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hichkok12 [17]
3 years ago
8

What is one benefit that nuclear power plants currently provide for our environment?

Physics
2 answers:
yan [13]3 years ago
6 0
The answer would B, trust me it’s right
allsm [11]3 years ago
6 0

Answer: A

Explanation:

Nuclear power plants deliver electricity 24 hours a day, 7 days a week, irrespective of weather and seasons. As well as reducing greenhouse gas emissions, nuclear generation helps reduce air pollution

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The energy carried by sound waves is called -
goldenfox [79]

Answer:

kinetic mechanical energy

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3 years ago
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1. How would you describe the area in which you live? Use terms such as desert, prairie, or
salantis [7]

Answer:

forest rual area

Explanation:

because thats where i live

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3 years ago
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Let w(x)=3x-7.If w(x)=14, find x
dlinn [17]

Answer:

7

Explanation:

We are given:

    w(x) = 3x   -   7

     w(x)  = 14

The problem here entails us to solve for x;

To solve for x; equate the two expressions:

         

          3x  - 7  = 14

           3x  = 14 + 7

           3x  = 21  

             x = 7

So the value of x  = 7

6 0
3 years ago
What is the value of work done on an object when a 70 newton force moves it 9.0 meters in the same direction as the force
olchik [2.2K]
<span>Work, very simply, equals force times distance (when the force and distance are in the same direction. otherwise you get a little bit of trig added on) \[W=F*\Delta x\] W=70N * 9.0 m = 630 Nm = 630 J</span>
3 0
3 years ago
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A thin ring of radius 73 cm carries a positive charge of 610 nC uniformly distributed over it. A point charge q is placed at the
kow [346]

Answer:

q = - 93.334 nC

Explanation:

GIVEN DATA:

Radius of ring  73 cm

charge on ring 610 nC

ELECTRIC FIELD p FROM CENTRE IS AT 70 CM

E  =  2000 N/C

Electric field due tor ring is guiven as

E = \frac{KQx}{[x^2+ R^2]^{3/2}}

E = \frac{9\time 10^9 \times 610\times 10^[-9} 0.70}{(0.70^2 + 0.73^2)^{3/2}}

E1 = 3714.672 N/C

electric field due to point charge q

E  =\frac[kq}{x^2}

E = \frac{9\times 10^9 \times q}{0.70^2}

E2 = 1.837\times 10^{10}\times q

now the eelctric charge at point P is

E = E1 + E22000 =  3714.672 + 1.837\times 10[10} \times q

solving for q

q = - 93.334 nC

7 0
4 years ago
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