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3 years ago

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

Physics

1 answer:

anygoal [31]3 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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When the atmosphere is not quite clear, one may sometimes see colored circles concentric with the Sun or the Moon. These are gen

stealth61 [152]

Answer:

D) diffraction

Explanation:

Corona is an optical phenomenon produced by the diffraction of sunlight or moonlight, as light moves through water droplets in the atmosphere.

This phenomenon produces one or more diffuse concentric rings of light around the Sun or Moon, usually seen as colored circles.

Therefore, the explanation for these phenomena of colored concentric circles, sometimes seen with the Sun or the Moon involves diffraction.

5 0

3 years ago

Can someone help me?!!!!!

german

<h2>Hello!</h2>

The answer is:

The first option, the walker traveled 360m more than the actual distance between the start and the end points.

Why?

Since each block is 180 m long, we need to calculate the vertical and the horizontal distance, in order to calculate how farther did the travel walk between the start and the end points (displacement).

So, calculating we have:

Traveler:

$Distance=NorthCoveredDistance+EastCoveredDistance$

$Distance=4*180m+3*180m=720m+540m=1260m$

Actual distance between the start and the end point (displacement):

$ActualDistance=\sqrt{NorthDistance+EastDistance}\\\\ActualDistance=\sqrt{NorthDistance^{2} +EastDistance^{2}}\\\\ActualDistance=\sqrt{(720m)^{2} +(540m)^{2}}\\\\ActualDistance=\sqrt{518400m^{2} +291600m^{2}}\\\\ActualDistance=\sqrt{810000m^{2}}=900m$

Now, to calculate how much farter did the traveler walk, we need to use the following equation:

$DistanceDifference=WalkerCoveredDistance-ActualDistance\\\\DistanceDifference=1260m-900m=360m$

Therefore, we have that distance differnce between the distance covered by the walker and the actual distance is 360m.

Hence, we have that the walker traveled 360m more than the actual distance between the start point and the end point.

Have a nice day!

3 0

3 years ago

A charge of 32.0 nC is placed in a uniform electric field that is directed vertically upward and has a magnitude of 4.30x 104 V/

hodyreva [135]

A) The work done by the electric field is zero

B) The work done by the electric field is $9.1\cdot 10^{-4} J$

C) The work done by the electric field is $-2.4\cdot 10^{-3} J$

Explanation:

The electric field applies a force on the charged particle: the direction of the force is the same as that of the electric field (for a positive charge).

The work done by a force is given by the equation

$W=Fd cos \theta$

where

F is the magnitude of the force

d is the displacement of the particle

$\theta$ is the angle between the direction of the force and the direction of the displacement

In this problem, we have:

The force is directed vertically upward (because the field is directed vertically upward)
The charge moves to the right, so its displacement is to the right

This means that force and displacement are perpendicular to each other, so

$\theta=90^{\circ}$

and $cos 90^{\circ}=0$ : therefore, the work done on the charge by the electric field is zero.

In this case, the charge move upward (same direction as the electric field), so

$\theta=0^{\circ}$

and

$cos 0^{\circ}=1$

Therefore, the work done by the electric force is

$W=Fd$

and we have:

$F=qE$ is the magnitude of the electric force. Since

$E=4.30\cdot 10^4 V/m$ is the magnitude of the electric field

$q=32.0 nC = 32.0\cdot 10^{-9}C$ is the charge

The electric force is

$F=(32.0\cdot 10^{-9})(4.30\cdot 10^4)=1.38\cdot 10^{-3} N$

The displacement of the particle is

d = 0.660 m

Therefore, the work done is

$W=Fd=(1.38\cdot 10^{-3})(0.660)=9.1\cdot 10^{-4} J$

In this case, the angle between the direction of the field (upward) and the displacement (45.0° downward from the horizontal) is

$\theta=90^{\circ}+45^{\circ}=135^{\circ}$

Moreover, we have:

$F=1.38\cdot 10^{-3} N$ (electric force calculated in part b)

While the displacement of the charge is

d = 2.50 m

Therefore, we can now calculate the work done by the electric force:

$W=Fdcos \theta = (1.38\cdot 10^{-3})(2.50)(cos 135.0^{\circ})=-2.4\cdot 10^{-3} J$

And the work is negative because the electric force is opposite direction to the displacement of the charge.

Learn more about work and electric force:

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5 0

3 years ago

If a charged particle is moving parallel to a magnetic field, it experiences

aleksandr82 [10.1K]

Answer:it experiences no force

Explanation:

a charge moving in a direction parallel to the magnetic field experience no force.since the angle e is 0,force would also be 0

3 0

3 years ago

The amount of friction divided by the weight of an object forms a unit less number called the

Romashka [77]

Answer:

Coefficient of friction.

Explanation:

The amount of friction divided by the weight of an object is equal to the coefficient of friction. It is a dimensional less number. It can be given by :

$F=\mu N$

N is normal force.

$\mu$ = coefficient of friction

$\mu=\dfrac{F}{N}$

3 0

3 years ago