1000 khz am radio station broadcasts with a power of 20 kw number of photon emitted per second is 30.16 x 10^30 photon/s.
The frequency of the radio station is:
f
=
1000
k
H
z
=
1
×
10^6Hz
The transmit power is: P = 20kW = 20 X 10^3 W
The transmit power is: h = 6.63 x 10 ^-34 m^2.kg/s
The number of photon emitted per second = N = P/hf = <u>30.16 x 10^30 </u>photon/s.
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Answer: Remain unchanged
Explanation:
The boat with water barrel overboard floats in swimming pool when weight of the water displaced by the boat is equal to the buoyant force acting on the boat.
When the water in the barrel is poured overboard, the level of the swimming pool level would remain unchanged as the weight of the boat with the water and barrel would remain unchanged ( as the density and volume of the whole system remains same) and hence, the weight of the water (of the swimming pool) displaced by the boat would remain same.
A boat loaded with a barrel of water floats in a swimming pool. When the water in the barrel is poured overboard, the swimming pool level will <u>remain unchanged. </u>
Answer:
Answer:B
Explanation:
Because it all stayed consistant
Complete Question
The diagram for this question is shown on the first uploaded image
Answer:
a E =
b E =
c E = 0 N/C
d 
e 
f V = 
g 
h 
i 
Explanation:
From the question we are given that
The first charge 
The second charge 
The first radius 
The second radius 

And ![Potential \ Difference = \frac{1}{4\pi \epsilon_0} [\frac{q_1 }{r}+\frac{q_2}{R_2} ]](https://tex.z-dn.net/?f=Potential%20%5C%20Difference%20%3D%20%5Cfrac%7B1%7D%7B4%5Cpi%20%5Cepsilon_0%7D%20%20%20%5B%5Cfrac%7Bq_1%20%7D%7Br%7D%2B%5Cfrac%7Bq_2%7D%7BR_2%7D%20%5D)
The objective is to obtain the the magnitude of electric for different cases
And the potential difference for other cases
Considering a
r = 4.00 m


Considering b

This implies that the electric field would be

This because it the electric filed of the charge which is below it in distance that it would feel

= 
Considering c
r = 0.200 m
=> 
The electric field = 0
This is because the both charge are above it in terms of distance so it wont feel the effect of their electric field
Considering d
r = 4.00 m
=> 
Now the potential difference is

This so because the distance between the charge we are considering is further than the two charges given
Considering e
r = 1.00 m 
![V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{r} +\frac{q_2}{R_2} ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{1.00} \frac{1.00*10^{-6}}{1.00} ] = 26.79 *10^3 V](https://tex.z-dn.net/?f=V%20%3D%20%5Cfrac%7B1%7D%7B4%5Cpi%20%5Cepsilon_0%7D%20%5B%5Cfrac%7Bq_1%7D%7Br%7D%20%2B%5Cfrac%7Bq_2%7D%7BR_2%7D%20%20%5D%20%3D%208.99%2A10%5E9%20%2A%20%5B%5Cfrac%7B2.00%2A10%5E%7B-6%7D%7D%7B1.00%7D%20%5Cfrac%7B1.00%2A10%5E%7B-6%7D%7D%7B1.00%7D%20%5D%20%3D%2026.79%20%2A10%5E3%20V)
Considering f

![V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{r} +\frac{q_2}{R_2} ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{0.700} \frac{1.0*10^{-6}}{1.00} ] = 34.67 *10^3 V](https://tex.z-dn.net/?f=V%20%3D%20%5Cfrac%7B1%7D%7B4%5Cpi%20%5Cepsilon_0%7D%20%5B%5Cfrac%7Bq_1%7D%7Br%7D%20%2B%5Cfrac%7Bq_2%7D%7BR_2%7D%20%20%5D%20%3D%208.99%2A10%5E9%20%2A%20%5B%5Cfrac%7B2.00%2A10%5E%7B-6%7D%7D%7B0.700%7D%20%5Cfrac%7B1.0%2A10%5E%7B-6%7D%7D%7B1.00%7D%20%5D%20%3D%2034.67%20%2A10%5E3%20V)
Considering g

![V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{r} +\frac{q_2}{R_2} ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{0.500} \frac{1.0*10^{-6}}{1.00} ] = 44.95 *10^3 V](https://tex.z-dn.net/?f=V%20%3D%20%5Cfrac%7B1%7D%7B4%5Cpi%20%5Cepsilon_0%7D%20%5B%5Cfrac%7Bq_1%7D%7Br%7D%20%2B%5Cfrac%7Bq_2%7D%7BR_2%7D%20%20%5D%20%3D%208.99%2A10%5E9%20%2A%20%5B%5Cfrac%7B2.00%2A10%5E%7B-6%7D%7D%7B0.500%7D%20%5Cfrac%7B1.0%2A10%5E%7B-6%7D%7D%7B1.00%7D%20%5D%20%3D%2044.95%20%2A10%5E3%20V)
Considering h

![V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{R_1} +\frac{q_2}{R_2} ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{0.500} \frac{1.0*10^{-6}}{1.00} ] = 44.95 *10^3 V](https://tex.z-dn.net/?f=V%20%3D%20%5Cfrac%7B1%7D%7B4%5Cpi%20%5Cepsilon_0%7D%20%5B%5Cfrac%7Bq_1%7D%7BR_1%7D%20%2B%5Cfrac%7Bq_2%7D%7BR_2%7D%20%20%5D%20%3D%208.99%2A10%5E9%20%2A%20%5B%5Cfrac%7B2.00%2A10%5E%7B-6%7D%7D%7B0.500%7D%20%5Cfrac%7B1.0%2A10%5E%7B-6%7D%7D%7B1.00%7D%20%5D%20%3D%2044.95%20%2A10%5E3%20V)
Considering i

![V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{R_1} +\frac{q_2}{R_2} ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{0.500} \frac{1.0*10^{-6}}{1.00} ] = 44.95 *10^3 V](https://tex.z-dn.net/?f=V%20%3D%20%5Cfrac%7B1%7D%7B4%5Cpi%20%5Cepsilon_0%7D%20%5B%5Cfrac%7Bq_1%7D%7BR_1%7D%20%2B%5Cfrac%7Bq_2%7D%7BR_2%7D%20%20%5D%20%3D%208.99%2A10%5E9%20%2A%20%5B%5Cfrac%7B2.00%2A10%5E%7B-6%7D%7D%7B0.500%7D%20%5Cfrac%7B1.0%2A10%5E%7B-6%7D%7D%7B1.00%7D%20%5D%20%3D%2044.95%20%2A10%5E3%20V)
Answer:
(A) Reading will be 65 N
(B) Net force on the elevator will be 49.076 N
Explanation:
We have given the balance force = 65 N
Acceleration due to gravity 
We know that W=mg
So 
m = 6.632 kg
(a) In first case as the as the speed is constant so the force on the elevator will be 65 N
(B) In second case as the elevator is decelerating at a rate of 
So net acceleration = 9.8-2.4=
So net force on elevator will be = m× net acceleration = 6.632×7.4 = 49.076 N