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Lunna [17]
3 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]3 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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a projectile is fired in such away that its horizontal range is equal to three times its naximum height.what is the angle of pro
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Answer:

\theta=53.13^o

Explanation:

<u>2-D Projectile Motion</u>

In 2-D motion, there are two separate components of the acceleration, velocity and displacement. The horizontal component has zero acceleration, while the acceleration in the vertical direction is always the acceleration due to gravity. The basic formulas for this type of movement are

V_x=V_{ox}=V_ocos\theta

V_y=V_{oy}-gt=V_osin\theta-gt

x=V_{ox}t

\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}

\displaystyle x_{max}=\frac{2V_{ox}V_{oy}}{g}

\displaystyle y_{max}=\frac{V_{oy}^2}{2g}

The projectile is fired in such a way that its horizontal range is equal to three times its maximum height. We need to find the angle \theta at which the object should be launched. The range is the maximum horizontal distance reached by the projectile, so we establish the base condition:

x_{max}=3y_{max}

\displaystyle \frac{2V_{ox}V_{oy}}{g}=3\frac{V_{oy}^2}{2g}

Using the formulas for V_{ox}, V_{oy}:

\displaystyle \frac{2V_{o}cos\theta V_{o}sin\theta}{g}=3\frac{V_{o}^2sin^2\theta}{2g}

Simplifying

4cos\theta sin\theta=3sin^2\theta

Dividing by sin\theta

4cos\theta=3sin\theta

Rearranging

tan\theta=\frac{4}{3}

\theta=arctan\frac{4}{3}

\theta=53.13^o

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3 years ago
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A 7300 kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.20 m/s2 and feels no appreci
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Answer:

Explanation:

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