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Genrish500 [490]
3 years ago
12

Voltage sources are used to sustain

Physics
2 answers:
Lelu [443]3 years ago
7 0

Answer:

d. the movement of electric charge

Explanation:

Voltage source is used to provide constant voltage difference across the ends of a given conductor where it is connected

Now when voltage source is connected across the conductor we can say that due to constant voltage difference there would be a constant current will maintain in the conductor

It is given by

i = \frac{\Delta V}{R}

now since the voltage difference and resistance both are constant so current will be same in the circuit

So here the flow of charge will remain constant due to voltage source

So correct answer would be

d. the movement of electric charge

Sindrei [870]3 years ago
5 0
The movement of electric charge
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The strong magnetic fields is Long rang attractive power Reuther then week magnetic field.
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Will it take longer for a train or a car traveling at 100 mi/hr to stop
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What is championship game of baseball called? The World Series The Super Bowl The World Cup​
emmasim [6.3K]

Answer:

The World Series

Explanation:

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American Football and Football are different things. The first is what Americans call football, while the other is what Americans call soccer. It is confusing.

6 0
2 years ago
Astronaut Sarah leaves Earth in a spaceship at a speed of 0.280c relative to Earth. Sarah's destination is a star-system 12.5 li
Usimov [2.4K]

Answer:

L= 12 light years

Explanation:

for length dilation we use the formula

L=L_0\sqrt{1-\frac{v^2}{c^2} }

now calculating Lo

Lo = 12.5×365×24×3600×3×10^8

= 1.183×10^17 m

now putting the values of v and Lo in the above equation we get

L=1.183\times10^{17}\sqrt{1-\frac{0.28c^2}{c^2} }

= 1.136×10^17 m

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8 0
3 years ago
In a particular case of an object in front of a spherical mirror with a focal length of +12.0 cm, the magnification is +4.00.(a)
salantis [7]

Answer:

9 cm

-36 cm

Explanation:

u = Object distance

v = Image distance

f = Focal length = 12

m = Magnification = 4

m=-\frac{v}{u}\\\Rightarrow 4=-\frac{v}{u}\\\Rightarrow v=-4u

Lens equation

\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{12}=\frac{1}{u}+\frac{1}{-4u}\\\Rightarrow \frac{1}{12}=\frac{3}{4u}\\\Rightarrow u=9\ cm

Object distance is 9 cm

v=-4\times 9=-36\ cm

Image distance is -36 cm (other side of object)

7 0
3 years ago
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