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3 years ago

5

Power is calculated by multiplying voltage by

1 answer:

butalik [34]3 years ago

5 0

If you're calculating the power of an athlete who is lifting weights,
or the power of a windmill that is pumping water from a well, or
the power of two horses turning millstones, then those electrical
units won't help at all.

But if you happen to be calculating the power delivered to an
electrical circuit or dissipated by an electrical device, then you
can use ...

(voltage) times (current)
or
(voltage)² divided by (resistance)
or
(current)² times (resistance) .

The choice just depends on which quantities you know
or can easily measure.

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By what factor is the self-inductance of an air solenoid changed if its length and number of coil turns are both tripled

fredd [130]

Answer:

The new self inductance is 3 times of the initial self inductance.

Explanation:

The self inductance of a solenoid is given by :

$L=\dfrac{\mu_oN^2 A}{L}$

Where

N is number of turns per unit length

A is area of cross section

l is length of solenoid

If length and number of coil turns are both tripled,

l' = 3l and N' = 3N

New self inductance is given by :

$L'=\dfrac{\mu_oN'^2 A}{L'}\\\\=\dfrac{\mu_o(3N)^2 A}{3L}\\\\=3\dfrac{\mu_oN^2 A}{L}\\\\=3L$

So, the new self inductance is 3 times of the initial self inductance.

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2 years ago

Why would an atomic nucleus give off a particle?

zmey [24]

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2 years ago

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Which part(s) of a hair can be analyzed for nuclear DNA?

kiruha [24]

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3 years ago

What is the scientific term for rocks formed from lava?

sergeinik [125]

Answer:

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3 years ago

A wire carries a steady current of 2.60 A. A straight section of the wire is 0.750 m long and lies along the x axis within a uni

ad-work [718]

Answer:

The magnetic force on the section of wire is $-2.925\hat{j}\ N$ .

Explanation:

Given that,

Current $I = 2.60\hat{i}\ A$

Length = 0.750 m

Magnetic field $B = 1.50\hat{k}\ T$

We need to calculate the magnetic force on the section of wire

Using formula of magnetic force

$\vec{F}=l\vec{I}\times\vec{B}$

$\vec{F}=0.750\times2.60\hat{i}\times1.50\hat{k}$

Since, $\hat{i}\times\hat{k}=-\hat{j}$

$\vec{F}=-2.925\hat{j}\ N$

Hence, The magnetic force on the section of wire is $-2.925\hat{j}\ N$ .

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3 years ago