)(0.11s ^2)
<span>distance = 0.18 m</span></span>
