Answer:
E= 2.158× 10*5N/C
Explanation:
K= 8.99×10*9, q= 6×10*-6C, d= 0.5m
E= kq/d*2
E= (8.99×10*9× 6×10*-6)/0.5*2
E= 215760
E= 2.158 ×10*5N/C
a. The force applied would be equal to the frictional
force.
F = us Fn
where, F = applied force = 35 N, us = coeff of static
friction, Fn = normal force = weight
35 N = us * (6 kg * 9.81 m/s^2)
us = 0.595
b. The force applied would now be the sum of the
frictional force and force due to acceleration
F = uk Fn + m a
where, uk = coeff of kinetic friction
35 N = uk * (6 kg * 9.81 m/s^2) + (6kg * 0.60 m/s^2)
uk = 0.533
Explanation:
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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.
1000 khz am radio station broadcasts with a photon of 20 kw1000 khz am radio station broadcasts with a power of 20 kw1000 khz am radio station broadcasts with a power of 20 kw1000 khz am radio station broadcasts with a power of 20 kw1000 khz am radio station broadcasts with a power of 20 kw.1000 khz am radio station broadcasts with a power of 20 kw1000 khz am radio station broadcasts with a power of 20 kw1000 khz am radio station broadcasts with a power of 20 kw.
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