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3 years ago

A 15 W electric shaver is used for 3 minutes. How much energy does it use in joules?

Physics

1 answer:

borishaifa [10]3 years ago

3 0

P=change in E/t
Change in E=p*t
=15*3
=45

The answer is 45J.

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A resistor with resistance R is connected to a battery that has emf 13.0 V and internal resistance r = 0.350 Ω . For what two va

makvit [3.9K]

Answer:

R = 1,746 Ω

Explanation:

The power dissipated in the circuit is

P = V I = V² / R

Let's find the current

R = V² / P

Let's calculate

R = 13²/81

R = 2,096 Ω

This is total resistance

R_total = R + r

R = R_total - r

R = 2,096 -0,350

R = 1,746 Ω

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4 years ago

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3 years ago

Read 2 more answers

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Arisa [49]

Answer:

Each type of electromagnetic radiation have been matched to their descriptions as follows:

Ultraviolet light A) Given off by very hot objects, such as the sun

Gamma rays B) Given off by radioactive substances

Microwaves C) Used in radar and to heat food

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Explanation:

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3 years ago

You create a ramp using two text books and a 0.50m board. Using a timer you determine that a cart can roll down the ramp in 0.55

ahrayia [7]

Answer:

The velocity of the cart at the bottom of the ramp is 1.81m/s, and the acceleration would be 3.30m/s^2.

Explanation:

Assuming the initial velocity to be zero, we can obtain the velocity at the bottom of the ramp using the kinematics equations:

$v=v_0+at\\\\v^2=v_0^2+2ad$

Dividing the second equation by the first one, we obtain:

$v=\frac{v_0^2+2ad}{v_0+at}$

And, since $v_0=0$ , then:

$v=\frac{2ad}{at}\\\\v=\frac{2d}{t}\\\\v=\frac{2(0.50m)}{0.55s}\\\\v=1.81m/s$

It means that the velocity at the bottom of the ramp is 1.81m/s.

We could use this data, plus any of the two initial equations, to determine the acceleration:

$v=v_0+at\\\\\implies a=\frac{v}{t}\\\\a=\frac{1.81m/s}{0.55s}\\\\a=3.30m/s^2$

So the acceleration is 3.30m/s^2.

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4 years ago

If the charge on the negative plate of the capacitor is 121 nano-Coulomb, how many excess electrons are on that plate? Write you

Julli [10]

Answer:

n = 756.25 giga electrons

Explanation:

It is given that,

If the charge on the negative plate of the capacitor, $Q=121\ nC=121\times 10^{-9}\ C$

Let n is the number of excess electrons are on that plate. Using the quantization of charges, the total charge on the negative plate is given by :

$Q=ne$

e is the charge on electron

$n=\dfrac{Q}{e}$

$n=\dfrac{121\times 10^{-9}}{1.6\times 10^{-19}}$

$n=7.5625\times 10^{11}$

n = 756.25 giga electrons

So, there are 756.25 giga electrons are on the plate. Hence, this is the required solution.

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4 years ago

A 15 W electric shaver is used for 3 minutes. How much energy does it use in joules?​

A 15 W electric shaver is used for 3 minutes. How much energy does it use in joules?