Answer:
0.34 sec
Explanation:
Low point of spring ( length of stretched spring ) = 5.8 cm
midpoint of spring = 5.8 / 2 = 2.9 cm
Determine the oscillation period
at equilibrum condition
Kx = Mg
g= 9.8 m/s^2
x = 2.9 * 10^-2 m
k / m = 9.8 / ( 2.9 * 10^-2 ) = 337.93
note : w =
=
= 18.38 rad/sec
Period of oscillation = 
= 0.34 sec
The answer is orbit, we are orbiting the sun as the moon orbits us
Answer:
it makes our work easy and
time saving
it multiply our force applied
it complete our work with high efficiency
by using this,less effort is required for the work.
Answer:
5.7 x 10^12 C
Explanation:
Let the charge on earth and moon is q.
mass of earth, Me = 5.972 x 10^24 kg
mass of moon, Mm = 7.35 x 10^22 kg
Let d be the distance between earth and moon.
the gravitational force between them is

The electrostatic force between them is

According to the question
1 % of Fg = Fe



q = 5.7 x 10^12 C
Thus, the charge on earth and the moon is 5.7 x 10^12 C.
It really all depends weather the temperature