(c) Due to up thrust
1 answer:
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The new gravitational attraction will be 1/4 as much
Explanation:
The magnitude of the gravitational force between two objects is given by
where
is the gravitational constant
m1, m2 are the masses of the two objects
r is the separation between them
In this problem, the original force between the two objects is F, when they are separated by a distance r.
Later, the distance between the two objects is doubled, so the new distance is
Therefore, the new force will be
Therefore, the new force will be one-fourth as much.
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Answer:
s = 1.7 m
Explanation:
from the question we are given the following:
Mass of package (m) = 5 kg
mass of the asteriod (M) = 7.6 x 10^{20} kg
radius = 8 x 10^5 m
velocity of package (v) = 170 m/s
spring constant (k) = 2.8 N/m
compression (s) = ?
Assuming that no non conservative force is acting on the system here, the initial and final energies of the system will be the same. Therefore
• Ei = Ef
• Ei = energy in the spring + gravitational potential energy of the system
• Ei = \frac{1}{2}ks^{2} + \frac{GMm}{r}
• Ef = kinetic energy of the object
• Ef = \frac{1}{2}mv^{2}
• \frac{1}{2}ks^{2} + (-\frac{GMm}{r}) = \frac{1}{2}mv^{2}
• s =
s =
s = 1.7 m