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notka56 [123]
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.

Physics
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 trip is taken that passes through the following points in order
riadik2000 [5.3K]

Answer:

35, I got you bro, i got you

8 0
3 years ago
(4.3 x 10-2 )4.950 x 105)<br> O 21.285 x 103<br> O 2.13 x 104<br> O 21 x 103<br> O 2.1 x 104
Natalka [10]

Answer:

The answer is:

It's 21.285 × 103

7 0
3 years ago
NEED ANSWER ASAP!!! Angela has a bucket of mass 2 kg tied to a string. She places a drinking glass of mass 0.5 kg in the bucket.
Schach [20]

a. The free-body diagram for the glass when it is at the top of the circle is attached below.

b. The equation for the net force on the glass at the top of the circle in terms of w, Fn, m, v, and r is mg x g + N -  mg x Vtop² /R =0

c. The glass will fall out of the bucket if the normal force between the glass and bucket equals zero. The speed with which she spin the bucket to prevent this from happening is 3.83 m/s.

d. The string will break if the tension on it is more than 100 N. The range of speeds can  prevent the string from breaking is 3.83< Vtop<4.99 m/s

<h3>What is Net force?</h3>

When two or more forces are acting on the system of objects, then the to attain equilibrium, net force must be zero.

Given, Angela has a bucket of mass 2 kg tied to a string. She places a drinking glass of mass 0.5 kg in the bucket. She spins the bucket in a vertical circle of radius 1.5 m. She must swing the bucket to keep the glass from falling out.

a. The free body diagram of the bucket and glass is attached below.

b. Bucket will undergo centrifugal force

Fb = mVtop² /R

From the equilibrium of forces, we have

For bucket,

T +mb xg - N =  mb x Vtop² /R..............(1)

For glass,

mg x g + N =  mg x Vtop² /R..............(2)

Thus, this is the net force equation on the glass.

c. On adding both the equations. we have

T + (mb + mg) xg = (mb + mg) Vtop² /R

Substituting the values, T = 0 and from the question, we get

0 + (2+0.5) 9.81 = (2+0.5)(Vtop²/0.5)

Vtop = 3.83 m/s

Thus, the speed of spin to prevent glass from falling out is 3.83 m/s

d. The string will break if the tension on it is more than 100 N

100 + (2+0.5) 9.81 = (2+0.5)(Vtop²/0.5)

Vtop = 4.99 m/s

Thus, the range of velocity is  3.83< Vtop<4.99 m/s

Learn more about net force.

brainly.com/question/18031889

#SPJ1

8 0
2 years ago
You wish to cool a 1.83 kg block of tin initially at 88.0°C to a temperature of 57.0°C by placing it in a container of kerosene
uranmaximum [27]

Answer:

0.273 liters are needed to accomplish this task without boiling.

Explanation:

The minimum boiling point of kerosene is 150\,^{\circ}C. According to this question, we need to determine the minimum volume of liquid such that heat received is entirely sensible, that is, with no phase change.

If we consider a steady state process and that energy interactions with surrounding are negligible, then we get the following formula by the Principle of Energy Conservation:

\rho_{k}\cdot V_{k}\cdot c_{k}\cdot (T-T_{k,o}) = m_{t}\cdot c_{t}\cdot (T_{t,o}-T) (1)

Where:

\rho_{k} - Density of kerosene, measured in kilograms per cubic meter.

V_{k} - Volume of kerosene, measured in cubic meters.

c_{k}, c_{t} - Specific heats of the kerosene and tin, measured in joule per kilogram-Celsius.

T_{k,o}, T_{t,o} - Initial temperatures of kerosene and tin, measured in degrees Celsius.

T - Final temperatures of the kerosene-tin system, measured in degrees Celsius.

Please notice that the block of tin is cooled at the expense of the temperature of the kerosene until thermal equilibrium is reached.

From (1), we clear the volume of kerosene:

V_{k} = \frac{m_{t}\cdot c_{t}\cdot (T_{t,o}-T)}{\rho_{k}\cdot c_{k}\cdot (T-T_{k,o})}

If we know that m_{t} = 1.83\,kg, c_{t} = 218\,\frac{J}{kg\cdot ^{\circ}C}, T_{t,o} = 88\,^{\circ}C, T_{k,o} = 24.0\,^{\circ}C, T = 57\,^{\circ}C, c_{k} = 2010\,\frac{J}{kg\cdot ^{\circ}C} and \rho_{k} = 820\,\frac{kg}{m^{3}}, then the volume of the liquid needed to accomplish this task without boiling is:

V_{k} = \frac{(1.83\,kg)\cdot \left(218\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (88\,^{\circ}C-57\,^{\circ}C)}{\left(820\,\frac{kg}{m^{3}} \right)\cdot \left(2010\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (57\,^{\circ}C-24\,^{\circ}C)}

V_{k} = 2.273\times 10^{-4}\,m^{3}

V_{k} = 0.273\,L

0.273 liters are needed to accomplish this task without boiling.

3 0
3 years ago
you see a lightning bolt in the sky. You hear a clap of thunder 3 seconds latter. The speed of 330 m/s. How far away was the lig
Naddika [18.5K]
The i got answer is 990m
8 0
3 years ago
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