plasma is a superheated liquid
So, a would-be the correct option.
Answer:
Explanation:
Given that,
Number of turn N = 40
Diameter of the coil d= 11cm = 0.11m
Then, radius = d/2 = 0.11/2 =0.055m
r = 0.055m
Then, the area is given as
A =πr²
A = π × 0.055²
A = 9.503 × 10^-3 m²
Magnetic Field B = 0.35T
Magnetic field reduce to zero in 0.1s, t = 0.1s
so we want to find induce electric field. To find the electric field,(E) we need to find the electric potential (V).
E.M.F is given as
ε = —N • dΦ/dt
Where magnetic flux is given as
Φ = BA
Then, ε = —N • dΦ/dt
ε = —N • dBA/dt
ε = —NBA/t
Then, its magnitude is
ε = NBA/t
Inserting the values of N, B, A and t
ε = 40×0.35×9.503×10^-3/0.1
ε = 1.33 V
Then, using the relationship between Electric field and electric potential
V = Ed
ε = E•d
E = ε/d
E = 1.33/0.11
E = 12.09 V/m
As an object falls from rest, its gravitational energy is converted to kinetic energy
G.P.E = K.E = mgh
K.E = (80 Kg)(9.8 m/s²)(30 m)
K.E. = 23,520 J
Answer:
The force applied on one wheel during braking = 6.8 lb
Explanation:
Area of the piston (A) = 0.4 
Force applied on the piston(F) = 6.4 lb
Pressure on the piston (P) = 
⇒ P = 
⇒ P = 16 
This is the pressure inside the cylinder.
Let force applied on the brake pad = 
Area of the brake pad (
)= 1.7
Thus the pressure on the brake pad (
) = 
When brake is applied on the vehicle the pressure on the piston is equal to pressure on the brake pad.
⇒ P =
⇒ 16 =
⇒ = 16 ×
Put the value of we get
⇒ = 16 × 1.7
⇒ = 27.2 lb
This the total force applied during braking.
The force applied on one wheel = 
= 
= 6.8 lb
⇒ The force applied on one wheel during braking.
