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Galina-37 [17]
3 years ago
14

The coil of an ac generator has 50 loops and a cross-sectional area of LaTeX: 0.2~m^20.2 m 2 . What is the maximum emf that can

be generated, if the generator is spinning with an angular speed of LaTeX: 5.0~rad/s5.0 r a d / s in a LaTeX: 2.0~T2.0 Tmagnetic field
Physics
1 answer:
lisov135 [29]3 years ago
7 0

Answer:

Maximum emf in the coil is 100 volt    

Explanation:

We have given number of loops in the coil N = 50

Cross sectional area A=0.2m^2

Angular speed is given \omega =5rad/sec

Magnetic field is given B=0.2T

We have to find the maximum emf in the coil

Maximum emf in the coil is given by e=NBA\omega

e=50\times 2\times 0.2\times 5=100volt

So maximum emf induced in the coil is 100 volt

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finlep [7]

Answer:

Stretch can be obtained using the Elastic potential energy formula.

The expression to find the stretch (x) is x=\sqrt{\frac{2\times EPE}{k}}

Explanation:

Given:

Elastic potential energy (EPE) of the spring mass system and the spring constant (k) are given.

To find: Elongation in the spring (x).

We can find the elongation or stretch of the spring using the formula for Elastic Potential Energy (EPE).

The formula to find EPE is given as:

EPE=\frac{1}{2}kx^2

Rewriting the above expression in terms of 'x', we get:

x=\sqrt{\frac{2\times EPE}{k}}

Example:

If EPE = 100 J and spring constant, k = 2 N/m.

Elongation or stretch is given as:

x=\sqrt{\frac{2\times EPE}{k}}\\\\x=\sqrt{\frac{2\times 100}{2}}\\\\x=\sqrt{100}=10\ m

Therefore, the stretch in the spring is 10 m.

So, stretch in the spring can be calculated using the formula for Elastic Potential Energy.

6 0
2 years ago
A charge of -3.02 μC is fixed in place. From a horizontal distance of 0.0377 m, a particle of mass 9.43 x 10^-3 kg and charge -9
Andreyy89

Answer:

d = 0.0306 m

Explanation:

Here we know that for the given system of charge we have no loss of energy as there is no friction force on it

So we will have

U + K = constant

\frac{kq_1q_2}{r_1} + \frac{1}{2}mv_1^2 = \frac{kq_1q_2}{r_2} + \frac{1}{2}mv_2^2

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So we have

\frac{(9 \times 10^9)(3.02 \mu C)(9.78 \mu C)}{0.0377} + \frac{1}{2}(9.43 \times 10^{-3})(80.4)^2 = \frac{(9 \times 10^9)(3.02 \mu C)(9.78 \mu C)}{r} + 0

7.05 + 30.5 = \frac{0.266}{r}

r = 7.08 \times 10^{-3} m

so distance moved by the particle is given as

d = r_1 - r_2

d = 0.0377 - 0.00708

d = 0.0306 m

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Answer:

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