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3 years ago

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A person pushes a 10 kg box from rest and accelerates it to a speed of 4 m/s with a constant force. If the box is pushed for a t

ime of 2.5 s, what is the force exerted by the person?

1 answer:

lutik1710 [3]3 years ago

7 0

The box is accelerated from rest to 4 m/s in a matter of 2.5 s, so its acceleration <em>a</em> is such that

4 m/s = <em>a</em> (2.5 s) → <em>a</em> = (4 m/s) / (2.5 s) = 1.6 m/s²

Then the force applied to the box has a magnitude <em>F</em> such that

<em>F</em> = (10 kg) (1.6 m/s²) = 16 N

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Answer:

$R = 24.3 m$

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As we know that the orbital speed is given as

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here we know that

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now we have

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now acceleration due to gravity of planet is given as

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4 years ago

How much voltage is required to run 0.64 A of current through a 240 resistor? Use V= IR.

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B. If you press that into a calculator it comes up with 153.6. You then shift the decimal point 2 times forward and you end up getting 1.5 x 10^2 V.

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An electric field of magnitude 2.35 V/m is oriented at an angle of 25.0° with respect to the positive z-direction. Determine the

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Answer:

The magnitude of the electric flux is $3.53\ N-m^2/C$

Explanation:

Given that,

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Area $A= 1.65 m^2$

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Using formula of the magnetic flux

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$\phi = EA\cos\theta$

Where,

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Put the value into the formula

$\phi=2.35\times1.65\times\cos 25^{\circ}$

$\phi=2.35\times1.65\times0.91$

$\phi=3.53\ N-m^2/C$

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3 years ago

A 10 gauge copper wire carries a current of 20 A. Assuming one free electron per copper atom, calculate the magnitude of the dri

nordsb [41]

Complete Question

A 10 gauge copper wire carries a current of 20 A. Assuming one free electron per copper atom, calculate the drift velocity of the electrons. (The cross-sectional area of a 10-gauge wire is 5.261 mm2.) mm/s

Answer:

The drift velocity is $v = 0.0002808 \ m/s$

Explanation:

From the question we are told that

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So

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$N = 8.46 * 10^{28} \ electrons$

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substituting values

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3 years ago

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liubo4ka [24]

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