Answer:
Explanation:
Given
Lowest four resonance frequencies are given with magnitude
50,100,150 and 200 Hz
The frequency of vibrating string is given by

where n=1,2,3 or ...n
L=Length of string
T=Tension
Mass per unit length
When string is clamped at mid-point
Effecting length becomes 
Thus new Frequency becomes

i.e. New frequency is double of old
so new lowest four resonant frequencies are 100,200,300 and 400 Hz
Work = Force x Distance
Assuming that this work is being done parallel to the displacement that is, but under that assumption:
W = (50)(10)
W = 500 J
The coefficient of static friction is 0.234.
Answer:
Explanation:
Frictional force is equal to the product of coefficient of friction and normal force acting on any object.
So here the mass of the object is given as 2 kg, so the normal force will be acting under the influence of acceleration due to gravity.
Normal force = mass * acceleration due to gravity
Normal force = 2 * 9.8 = 19.6 N.
And the frictional force is given as 4.6 N, then

Coefficient of static friction = 4.6 N / 19.6 N = 0.234
So the coefficient of static friction is 0.234.
Answer:
Constructive interference
Explanation:
The answer is D. centripetal force
This definition of centripetal force is a force acting towards the center of an object moving in a circular motion. Acceleration is the rate of which velocity changes over a period of time. If you pick a point on say a rotating circle and find the tangential velocity at each point it goes through, the acceleration would be towards the middle.