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sammy [17]
3 years ago
8

The product of voltage times amperage is known as what?

Physics
1 answer:
Usimov [2.4K]3 years ago
5 0

Answer:

Power=V*I which corresponds to the second option shown: "voltage times amperage"

Explanation:

The electric power is the work done to move a charge Q across a given difference of potential V per unit of time.

Since such electrical work is the product of the potential difference V times the charge that moves through that potential, and this work is to be calculated by the unit of time, we need to divide the product by time (t) which leads to the following final simple equation

Power=\frac{V\,Q}{t} =V\,\frac{Q}{t} = V\, I

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Does anyone know the answer? I’ll mark you as a brainliest if u get it right. thank u :)
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Answer: ohdela student i will be calling home.

Explanation:

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3 years ago
A 900 kg car pushes a 2400 kg truck that has a dead battery. when the driver steps on the accelerator, the drive wheels of the c
Wewaii [24]
A)0
b)magnitude of the force on the car is 500n
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3 years ago
What happens overtime to rocks that are stressed
malfutka [58]

Answer:

Stress is the force applied to an object. In geology, stress is the force per unit area that is placed on a rock. Four types of stresses act on materials.

A deeply buried rock is pushed down by the weight of all the material above it. Since the rock cannot move, it cannot deform. This is called confining stress.

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Explanation:

4 0
3 years ago
A convex security mirror has a radius of curvature of 12.0 cm. What is the magnification of a pare 3.0 m from the mirror?
Makovka662 [10]

Answer:

magnification will be -0.025

Explanation:

We have given the radius of curvature = 12 cm

And object distance = 3 m

So focal length f=\frac{R}{2}=\frac{12}{2}=6cm

Now for mirror we know that \frac{1}{f}=\frac{1}{u}+\frac{1}{v}

So \frac{1}{0.06}=\frac{1}{3}+\frac{1}{v}

16.66-0.333=\frac{1}{v}

v = 0.750 m

Now magnification of the mirror is m=\frac{-v}{u}=\frac{-0.750}{3}=-0.025

5 0
4 years ago
Two coherent sources of radio waves, A and B, are 5.00 metersapart. Each source emits waves with wavelength 6.00 meters.Consider
alexgriva [62]

Answer

given,

distance between A and B is equal to 5 m

let x be the distance of loud speaker from the source A

For constructive interference, path difference = mλ

x - (5 - x ) = 0           , for m = 0

2 x - 5 = 0

2 x = 5

  x = 2.5 cm

For destructive interference

|x - (5 - x )| = (2 m + 1)\dfrac{\lambda}{2}

m = 0

|2x - 5| = \dfrac{\lambda}{2}

|2x - 5| = \dfrac{6}{2}

\pm(2x - 5) = 3

x = 4 , 1 cm

3 0
3 years ago
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