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.
The law<span> of conservation of </span><span>energy.</span>
The formula for the energy stored in the magnetic field of an inductor is:
E = (1/2) (inductance) (current)² .
In the present situation:
Energy = (3 kilo-watt-hour) x (1,000 / kilo) x (joule/watt-sec) x (3,600 sec/hr)
= (3 · 1000 · 3,600) (kilo·watt·hr·joule·sec / kilo·watt·sec·hr)
= 1.08 x 10⁷ joules .
Now to find the inductance:
E = (1/2) (inductance) (current)²
(1.08 x 10⁷ joules) = (1/2) (inductance) (300 Amp)²
(2.16 x10⁷ joules) = (inductance) (300 Amp)²
Inductance = (2.16 x10⁷ joules) / (300 Amp)²
= 2.16 x10⁷ / 90,000 Henrys
I get 240 Henrys .
This is a big inductance. Possibly the size of your house.
To get a big inductance, you want to wind the coil
with a huge number of turns of very fine wire, in
a small space.
In this case, however, if you plan on running 300A through
your coil, it'll have to be wound with a very thick conductor ...
like maybe 1/4-inch solid copper wire, or even copper tubing,
You have competing requirements.
There are cheaper, easier, better ways to store 3 kWh of energy.
In fact, a quick back-of-the-napkin calculation says that
3 or 4 car batteries will do the job nicely.
Answer:
7.35 m/s
Explanation:
Using y - y' = ut - 1/2gt², we find the time it takes the ball to fall from the 1.2 m table top and hit the floor.
y' = initial position of ball = 1.2 m, y = final position of ball = 0 m, u = initial vertical velocity of ball = 0 m/s, g = acceleration due to gravity = 9.8 m/s² and t = time taken for ball to hit the ground.
So, substituting the values of the variables into the equation, we have
y - y' = ut - 1/2gt²
0 - 1.2 m = (0 m/s)t - 1/2(9.8 m/s²)t²
- 1.2 m = 0 - (4.9 m/s²)t²
- 1.2 m = - (4.9 m/s²)t²
t² = - 1.2 m/- (4.9 m/s²)
t² = 0.245 s²
t = √(0.245 s²)
t = 0.49 s
Since d = vt where d = horizontal distance ball moves = 3.6 m, v = horizontal velocity of ball = unknown and t = time it takes ball to land = 0.49 s.
So, d = vt
v = d/t
= 3.6 m/0.49 s
= 7.35 m/s
Since the initial velocity of the ball is 7.35 m/s since the initial vertical velocity is 0 m/s.
It is shown thus V = √(u² + v²)
= √(0² + v²)
= √(0 + v²)
= √v²
= v
= 7.35 m/s
