Explanation:
The size of the force varies inversely as the square of the distance between the two charges. Therefore, if the distance between the two charges is doubled, the attraction or repulsion becomes weaker, decreasing to one-fourth of the original value.
A wave is characterized by the cyclic occurrences of crests and troughs. Wavelengthis defined as the distance between two consecutive troughs or two crests and the Frequency is defined as the number of cycles that pass through a point per second
Work = (force) x (distance)
The work he did: Work = (700 N) x (4m) = 2,800 joules
The rate at which
he did it (power): Work/time = 2,800 joules/2 sec
= 1,400 joules/sec
= 1,400 watts
= 1.877... horsepower (rounded)
Answer:
Explanation:
Applied force, F = 18 N
Coefficient of static friction, μs = 0.4
Coefficient of kinetic friction, μs = 0.3
θ = 27°
Let N be the normal reaction of the wall acting on the block and m be the mass of block.
Resolve the components of force F.
As the block is in the horizontal equilibrium, so
F Cos 27° = N
N = 18 Cos 27° = 16.04 N
As the block does not slide so it means that the syatic friction force acting on the block balances the downwards forces acting on the block .
The force of static friction is μs x N = 0.4 x 16.04 = 6.42 N .... (1)
The vertically downward force acting on the block is mg - F Sin 27°
= mg - 18 Sin 27° = mg - 8.172 ... (2)
Now by equating the forces from equation (1) and (2), we get
mg - 8.172 = 6.42
mg = 14.592
m x 9.8 = 14.592
m = 1.49 kg
Thus, the mass of block is 1.5 kg.