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3 years ago

6

Derive an approximate formula for the mutual inductance of two circular rings of the same radius a, arranged like wheels on the

same axle with their centers a distance b apart. Use an approximation good for b≫a.

1 answer:

son4ous [18]3 years ago

7 0

Answer:

Please check the attached file for the derivation

Explanation:

Check the attached file

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A place that limits a wave’s motion.

Yuki888 [10]

Answer:

boundary

Explanation:

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3 years ago

You need to fill a basketball with air to play with it. You know that your pump expels air at a velocity of 6 cm/s. You know tha

Scrat [10]

Answer:

The diameter of the needle is <u>4.675 cm</u>.

Explanation:

Given:

Volume flow rate is, $\dot V=103\ cm^3$

Velocity of air expelled by pump is, $v=6\ cm/s$

Let the area of the needle be 'A' cm² and the diameter be 'd' cm.

We know that, volume flow rate of the air expelled by pump is given as the product of the needle's area and velocity of air flowing through that area.

Therefore, volume flow rate is given as:

$\dot V=A\times v\\\\103=A\times 6\\\\6A=103\\\\A=\frac{103}{6}\ cm^2$

Now, considering the needle to be circular, area of the needle can be written as:

$A=\frac{\pi}{4}\times d^2\\\\\frac{103}{6}=\frac{\pi}{4}\times d^2\\\\d^2=\frac{103\times 4}{6\pi}\\\\d=\sqrt{\frac{103\times 4}{6\pi}}\\\\d=4.675\ cm$

Therefore, the diameter of the needle is 4.675 cm.

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3 years ago

a 10-N force is exerted on a box, moving it 20 m in the same direction. What is the magnitude of work done on the box?

-Dominant- [34]

F = 10 N

d = 20 m

θ = 0°

W = F (dot product) D = F * D * cos(angle between them)

W = FDcosθ

W = 10 * 20 * cos0 = 200 J

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3 years ago

Situation 6.1 A 13.5-kg box slides over a rough patch 1.75 m long on a horizontal floor. Just before entering the rough patch, t

GenaCL600 [577]

B) 14.0 N

The way to solve this problem is to determine the kinetic energy the box had before and after the rough patch of floor. The equation for kinetic energy is:

E = 0.5 M V^2

where

E = Energy

M = Mass

V = velocity

Substituting the known values, let's calculate the before and after energy.

Before:

E = 0.5 M V^2

E = 0.5 13.5kg (2.25 m/s)^2

E = 6.75 kg 5.0625 m^2/s^2

E = 34.17188 kg*m^2/s^2 = 34.17188 joules

After:

E = 0.5 M V^2

E = 0.5 13.5kg (1.2 m/s)^2

E = 6.75 kg 1.44 m^2/s^2

E = 9.72 kg*m^2/s^2 = 9.72 Joules

So the box lost 34.17188 J - 9.72 J = 24.451875 J of energy over a distance of 1.75 meters. Let's calculate the loss per meter by dividing the loss by the distance.

24.451875 J / 1.75 m = 13.9725 J/m = 13.9725 N

Rounding to 1 decimal place gives 14.0 N which matches option "B".

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3 years ago

Oersted found the direction of the magnetic field around a wire carrying current by placing a (fill in the blank)

jolli1 [7]

Answer: he used a compass to find the direction of the magnetic field

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2 years ago