Answer:
(a) surface area of the plate will be equal to 
(b) Charge on the capacitor is equal to 
Explanation:
We have given spacing between the plates d = 0.05 mm = 
Value of capacitance 
(A) Capacitance of a parallel plate capacitor is equal to 
So 

So surface area of the plate will be equal to 
(B) It is given that capacitor is charged by 1.5 volt
So voltage V = 1.5 volt
Charge on the capacitor is equal to 
So 
Answer:
Please help on any part you can. I know it is a lot but any help I’d greatly appreciate. I attempted the problem but still do not understand. Thank you so much!
Explanation:
Please help on any part you can. I know it is a lot but any help I’d greatly appreciate. I attempted the problem but still do not understand. Thank you so much!

Answer:
12.5 m/s
Explanation:
From the question given above, the following data were obtained:
Initial velocity (u) = 0 m/s
Height (h) = 8 m
Final velocity (v) at 8 m above the lowest point =?
NOTE: Acceleration due to gravity (g) = 9.8 m/s²
The velocity of the roller coaster at 8 m above the lowest point can be obtained as follow:
v² = u² + 2gh
v² = 0² + (2 × 9.8 × 8)
v² = 0 + 156.8
v² = 156.8
Take the square root of both side
v = √156.8
v = 12.5 m/s
Therefore, the velocity of the roller coaster at 8 m above the lowest point is 12.5 m/s.
Answer:
123.30 m
Explanation:
Given
Speed, u = 22 m/s
acceleration, a = 1.40 m/s²
time, t = 7.30 s
From equation of motion,
v = u + at
where,
v is the final velocity
u is the initial velocity
a is the acceleration
t is time
V = at + U
using equation v - u = at to get line equation for the graph of the motion of the train on the incline plane
where m is the slope
Comparing equation (1) and (2)

a = m
Since the train slows down with a constant acceleration of magnitude 1.40 m/s² when going up the incline plane. This implies the train is decelerating. Therefore, the train is experiencing negative acceleration.
a = - 1.40 m/s²
Sunstituting a = - 1.40 m/s² and u = 22 m/s


The speed of the train at 7.30 s is 11.78 m/s.
The distance traveled after 7.30 sec on the incline is the area cover on the incline under the specific interval.
Area of triangle + Area of rectangle
![[\frac{1}{2} * (22 - 11.78) * (7.30)] + [(11.78 - 0) * (7.30)]](https://tex.z-dn.net/?f=%5B%5Cfrac%7B1%7D%7B2%7D%20%2A%20%2822%20-%2011.78%29%20%2A%20%287.30%29%5D%20%20%2B%20%5B%2811.78%20-%200%29%20%2A%20%287.30%29%5D)
= 37.303 + 85.994
= 123. 297 m
≈ 123. 30 m
Answer:
LAW 1 : For a given metal and frequency, the number of photoelectrons emitted is directly proportional to the intensity of the incident radiation.
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LAW 2: For a given metal, there exists a certain frequency below which the photoelectric emission does not take place. This frequency is called threshold frequency.
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LAW 3: For a frequency greater than the threshold frequency, the kinetic energy of photoelectrons is dependent upon frequency or wavelength but not on the intensity of light.
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LAW 4: Photoelectric emission is an instantaneous process. The time lag between incidence of radiations and emission of electron is 10^-9 seconds.
Explanation: