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SVEN [57.7K]
2 years ago
15

What would be the relative difference between the currents required by a 60W lightbulb and a

Physics
1 answer:
anygoal [31]2 years ago
4 0

The current required by the 60W bulb is three-fifths that required by the 110W bulb

<h3>Electrical Power</h3>
  • Electrical power is defined as the amount of electrical work done per second
  • Power = electrical work done in joules/time taken in seconds

The formula for calculating power is as follows:

  • P = IV = V²/R = I²R

Assuming the voltage across the light bulbs is the same

From the formula, P = IV;

I = P/V

For the 60 W bulb

I₁ = 60/V

For the 100W bulb

I₂ = 100/V

Taking ratios of the currents:

I₁/I₂ = {60/V}/{`100/V}

I₁/I₂ = 3/5

Therefore, the current required by the 60W bulb is three-fifths that required by the 110W bulb

Learn more about electrical power and currents at: brainly.com/question/12822995

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A motorcycle starts to move from rest. If the velocity of the motorcycle becomes 90 km/hr
bagirrra123 [75]

Explanation:

90 kmhr—1 x 1000/3600 = 25ms—1

U = 0 ms—1

V = 25ms—1

t = 10 s

a = ?

a = V - U/t

a = 25 - 0/10

a = 25/10

a = 2.5 ms—1

6 0
2 years ago
A roadway for stunt drivers is designed for racecars moving at a speed of 40 m/s. A curved section of the roadway is a circular
Fynjy0 [20]

Answer:

Bank angle = 35.34o

Explanation:

Since the road is frictionless,

Tan (bank angle) = V^2/r*g

Where V = speed of the racing car in m/s, r = radius of the arc in metres and g = acceleration due to gravity in m/s^2

Tan ( bank angle) = 40^2/(230*9.81)

Tan (bank angle) = 0.7091

Bank angle = tan inverse (0.7091)

Bank angle = 35.34o

3 0
2 years ago
How long does it take for mars to rotate on its axis
Svetllana [295]

Hello There!

It takes the planet Mars around 24 hours, 37 minutes, 23 seconds to rotate on its axis. This is around the same amount of time that it takes our planet to rotate once on its axis.

8 0
3 years ago
Read 2 more answers
when the driver of a car applies the brakes a force of 4000N brings the car to a stop over a distance of 50m.calculate the work
Dimas [21]

Answer:

W = F*d

W = 4000*50

W = 200000J

W = 200KJ

4 0
3 years ago
What is the practical applications of radio waves
Keith_Richards [23]
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The lowest commonly encountered radio frequencies are produced by high-voltage AC power transmission lines at frequencies of 50 or 60 Hz. These extremely long wavelength electromagnetic waves (about 6000 km) are one means of energy loss in long-distance power transmission.

Extremely low frequency (ELF) radio waves of about 1 kHz are used to communicate with submerged submarines. The ability of radio waves to penetrate salt water is related to their wavelength (much like ultrasound penetrating tissue)—the longer the wavelength, the farther they penetrate. Since salt water is a good conductor, radio waves are strongly absorbed by it; very long wavelengths are needed to reach a submarine under the surface.



HOPE THIS REALLY HELPS YOU.
THANK YOU.
5 0
2 years ago
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