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3 years ago

I will mark brainliest if your answer is correct!

Physics

2 answers:

Strike441 [17]3 years ago

6 0

Answer:

The electromotive force (∈) = 2 V

Further explanation

Given

I=2 A

t = 30 s

E = 120 J

Required

The electromotive force (∈)

Solution

The electromotive force : the potential difference between two electrodes in galvanic cell

Can be formulated :

E = energy, J

Q = charge , Coulomb = i.t

Explanation:

pochemuha3 years ago

3 0

The electromotive force (∈) = 2 V

<h3>Further explanation</h3>

Given

I=2 A

t = 30 s

E = 120 J

Required

The electromotive force (∈)

Solution

The electromotive force : the potential difference between two electrodes in galvanic cell

Can be formulated :

$\tt \epsilon=\dfrac{E}{Q}$

E = energy, J

Q = charge , Coulomb = i.t

$\tt \epsilon=\dfrac{120~J}{2~A\times 30~s}\\\\\epsilon=2~V$

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4 years ago

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3 years ago

Two coils are wound around the same cylindrical form. When the current in the first coil is decreasing at a rate of -0.245 A/s ,

Airida [17]

Answer:

Complete question:

c.If the current in the second coil increases at a rate of 0.365 A/s , what is the magnitude of the induced emf in the first coil?

a. $M= 6.53\times10^{-3} H$

b.flux through each turn = Ф = $4.08\times10^{-4} Wb$

c.magnitude of the induced emf in the first coil = e= $2.38\times10^{-3} V$

Explanation:

a. rate of current changing = $\frac{di}{dt}=[tex]M=\frac{1.60\times10^{-3} V}{0.240\frac{A}{s} }}$ [/tex]

Induced emf in the coil =e= $1.60\times10^{-3} V$

For mutual inductance in which change in flux in one coil induces emf in the second coil given by the farmula based on farady law

$e=-M\frac{di}{dt}$

$M=\frac{e}{\frac{di}{dt} }$

$M=\frac{1.60\times10^{-3} }{-0.245}$

$M= 6.53\times10^{-3} H$

Flux through each turn=?

Current in the first coil =1.25 A

Number of turns = 20

using MI = NФ

flux through each turn = Ф = $\frac{6.53\times10^{-3}\times1.25}{20}$

flux through each turn = Ф = $4.08\times10^{-4} Wb$

second coil increase at a rate = 0.365 A/s

magnitude of the induced emf in the first coil =?

using $e=-M\frac{di_{2} }{dt}$

$e= 6.53\times10^{-3} \times 0.365$

magnitude of the induced emf in the first coil = e= $2.38\times10^{-3} V$

4 0

3 years ago

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frozen [14]

Answer:

$\theta = 49.81^0$

Explanation:

Given that:

$\omega = 41.5 \ rev/min\\\\\omega = 41.5 *\frac{1}{60}* 2 \pi\\\\\omega = 4.45 \ rad/s\\\\\\diameter = 0.748 m$

If we let the piece of the close lose contact at ∠θ;

Then ; from force balance;

we have:

$\\\\mg sin \theta = \frac{mv^2}{r}\\\\sin \theta = \frac{2v^2}{dg}\\\\\theta = sin^{-1} (\frac{2v^2}{dg})$

where;

$v = \frac{\omega d}{2}\\\\v = \frac{4.45 *0.748}{2}\\\\v = 1.6643\\\\v^2 = 2.77$

Again:

$\theta = sin^{-1}(\frac{2v^2}{dg})\\\\\theta = sin^{-1}( \frac{2*2.77}{0.74*9.8})\\\\\theta = 49.81^0$

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3 years ago

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hram777 [196]

Answer:

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Explanation:

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4 years ago

I will mark brainliest if your answer is correct!​

I will mark brainliest if your answer is correct!