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lisov135 [29]
2 years ago
10

On Earth, plasma exists in the ionosphere, in flames, and in chemical and nuclearexplosions. Matter in a controlled thermonuclea

r reactor also exists in a plasma state.Plasma is most like a gas. However, it differs from un-ionized gas in that it is a goodconductor of electricity and heat. Scientists hope to understand the occurrence of plasmain nature and to harness it as an inexpensive energy source.Based on the passage, how can the cost of developing thermonuclear power with plasmabe defended?A. It is almost perfected already.B. Current nuclear power is too dangerous.C. A plasma power source can be used without special equipment.D. It can provide an inexpensive power source.OE. It is applicable to all technologies.
Physics
1 answer:
nasty-shy [4]2 years ago
3 0

The cost of developing thermonuclear power with plasmabe defended because D. It can provide an inexpensive power source.

<h3>How did the cost of developing thermonuclear power defended?</h3>

The  cost of developing thermonuclear power defended becvause we can see in the paragraph how it was told that the generation of ths power can be donee through the understanding of  the occurrence of plasmain nature,

It should be noted that this  thermonuclear power with plasmabe  posses the characteristics which make it to exist in the ionosphere, and it can be felt in the flames as well; as in the chemical and nuclearexplosions.

In conclusion the power can be seen as an inexpensive source power because the p[roduction of this power cn be found in most of the thing that can be found around us as discused above.

Therefore, option D is correct.

Read more about cost at:

brainly.com/question/25109150

#SPJ1

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When the electrons reach the collector, they flow towards the positivly charged grid. The resulting current is measured. Note th
Masja [62]

Answer:

We can conclude by saying that in the beginning current will increase but after sometime, it becomes saturated.

Explanation:

Note: No information on change in number of electron generated.

Since there is a collision, the electrons emitted will not reach the collector at same time. As the voltage is increased, the the speed with which the electrons will reach the collector starts to increase. Due to this, electric current will first increases till all the emitted electrons reach the collector. Since we are not provided with the information that number of electrons generated are changing, after increasing voltage current will increase for some time and then reaches a saturated state.

We can conclude by saying that in the beginning current will increase but after sometime it becomes saturated.

4 0
3 years ago
Which two factors affect the size of the gravitational field?
aliina [53]

Answer:

Explanation:

mass and distance

8 0
3 years ago
g A particle (charge = +40 mC) is located on the x axis at the point x = -20 cm, and a second particle (charge = -50 mC) is plac
krok68 [10]

Answer:

Explanation:

We shall find electric field at origin due to two given charges sitting   on the either side of origin .

Total field will add up due to their same direction .

Field due to a charge Q

= 9 x 10⁹ x Q / R²  ;  R is distance of point , Q is charge

Field due to first charge

= 9 x 10⁹ x 40 x 10⁻³ / 2² x 10⁻⁴

= 90 x 10¹⁰ N/C

Field due to second  charge

= 9 x 10⁹ x 50 x 10⁻³ / 2² x 10⁻⁴

= 112.5 x 10¹⁰ N/C

Total field

= 202.5 x 10¹⁰ N/C

Force on given charge at origin

= charge x field

= 4 x 10⁻³ x 202.5 x 10¹⁰

= 810 x 10⁷ N .

4 0
4 years ago
Help with this question.....​
mariarad [96]

Answer:

g_{moon}=1.67 [m/s^{2} ]

Explanation:

The weight of some mass is defined as the product of mass by gravitational acceleration. In this way using the following formula we can find the weight.

w =m*g\\

where:

w = weight [N]

m = mass = 0.06 [kg]

g = gravity acceleration = 10 [N/kg]

Therefore:

w=0.06*10\\w=0.6[N]

By Hooke's law we know that the force in a spring can be calculated by means of the following expression.

F=W\\F = k*x

where:

k = spring constant [N/m]

x = deformed distance = 6 [cm] = 0.06 [m]

We can find the spring constant.

k= F/x\\k=0.6/0.06\\k=10 [N/m]

Since we use the same spring on the moon and the same mass, the constant of the spring does not change, the same goes for the mass.

F_{moon}=k*0.01\\F = 10*0.01\\F=0.1[N]

Since this force is equal to the weight, we can now determine the gravitational acceleration.

F=m*g_{moon}\\g=F/m\\g = 0.1/0.06\\g_{moon} = 1.67[m/s^{2} ]

6 0
3 years ago
If the moon was closer to earth how long would it take to orbit the earth? It is only right before the Roche limit, it’s theoret
Alex_Xolod [135]

Answer:

it would probably be the same but if the moon is closer gravity would infect us humans

Explanation:

6 0
3 years ago
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