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sertanlavr [38]

2 years ago

14

Which equation is used to calculate the electric potential in an electric field from a point charge? V = kq over d squared V = k

q over d V =delta U over q V = delta U over E

2 answers:

Contact [7]2 years ago

4 0

Answer:

The answer is V =delta U over q

Explanation:

Electric potential is defined as the magnitude of the electric field through the potential energy that a charge would have if placed at that point. Mathematically, the potential is defined with the following expression:

$V=\frac{E_{p} }{q}$

where:

V is the electric potential. Its unit is Julius by Coulomb (J/C).

Ep is the electric potential energy that has a charge

q is the charge

In the question Ep = ΔU

Oduvanchick [21]2 years ago

4 0

Answer:

The answer is V=kq/d

Explanation:

The electrostatic potential is defined in terms of work done in taking a unit positive charge from infinity to the point in question.

V=kq/d

Where d= distance from infinity to the point in question

k= constant of proportionality

q= attracting charge

The electric potential is measured in volts

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1 year ago

A 40-kg box is being pushed along a horizontal smooth surface. The pushing force is 15 n directed at an angle of 15Â° below the

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Answer:

Acceleration of the crate is 0.362 m/s^2.

Explanation:

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Let the acceleration be "a".

FBD is attached with where we can see the horizontal and vertical component of force.

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2 years ago

The component of the external magnetic field along the central axis of a 78-turn circular coil of radius 34.0 cm decreases from

grigory [225]

Answer:

Induced current, I = 18.88 A

Explanation:

It is given that,

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Radius of the circular coil, r = 34 cm = 0.34 m

Magnetic field changes from 2.4 T to 0.4 T in 2 s.

Resistance of the coil, R = 1.5 ohms

We need to find the magnitude of the induced current in the coil. The induced emf is given by :

$\epsilon=-N\dfrac{d\phi}{dt}$

Where

$\dfrac{d\phi}{dt}$ is the rate of change of magnetic flux,

And $\phi=BA$

$\epsilon=-NA\dfrac{dB}{dt}$

$\epsilon=-78\times \pi (0.34)^2\dfrac{(0.4-2.4)}{2}$

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Using Ohm's law, $\epsilon=I\times R$

Induced current, $I=\dfrac{\epsilon}{R}$

$I=\dfrac{28.32}{1.5}$

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So, the magnitude of the induced current in the coil is 18.88 A. Hence, this is the required solution.

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riadik2000 [5.3K]

Answer:

Plss see attached file

Explanation:

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