Answer:
The coefficient of friction in the hall is 0.038
Explanation:
Given;
mass of the Parker, m = 73.2 kg
applied force on the parker, F = 123 N
frictional force, Fs = 27.4 N
the coefficient of friction in the hall = ?
frictional force is given by;
Fs = μN
Where;
μ is the coefficient of friction
N is normal reaction = mg
Fs = μmg
μ = Fs / mg
μ = (27.4) / (73.2 x 9.8)
μ = 0.038
Therefore, the coefficient of friction in the hall is 0.038
Answer:
3.71 m/s in the negative direction
Explanation:
From collisions in momentum, we can establish the formula required here which is;
m1•u1 + m2•v2 = m1•v1 + m2•v2
Now, we are given;
m1 = 1.5 kg
m2 = 14 kg
u1 = 11 m/s
v1 = -1 m/s (negative due to the negative direction it is approaching)
u2 = -5 m/s (negative due to the negative direction it is moving)
Thus;
(1.5 × 11) + (14 × -5) = (1.5 × -1) + (14 × v2)
This gives;
16.5 - 70 = -1.5 + 14v2
Rearranging, we have;
16.5 + 1.5 - 70 = 14v2
-52 = 14v2
v2 = - 52/14
v2 = 3.71 m/s in the negative direction
Answer:
7200 N/m
Explanation:
Metric unit conversion
100g = 0.1 kg
5 cm = 0.05 m
50 cm = 0.5 m
As the block is released from the spring and travelling to height h = 1.5m off the ground, the elastics energy is converted to work of friction force and the potential energy at 1.5 m off the ground
The work by friction force is the product of the force F = 15N itself and the distance s = 0.5 m

Let g = 10 m/s2. The change in potential energy can be calculated as the following:

Therefore, as elastic energy is converted to potential energy and work of friction:



1) 
The capacitance of a parallel-plate capacitor is given by:

where
is the vacuum permittivity
A is the area of each plate
d is the distance between the plates
Here, the radius of each plate is

so the area is

While the separation between the plates is

So the capacitance is

And now we can find the energy stored,which is given by:

2) 0.71 J/m^3
The magnitude of the electric field is given by

and the energy density of the electric field is given by

and using
, we find
