Electric field strength near the surface of sphere is given by
Now we have
Q = 1 C
R = 5 cm
now electric field is given by
Part B)
Since charge on the sphere is very large to electric field near its surface is too high. So this is reasonable result near it.
Part c)
As per our assumption the charge is uniformly distributed on the surface and there is no effect of surrounding charge or surrounding electric field.
Answer:
The ball land at 3.00 m.
Explanation:
Given that,
Speed = 40 m/s
Angle = 35°
Height h = 1 m
Height of fence h'= 12 m
We need to calculate the horizontal velocity
Using formula of horizontal velocity
We need to calculate the time
Using formula of time
We need to calculate the vertical velocity
We need to calculate the vertical position
Using formula of distance
Put the value into the formula
We need to calculate the distance
Hence, The ball land at 3.00 m.
<span>Temperature is measured via
thermometer. It also measures the hotness or coldness of a system or
surroundings. It is based on the physical and chemical properties of a fluid
such as thermal expansion of fluids (fluids expand when heated and compress at
a low temperature) by studying the kinetic energy of the molecules of the
fluid.</span>
Answer:
(a) p = 3.4 kg-m/s (b) 37.78 N.
Explanation:
Mass of a basketball, m = 0.4 kg
Initial velocity of the ball, u = -5.7 m/s (as it comes down so it is negative)
It rebounds upward at a speed of 2.8 m/s (as it rebounds so positive)
(a) Change in momentum = final momentum - initial momentum
p = m(v-u)
p = 0.4 (2.8-(-5.7))
p = 3.4 kg-m/s
(b) Impulse = change in momentum
Ft = 3.4
We have, t = 0.09 s
Hence, this is the required solution.
Answer:
D: When one bulb burns out, all the other lights stay on
Explanation:
In series combination of light bulbs same current must flow through all the bulbs and hence if one bulb is burn out then current through all bulbs tripped to zero and all bulbs will turn off.
Now in parallel combination all bulbs are connected parallel to the source of energy due to which the bulbs will remain in circuit if any one bulb is burn out.
So here if we used combined circuit of the bulbs i.e. parallel then in that case if one of the bulb is burn out then it must show that rest of the other bulbs must glow.
so here correct answer would be
D: When one bulb burns out, all the other lights stay on