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Ganezh [65]
3 years ago
6

The children at the local park think the slide is too slow. There is too much fiction. What could you do to decrease the amount

of friction on the slide?
Physics
1 answer:
igor_vitrenko [27]3 years ago
7 0
In order to decrease the friction on the slide,
we could try some of these:

-- Install a drippy pipe across the top that keeps continuously
dripping olive oil on the top end of the slide.  The oil oozes
down the slide and keeps the whole slide greased.

-- Hire a man to spread a coat of butter on the whole slide,
every 30 minutes.

-- Spray the whole slide with soapy sudsy water, every 30 minutes.

-- Drill a million holes in the slide,and pump high-pressure air
through the holes.  Make the slide like an air hockey table.

-- Keep the slide very cold, and keep spraying it with a fine mist
of water.  The water freezes, and a thin coating of ice stays on
the slide.

-- Ask a local auto mechanic to please, every time he changes
the oil in somebody's car, to keep all the old oil, and once a week
to bring his old oil to the park, to spread on the slide.  If it keeps
the inside of a hot car engine slippery, it should do a great job
keeping a simple park slide slippery.

-- Keep a thousand pairs of teflon pants near the bottom of the ladder
at the beginning of the slide.  Anybody who wants to slide faster can
borrow a set of teflon pants, put them on before he uses the slide, and
return them when he's ready to go home from the park.
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A coil has 400 turns and self-inductance 7.50 mH. The current in the coil varies with time according to i = 1680 mA2 cos [πt/(0.
zhannawk [14.2K]

Answer:

(a) 1.58 V

(b) 0.0126 Wb

(c) 0.0493 V

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As per the question:

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Self Inductance of the coil, L = 7.50 mH = 7.50\times 10^{- 3}\ H

Current in the coil, i = 1680cos[\frac{\pi t}{0.0250}] A

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i_{max} = 1680\ mA = 1.680\ A

Now,

(a)  To calculate the maximum emf:

We know that maximum emf induced in the coil is given by:

e = \frac{Ldi}{dt}

e = L\frac{d}{dt}(1680)cos[\frac{\pi t}{0.0250}]

e = - 7.50\times 10^{- 3}\times \frac{\pi}{0.0250}\times \frac{d}{dt}(1680)sin[\frac{\pi t}{0.0250}]

For maximum emf, sin\theta should be maximum, i.e., 1

Now, the magnitude of the maximum emf is given by:

|e| = 7.50\times 10^{- 3}\times 1680\times 10^{- 3}\times \frac{\pi}{0.0250} = 1.58\ V

(b) To calculate the maximum average flux,we know that:

\phi_{m, avg} = L\times i_{max} = 7.50\times 10^{- 3}\times 1.680 = 0.0126\ Wb

(c) To calculate the magnitude of the induced emf at t = 0.0180 s:

e = e_{o}sin{\pi t}{0.0250}

e = 7.50\times 10^{- 3}\times sin{\pi \times 0.0180}{0.0250} = 2.96\times 10^{- 4} =0.0493\ V

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3 years ago
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Answer:

Δv/Δt

Explanation:

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