Let the sphere is having charge Q and radius R
Now if the proton is released from rest
By energy conservation we can say



now take square root of both sides

so the proton will move by above speed and
here Q = charge on the sphere
R = radius of sphere

Answer:
a) 24.4 Ω
b) 4.92 A
c) 495.9 W
d)
c. It will be larger. The resistance will be smaller so the current drawn will increase, increasing the power.
Explanation:
b)
The formula for power is:
P = IV
where,
P = Power of heater = 590 W
V = Voltage it takes = 120 V
I = Current Drawn = ?
Therefore,
590 W = (I)(120 V)
I = 590 W/120 V
<u>I = 4.92 A</u>
<u></u>
a)
From Ohm's Law:
V = IR
R = V/I
Therefore,
R = 120 V/4.92 A
<u>R = 24.4 Ω</u>
<u></u>
c)
For constant resistance and 110 V the power becomes:
P = V²/R
Therefore,
P = (110 V)²/24.4 Ω
<u>P = 495.9 W</u>
<u></u>
d)
If the resistance decreases, it will increase the current according to Ohm's Law. As a result of increase in current the power shall increase according to formula (P = VI). Therefore, correct option is:
<u>c. It will be larger. The resistance will be smaller so the current drawn will increase, increasing the power.</u>
The answer is d I took the test
Answer:
d because the proton would move towards the negative plate
Explanation: