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3 years ago

What do nuclear plants use instead of fossil fuels? i need help please

Physics

1 answer:

Leona [35]3 years ago

6 0

Answer:

Nuclear energy is cleaner while generating electricity. Nuclear fission provides energy without releasing greenhouse gases such as carbon dioxide. However, nuclear power plants generate radioactive waste, a critical factor when doing a fossil fuel to nuclear power pollution comparison.

Explanation:

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The length of the mercury thread is found to be 4cm and 24cm at ice point and steam point respectively on an ungraduated thermom

BabaBlast [244]

Answer:

The difference between ice and steam in Celsius (Centigrade) is 100 deg.

So the difference between and 4 cm and 24 cm of the thread corresponds to 100 deg C.

So 8 cm is 4 cm greater than the ice point

4 cm / 20 cm = 1/5 since the steam point and the ice point are 20 cm apart

Then 1/5 * 100 deg C = 20 deg C the requested temperature

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3 years ago

What are the three points of the fire triangle

prisoha [69]

1-fuel
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3-oxygen

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3 years ago

Help with this question please :>

zheka24 [161]

Answer: A if thats not right its C

Explanation:

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3 years ago

Read 2 more answers

When you have two like charges in a line â where is the electric field the greatest? is there ever a point where the field will

Vesna [10]

The magnitude of the electric field will be the greatest at the point where it is closest,to its charges.

Yes ,there is a point where the field will be zero.

what is an electric field?

The region where an electrostatic force is experienced by a charged entity is known as the electric field at a point.

As per the principle of field lines and vectors,where the field lines are in a close manner together,the field will be strongest.However ,where the field lines are in a manner apart,the field will be the weakest.

As per the concept,the electric field will be the greatest at the point where it is closest to its charges.For like charges, the electric field will be zero closer to the smaller charge and will be along the line joining the two charges. For opposite charges of equal magnitude, there will not be any zero electric fields.

Thus,we can conclude that there will be a point where the electric field is zero

learn more about electric field from here: brainly.com/question/28197462

#SPJ4

4 0

2 years ago

A particle has a charge of q = +4.9 μC and is located at the origin. As the drawing shows, an electric field of Ex = +242 N/C ex

irina1246 [14]

$F_{E_x}=1.19\cdot 10^{-3}N$ (+x axis)

$F_{B_x}=0$

$F_{B_y}=0$

$F_{E_x}=1.19\cdot 10^{-3} N$ (+x axis)

$F_{B_x}=0$

$F_{B_y}=3.21\cdot 10^{-3}N$ (+z axis)

$F_{E_x}=1.19\cdot 10^{-3} N$ (+x axis)

$F_{B_x}=3.21\cdot 10^{-3} N$ (+y axis)

$F_{B_y}=3.21\cdot 10^{-3}N$ (-x axis)

Explanation:

The electric force exerted on a charged particle is given by

$F=qE$

where

q is the charge

E is the electric field

For a positive charge, the direction of the force is the same as the electric field.

In this problem:

$q=+4.9\mu C=+4.9\cdot 10^{-6}C$ is the charge

$E_x=+242 N/C$ is the electric field, along the x-direction

So the electric force (along the x-direction) is:

$F_{E_x}=(4.9\cdot 10^{-6})(242)=1.19\cdot 10^{-3} N$

towards positive x-direction.

The magnetic force instead is given by

$F=qvB sin \theta$

where

q is the charge

v is the velocity of the charge

B is the magnetic field

$\theta$ is the angle between the directions of v and B

Here the charge is stationary: this means $v=0$ , therefore the magnetic force due to each component of the magnetic field is zero.

In this case, the particle is moving along the +x axis.

The magnitude of the electric force does not depend on the speed: therefore, the electric force on the particle here is the same as in part a,

$F_{E_x}=1.19\cdot 10^{-3} N$ (towards positive x-direction)

Concerning the magnetic force, we have to analyze the two different fields:

- $B_x$ : this field is parallel to the velocity of the particle, which is moving along the +x axis. Therefore, $\theta=0^{\circ}$ , so the force due to this field is zero.

$- B_y$ : this field is perpendicular to the velocity of the particle, which is moving along the +x axis. Therefore, $\theta=90^{\circ}$ . Therefore, $\theta=90^{\circ}$ , so the force due to this field is: