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Lunna [17]
4 years ago
13

Identify the procedure to determine a formula for self-inductance, or inductance for short. Using the formula derived in the tex

t, find the inductance in henries for a solenoid with 1500 loops of wire wound on a rod 13 cm long with radius 4 cm.
Physics
1 answer:
OlgaM077 [116]4 years ago
3 0

Answer:

L = 0.109 H

Explanation:

Given that,

Number of loops in the solenoid, N = 1500

Radius of the wire, r = 4 cm = 0.04 m

Length of the rod, l = 13 cm = 0.13 m

To find,

Self inductance in the solenoid

Solution,

The expression for the self inductance of the solenoid is given by :

L=\dfrac{\mu_o N^2 A}{l}

L=\dfrac{4\pi \times 10^{-7}\times (1500)^2\times \pi (0.04)^2}{0.13}

L = 0.109 H

So, the self inductance of the solenoid is 0.109 henries.

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RUDIKE [14]
According to the conservation of mechanical energy, the kinetic energy just before the ball strikes the ground is equal to the potential energy just before it fell. 

Therefore, we can say KE = PE
We know that PE = m·g·h

Which means KE = m·g·h

We can solve for h:

h = KE / m·g
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The correct answer is: the ball has fallen from a height of 13.6m.

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3 years ago
A Carnot engine operates with an efficiency of 26.0% when the temperature of its cold reservoir is 281 K. Assuming that the temp
VLD [36.1K]

Answer:

The temperature of cold reservoir should be 246.818 K for efficiency of 35%

Explanation:

In first case we have given efficiency of Carnot engine = 26 % = 0.26

Temperature of cold reservoir T_L=281K

We know that efficiency of Carnot engine is given by

\eta =1-\frac{T_L}{T_H}

0.26 =1-\frac{281}{T_H}

T_H=379.72K

For second Carnot engine efficiency is given as 35% = 0.35

And temperature of hot reservoir is same so T_H=379.72K

So 0.35=1-\frac{T_L}{379.72}

T_L=246.818K

So the temperature of cold reservoir should be 246.818 K for efficiency of 35%

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