Explanation:
Given that,
Displacement in ice block, 
Force exerted by water, 
To find,
Work done by the force during the displacement.
Solve,
We know that the product of force and displacement is called work done. It is also equal to the dot product of force and displacement as :


We know that, i.i = j.j = k.k = 1

So, the work done by the force on the block during the displacement is 4181 Joules.
Answer:
resonance
Explanation:
The particles of substance B will cause the particle of substance A to vibrate at the same frequency
Answer:
Their speed in a vacuum is a constant value.
Explanation:
Electromagnetic waves consits of oscillations of electric field and magnetic field. The oscillations of these fields occur in a direction perpendicular to the direction of propagation of the waves, so they are transverse wave. Electromagnetic waves, contrary to mechanical waves, do not need a medium to propagate, so they can also travel through a vacuum. In a vacuum, their speed is constant and has always the same value, the speed of light:

Answer:
The frictional force
6.446 N
The acceleration of the block a = 6.04 
Explanation:
Mass of the block = 3.9 kg
°
= 0.22
(a). The frictional force is given by


3.9 × 9.81 × 
29.3 N
Therefore the frictional force
0.22 × 29.3
6.446 N
(b). Block acceleration is given by

F = 30 N
= 6.446 N
= 30 - 6.446
= 23.554 N
The net force acting on the block is given by

23.554 = 3.9 × a
a = 6.04 
This is the acceleration of the block.