Answer:
The voltage will quadruple
Explanation:
The power dissipated in a circuit is given by
where
V is the voltage
R is the resistance
In this problem, the voltage across the circuit is doubled:
V' = 2V
So the new power dissipated is
so, the power dissipated will quadruple.
The battery doesn't 'use' power. The battery <em>produces</em> the power that all the other electrical devices use.
If the starter motor is using 2,520 watts, then the battery is producing energy at the rate of 2,520 watts. That means <em>2,520 Joules</em> of energy every second.
Thanks for giving us the formula.
E = P x t
Energy = Power x Time
Energy = (2,520 watts) x (1 second)
Energy = 2,520 Joules
V = I · R
Voltage = (current) · (Resistance)
Voltage = (250 A) · (2.09 x 10⁴)
Voltage = 5,225,000 volts .
I may be out of line here, but I'm pretty sure
that the resistance is 2.09 x 10⁻⁴ .
Then
Voltage = 0.05225 volt (not 5 million and something)
Given Information:
Power of bulb = w = 25 W
atts
distance = d = 9.5 cm = 0.095 m
Required Information:
Radiation Pressure = ?
Answer:
Radiation Pressure =7.34x10⁻⁷ N/m²
Explanation:
We know that radiation pressure is given by
P = I/c
Where I is the intensity of radiation and is given by
I = w/4πd²
Where w is the power of the bulb in watts and d is the distance from the center of the bulb.
So the radiation pressure becomes
P = w/c4πd²
Where c = 3x10⁸ m/s is the speed of light
P = 25/(3x10⁸*4*π*0.095²)
P = 7.34x10⁻⁷ N/m²
Therefore, the radiation pressure due to a 25 W bulb at a distance of 9.5 cm from the center of the bulb is 7.34x10⁻⁷ N/m²
