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This is a machine that converts electrical energy into mechanical energy.

Yuri [45]

<h2>Answer:</h2>

Motor

<h2>Explanation:</h2>

A motor is a machine that converts electrical energy into mechanical energy. In motors, electric energy is converted into mechanic energy when a magnetic torque acts on a conductor that carries a current. There are different types of motors like DC and AC motors. The moving part of a motor is called the rotor while the stationary part is called stator

8 0

3 years ago

The electron dot diagram shows the arrangement of dots without identifying the element.

olasank [31]

<em>Answer: </em>tellurium (Te)

<em>atomic number = 52 ,</em>

<em>Number of energy levels = 5;</em>

First energy level = 2

Second energy level = 8

Third energy level = 18

Fourth energy level = 18

Fifth energy level = 6

<em>In this electron configuration, 0uter most electrons are 6.</em>

5 0

2 years ago

Read 2 more answers

Initial velocity 10 m/s accelerates at 5 m/s for 2 seconds whats the final velocity

stiks02 [169]

Answer:

<em>The final velocity is 20 m/s.</em>

Explanation:

<u>Constant Acceleration Motion</u>

It's a type of motion in which the velocity of an object changes by an equal amount in every equal period of time.

Being a the constant acceleration, vo the initial speed, and t the time, the final speed can be calculated as follows:

$v_f=v_o+at$

The provided data is: vo=10 m/s, $a=5\ m/s^2$ , t=2 s. The final velocity is:

$v_f=10~m/s+5\ m/s^2\cdot 2\ s$

$v_f=20\ m/s$

The final velocity is 20 m/s.

8 0

3 years ago

Two charges, each of 2.9 microC are placed at two corners of a square 50cm on a side, If the charges are on one side of the squa

anyanavicka [17]

Answer:

The magnitude of the electric field and direction of electric field are $146.03\times10^{3}\ N/C$ and 75.36°.

Explanation:

Given that,

First charge $q_{1}= 2.9\mu C$

Second charge $q_{2}= 2.9\mu C$

Distance between two corners r= 50 cm

We need to calculate the electric field due to other charges at one corner

For E₁

Using formula of electric field

$E_{1}=\dfrac{kq}{r'^2}$

Put the value into the formula

$E_{1}=\dfrac{9\times10^{9}\times2.9\times10^{-6}}{(50\sqrt{2}\times10^{-2})^2}$

$E_{1}=52200=52.2\times10^{3}\ N/C$

For E₂,

Using formula of electric field

$E_{1}=\dfrac{kq}{r^2}$

Put the value into the formula

$E_{2}=\dfrac{9\times10^{9}\times2.9\times10^{-6}}{(50\times10^{-2})^2}$

$E_{2}=104400=104.4\times10^{3}\ N/C$

We need to calculate the horizontal electric field

$E_{x}=E_{1}\cos\theta$

$E_{x}=52.2\times10^{3}\times\cos45$

$E_{x}=36910.97=36.9\times10^{3}\ N/C$

We need to calculate the vertical electric field

$E_{y}=E_{2}+E_{1}\sin\theta$

$E_{y}=104.4\times10^{3}+52.2\times10^{3}\sin45$

$E_{y}=141310.97=141.3\times10^{3}\ N/C$

We need to calculate the net electric field

$E_{net}=\sqrt{E_{x}^2+E_{y}^2}$

Put the value into the formula

$E_{net}=\sqrt{(36.9\times10^{3})^2+(141.3\times10^{3})^2}$

$E_{net}=146038.69\ N/C$

$E_{net}=146.03\times10^{3}\ N/C$

We need to calculate the direction of electric field

Using formula of direction

$\tan\theta=\dfrac{141.3\times10^{3}}{36.9\times10^{3}}$

$\theta=\tan^{-1}(\dfrac{141.3\times10^{3}}{36.9\times10^{3}})$

$\theta=75.36^{\circ}$

Hence, The magnitude of the electric field and direction of electric field are $146.03\times10^{3}\ N/C$ and 75.36°.

4 0

2 years ago

A student touches sphere x and moves it close to, but not touching sphere y. What are the natures of the charges left on the two

e-lub [12.9K]

No charge I know this because

7 0

3 years ago