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3 years ago

A device that changes mechanical energy to electrical energy by rotating a coil of wire through a magnetic field is called a(an)

Physics

2 answers:

fredd [130]3 years ago

8 0

The correct answer to the question is : B) Generator.

EXPLANATION:

As per the question, the mechanical energy is converted into electrical energy.

All the power plants like thermal power plant, nuclear power plant and hydroelectric power plant use the same concept to produce electric energy.

Let us consider hydro power plant.

In hydro power plant, the water falling with high speed is allowed to fall on a turbine. The kinetic energy of the water rotates the turbine. The turbine is attached to a generator.

The coil of wire attached to it is rotated through a magnetic field and electricity is produced due to electromagnetic induction.

Hence, the correct answer to the question is generator which converts mechanical energy into electrical energy.

tigry1 [53]3 years ago

4 0

A device that changes mechanical energy to electrical energy by rotatin a coil of wire through a magnetic field is called B. Generator

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3 years ago

a golfer hits a 0.05 kg golf ball with an impulse of 3 N-s what is the change in velocity of the ball

Rainbow [258]

so your saying the start is 0 N and when he/she hits the ball its inertia is 3 N. if that is so m*v=

.05*3=.15

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2 years ago

A solid, horizontal cylinder of mass 18.0 kg and radius 1.70.0 m rotates with an angular speed of 40 rad/s about a fixed vertica

Radda [10]

Answer:39.88 rad/s

Explanation:

Given

mass of cylinder m_1=18 kg

radius R=1.7 m

angular speed $\omega =40rad/s$

mass of $m_2=0.8 kg$ dropped at r=0.3 m from center

let $\omega _2$ be the final angular velocity of cylinder

Conserving Angular momentum

$L_1=L_2$

$\left ( \frac{m_1R^2}{2}\right )\omega =\left ( \frac{m_1R^2}{2}+m_2r^2\right )\omega _2$

$\left ( \frac{18\cdot 1.7^2}{2}\right )\cdot 40=\left ( \frac{18\cdot 1.7^2}{2}+0.8\cdot 0.3^2\right )\omega _2$

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3 years ago

An industrial laser is used to burn a hole through a piece of metal. The average intensity of the light is W/m². What is the rms

inn [45]

I think your question should be:

An industrial laser is used to burn a hole through a piece of metal. The average intensity of the light is

$S = 1.23*10^9 W/m^2$

What is the rms value of (a) the electric field and

(b) the magnetic field in the electromagnetic wave emitted by the laser

Answer:

a) $6.81*10^5 N/c$

b) $2.27*10^3 T$

Explanation:

To find the RMS value of the electric field, let's use the formula:

$E_r_m_s = sqrt*(S / CE_o)$

Where

$C = 3.00 * 10^-^8 m/s$ ;

$E_o = 8.85*10^-^1^2 C^2/N.m^2$ ;

$S = 1.23*10^9 W/m^2$

Therefore

$E_r_m_s = sqrt*{(1.239*10^9W/m^2) / [(3.00*10^8m/s)*(8.85*10^-^1^2C^2/N.m^2)]}$

$E_r_m_s= 6.81 *10^5N/c$

b) to find the magnetic field in the electromagnetic wave emitted by the laser we use:

$B_r_m_s = E_r_m_s / C$ ;

$= 6.81*10^5 N/c / 3*10^8m/s$ ;

$B_r_m_s = 2.27*10^3 T$

8 0

3 years ago

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