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3 years ago

Please explain the method of making electricity from non renewable sources like coal?

1 answer:

7 0

The coal is urned to heat up water. this produces steam. the steam turns a turbine that turns a generator which provided energy yhat can be transferred into electrisity

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Two boats start together and race across a 48-km-wide lake and back. Boat A goes across at 48 km/h and returns at 48 km/h. Boat

nika2105 [10]

Answer:

Part 1)

Boat A will win the race

Part 2)

Boat A will win the race by 48 km as the 2nd boat will reach the other end while boat A will just touches the finish line

Part 3)

average velocity must be zero

Explanation:

As we know that the distance moved by the boat is given as

$d = 48 km$

now the time taken by the boat to move to and fro is given as

$t = \frac{d}{v}$

$t = \frac{48 + 48}{48}$

$t = 2 hrs$

Time taken by Boat B to cover the distance

$t = \frac{48}{24} + \frac{48}{72}$

$t = 2.66 h$

Part 1)

Boat A will win the race

Part 2)

Boat A will win the race by 48 km as the 2nd boat will reach the other end while boat A will just touches the finish line

Part 3)

Since the displacement of Boat A is zero

so average velocity must be zero

3 0

3 years ago

You wrap a wire around a piece of iron. If you slowly increase the strength of an electric current flowing through the wire, you

Maksim231197 [3]

The correct answer is A. the magnet to become stronger

The stronger the electric current in the piece of metal, the stronger the magnetic field will be.

4 0

3 years ago

Read 2 more answers

Two charges are located in the x – y plane. If ????1=−4.10 nC and is located at (x=0.00 m,y=0.600 m) , and the second charge has

faust18 [17]

Answer:

The x-component of the electric field at the origin = -11.74 N/C.

The y-component of the electric field at the origin = 97.41 N/C.

Explanation:

<u>Given:</u>

Charge on first charged particle, $q_1=-4.10\ nC=-4.10\times 10^{-9}\ C.$
Charge on the second charged particle, $q_2=3.80\ nC=3.80\times 10^{-9}\ C.$

Position of the first charge = $(x_1=0.00\ m,\ y_1=0.600\ m).$
Position of the second charge = $(x_2=1.50\ m,\ y_2=0.650\ m).$

The electric field at a point due to a charge $q$ at a point $r$ distance away is given by

$\vec E = \dfrac{kq}{|\vec r|^2}\ \hat r.$

where,

$k$ = Coulomb's constant, having value $\rm 8.99\times 10^9\ Nm^2/C^2.$

$\vec r$ = position vector of the point where the electric field is to be found with respect to the position of the charge $q$ .

$\hat r$ = unit vector along $\vec r$ .

The electric field at the origin due to first charge is given by

$\vec E_1 = \dfrac{kq_1}{|\vec r_1|^2}\ \hat r_1.$

$\vec r_1$ is the position vector of the origin with respect to the position of the first charge.

Assuming, $\hat i,\ \hat j$ are the units vectors along x and y axes respectively.

$\vec r_1=(0-x_1)\hat i+(0-y_1)\hat j\\=(0-0)\hat i+(0-0.6)\hat j\\=-0.6\hat j.\\\\|\vec r_1| = 0.6\ m.\\\hat r_1=\dfrac{\vec r_1}{|\vec r_1|}=\dfrac{0.6\ \hat j}{0.6}=-\hat j.$

Using these values,

$\vec E_1 = \dfrac{(8.99\times 10^9)\times (-4.10\times 10^{-9})}{(0.6)^2}\ (-\hat j)=1.025\times 10^2\ N/C\ \hat j.$

The electric field at the origin due to the second charge is given by

$\vec E_2 = \dfrac{kq_2}{|\vec r_2|^2}\ \hat r_2.$

$\vec r_2$ is the position vector of the origin with respect to the position of the second charge.

$\vec r_2=(0-x_2)\hat i+(0-y_2)\hat j\\=(0-1.50)\hat i+(0-0.650)\hat j\\=-1.5\hat i-0.65\hat j.\\\\|\vec r_2| = \sqrt{(-1.5)^2+(-0.65)^2}=1.635\ m.\\\hat r_2=\dfrac{\vec r_2}{|\vec r_2|}=\dfrac{-1.5\hat i-0.65\hat j}{1.634}=-0.918\ \hat i-0.398\hat j.$

Using these values,

$\vec E_2= \dfrac{(8.99\times 10^9)\times (3.80\times 10^{-9})}{(1.635)^2}(-0.918\ \hat i-0.398\hat j) =-11.74\ \hat i-5.09\ \hat j\ N/C.$

The net electric field at the origin due to both the charges is given by

$\vec E = \vec E_1+\vec E_2\\=(102.5\ \hat j)+(-11.74\ \hat i-5.09\ \hat j)\\=-11.74\ \hat i+(102.5-5.09)\hat j\\=(-11.74\ \hat i+97.41\ \hat j)\ N/C.$

Thus,

x-component of the electric field at the origin = -11.74 N/C.

y-component of the electric field at the origin = 97.41 N/C.

4 0

3 years ago

The mass of an object with 500 J of kinetic energy moving with a velocity of 5 m/s is/

antiseptic1488 [7]

Answer:

495

Explanation:

subtract 500 and 5

7 0

3 years ago

A crane carries a 1600 kg car at 1.5 m/s^2 with a chain that has a negligible mass. If the coefficient of kinetic friction betwe

padilas [110]

Answer:

option (d) 7.1 kN

Explanation:

Given:

Mass of the car, m = 1600 kg

Acceleration of the car, a = 1.5 m/s²

Coefficient of kinetic friction = 0.3

let the tension be 'T'

Now,

ma = T - f .................(1)

where f is the frictional force

also,

f = 0.3 × mg

where g is the acceleration due to the gravity

thus,

f = 0.3 × 1600 × 9.81 =

therefore,

equation 1 becomes

1600 × 1.5 = T - 4708.8

T = 2400 + 4708.8

T = 7108.8 N

T = 7.108 kN

Hence,

The correct answer is option (d) 7.1 kN

6 0

3 years ago