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3 years ago

5

A big box of sausages (30 kg) is lifted from the ground to the top shelf of the freezer. If the box is lifted at a constant spee

d, a distance of 1.75 m, what work is done against gravity?

1 answer:

Lubov Fominskaja [6]3 years ago

5 0

Answer:

Work done to lift the box is 515.03 J

Explanation:

By work energy theorem we know that work done by all forces is equal to change in kinetic energy

So we have

$W_g + W_{ex} = \Delta K$

so we have

$-mgh + W_{ex} = 0$

so we have

$W_{ex} = mgh$

$W_{ex} = 30(9.8)1.75$

$W_{ex} = 515.03 J$

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A Parallel circuit has certain characteristics and basic rules: A parallel circuit has two or more paths for current to flow through. Voltage is the same across each component of the parallel circuit. The sum of the currents through each path is equal to the total current that flows from the source.

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Explanation:

Given that,

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A long solenoid with 1.65 103 turns per meter and radius 2.00 cm carries an oscillating current I = 6.00 sin 90πt, where I is in

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Answer:

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Explanation:

Given that,

Radius = 2.00 cm

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$\epsilon =\mu_{0}nA\dfrac{dI}{dt}$

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A = area of cross section

$\dfrac{dI}{dt}$ =rate of current

Formula of electric field is defined as,

$E=\dfrac{\epsilon}{2\pi r}$

Where, r = radius

Put the value of emf in equation (I)

$E=\dfrac{\mu_{0}nA\dfrac{dI}{dt}}{2\pi r}$ ....(II)

We need to calculate the rate of current

$I=6.00\sin 90\pi t$ ....(III)

On differentiating equation (III)

$\dfrac{dI}{dt}=90\pi\times6.00\cos(90\pi t)$

Now, put the value of rate of current in equation (II)

$E=\dfrac{4\pi\times10^{-7}\times1.65\times10^{3}\times\pi\times(2.00\times10^{-2})^2\times90\pi\times6.00\cos(90\pi t)}{2\pi\times 2.00\times10^{-2}}$

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Hence, The electric field is $35\cos(90\pi t)\ mV/m$

7 0

3 years ago

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