What are the requirements for fine coating in electroplating method ?
1 answer:
You might be interested in
Answer:
a = 1.5*10^-3 m/s^2
x = 0.033m = 3.3cm
Explanation:
To calculate the acceleration and the distance traveled by the car you use the following formulas:
(1)
(2)
v: final velocity = 0,255 km/h
vo: initial velocity = 0 m/s
t: time = 3/4 min
a: acceleration = ?
x: distance
In order to use the equations (1) and (2) you first convert the units of the final velocity to m/s, and the time to seconds.
Next, you solve the equation (1) for the acceleration a:
With this value of a you can calculate the distance traveled by the car, by using the equation (2):
hence, the acceleration of the car is 1.5*10^-3 m/s^2 and the distance traveled in 3/4 min is 0.033m
The additional charge transferred to the capacitors by the battery is 60 μC.
The increase in the total charge stored on the capacitors is 60 μC.
<u>Explanation</u>:
C = εоA / d
If the separation is halved, then the capacitance will be doubled and according to the equation q = CV, the charge will be doubled too.
- Initial charge of the capacitor is:
q = CV
= (6e - 6) (10)
= 60 μC
Final charge of the capacitor:
q = (2C)V
= (2 6e - 6)
(10)
= 120 μC
additional charge transmitted is:
q' = 120 - 60
= 60 μC
- initial total charge:
= (C1 + C2) V
= (6 + 6) (10)
= 120 μF
final total charge:
= (C1 + C2) V
= (2 6 + 6)
(10)
= 180 μF
Increase in the charge:
q' = 180 - 120
= 60 μC