At some of the greatest distances probed by telescopes, where many ordinary galaxies are too dim to be seen, we find the brillia
1 answer:
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Answer:
T = 0.71 seconds
Explanation:
Given data:
mass m = 1Kg, spring constant K = 78 N/m, time period of oscillation T = 0.71 seconds.
We have to calculate time period when this same spring-mass system oscillates vertically.
As we know
This relation of time period is true under every orientation of the spring-mass system, whether horizontal, vertical, angled or inclined. Therefore, time period of the same spring-mass system oscillating vertically too remains the same.
Therefore, T = 0.71 seconds
Answer:
Magnitude of the resultant vector is R = 6.81 m
Explanation:
Given :
Vector A having magnitude of 2.5 m
Vector A having direction 37 degree south of east.
Vector B having magnitude of 3.5 m
Vector B having direction 20 degree north of east.
Therefore, the angle between the two vectors is, θ = 37+20 = 57 degree
So, the resultant of the two vectors are given by
R = 6.81 m
The force on the fry is
Explanation:
We can find the force acting on the fry by using Newton's second law:
where
F is the net force on the fry
m is its mass
a is its acceleration
For the fry in this problem,
Therefore, the force exerted on the fry is
Learn more about Newton's second law here:
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Answer:
(a)
(b)
Explanation:
According to Newton's second law:
Recall that the frictional force is related jointly with the coefficient of friction and normal force . Replacing in the above equation, we get the coefficient of friction:
(a) The coefficient of static friction is related with the force required to set the block in motion:
(b) The coefficient of kinetic friction is related with the force required to keep the block moving with constant speed: