As the radial distance between two objects increases, the gravitational force which pulls the two towards each other decreases following the inverse square law. This is based on Newton's Law of Universal Gravitation, which as an equation states . If the distance between the two is doubled, the force of gravity is only 1/4 as strong. If it is tripled, it is only 1/9 as strong, and so on.
Answer:
what is the question? there isn't one
Answer:
ω= 2.095rad/sec
Explanation:
angular frequency(ω)=2πf
F is the frequency
Frequency of an oscillation is the number of cycles completed per second. One oscillation represents 1cycle. Thus for 13 oscillations, the glider completes 13 cycles.
F= 13/39 =Hz
angular frequency(ω)=2πf
angular frequency(ω)=2π *
= 2**rad/sec
ω= 2.095rad/sec
Given:
Acceleration is uniform and acceleration (a) = 10 m/s^2
Now it has been mentioned for the first 2 secs the acceleration is 10m/s^2 .
Hence velocity= acceleration x time
Velocity= 10 x 2 = 20 m/s
Consider s as the distance traveled in the 3rd second.
Now we know s= ut+1/2(at^2)
Where s is the distance measured in m.
u is the initial velocity measured in m/sec
t is the time taken for the object to travel the above distance. This is equal to one second as we need to calculate the distance traveled between 3rd and 2nd second.
t = (3-2)= 1 sec
Substituting the given values in the above formula we get
s = 20 x1 + 1/2 (10 x 1 x 1)
s = 25 m
Thus the distance traveled by the object in the 3rd second is 25 m
I think answer is A. The Brightness of a Star