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.
Answer:
s = 6.25 10⁻²² m
Explanation:
Polarizability is the separation of electric charges in a structure, in the case of the atom it is the result of the separation of positive charges in the nucleus and the electrons in their orbits, macroscopically it is approximated by
p = q s
s = p / q
let's calculate
s = 1 10⁻⁴⁰ / 1.6 10⁻¹⁹
s = 0.625 10⁻²¹ m
s = 6.25 10⁻²² m
We see that the result is much smaller than the size of the atom, therefore this simplistic model cannot be taken to an atomic scale.
Answer:
a motor is used to covert electrical energy to mechanical energy
The formula of net Force is:F = mawhere m is the mass of the objecta is the acceleration of the object
thus, if we triple the net force applied to the object:
3F = maa = 3F / m
The acceleration is also tripled since the force is directly proportional to the acceleration.
Answer:
88.3
Explanation:
Emf in a rotating coil is given by rate of change of flux:
E= dФ/dt=(NABcos∅)/ dt
N: number of turns in the coil= 80
A: area of the coil= 0.25×0.40= 0.1
B: magnetic field strength= 1.1
Ф: angle of rotation= 90- 37= 53
dt= 0.06s
E= (80 × 0.4× 0.25×1.10 × cos53)/0.06= 88.3V
