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3 years ago

A current of 1.70 A flows in a wire. How many electrons are flowing past any point in the wire per second

Physics

1 answer:

castortr0y [4]3 years ago

8 0

Answer:

1.0625×10¹⁹ electrons

Explanation:

From the question,

Using,

Q = It........................ Equation 1

Where Q = charge, I = current flowing in the wire, t = time (seconds)

Given: I = 1.70 A, t = 1 Seconds.

Substitute these values into equation 1

Q = 1.7(1)

Q = 1.7 C.

But,

1.6×10⁻¹⁹ C = 1 electron

Therefore,

1.7 C = 1.7/(1.6×10⁻¹⁹) electrons

= 1.0625×10¹⁹ electrons

Hence, 1.0625×10¹⁹ electrons flows through the wire per seconds.

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(b). The angle of the total acceleration relative to the radial direction is 11.0°

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Using formula of centripetal acceleration

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$a_{c}=235.5\ m/s^2$

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Using formula of tangential acceleration

$a_{t}=r\alpha$

Put the value into the formula

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(a). We need to calculate the magnitude of the total acceleration of the ball

Using formula of total acceleration

$a=\sqrt{a_{c}^2+a_{t}^2}$

Put the value into the formula

$a=\sqrt{(235.5)^2+(46.104)^2}$

$a=239.97\ m/s^2$

(b). We need to calculate the angle of the total acceleration relative to the radial direction

Using formula of the direction

$\theat=\tan^{-1}(\dfrac{a_{t}}{a_{c}})$

Put the value into the formula

$\theta=\tan^{-1}(\dfrac{46.104}{235.5})$

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4 years ago

Two football players move in a straight line directly toward one another. Their equations of motion are as follows:

iren [92.7K]

The time when the two players will collide is 0.96 s.

The equation of motion of the two players is given as;

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x2 = –6.3 m + (2.8 m/s )t

The time when the two players collide, their displacement is equal or the difference in their position will be zero.

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