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igomit [66]
3 years ago
11

I think I know the answer but I am not sure

Physics
2 answers:
Romashka [77]3 years ago
5 0
The answer is a

Hope this helps!
Diano4ka-milaya [45]3 years ago
5 0
I wanna bang my head if i see a mathematics
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The earth has a vertical electric field at the surface, pointing down, that averages 100 N/C. This field is maintained by variou
zlopas [31]

Answer:

q = 3.6 10⁵  C

Explanation:

To solve this exercise, let's use one of the consequences of Gauss's law, that all the charge on a body can be considered at its center, therefore we calculate the electric field on the surface of a sphere with the radius of the Earth

          r = 6 , 37 106 m

          E = k q / r²

          q = E r² / k

          q = \frac{100 \ (6.37 \ 10^6)^2}{9 \ 10^9}

          q = 4.5 10⁵ C

Now let's calculate the charge on the planet with E = 222 N / c and radius

           r = 0.6 r_ Earth

           r = 0.6 6.37 10⁶ = 3.822 10⁶ m

           E = k q / r²

            q = E r² / k

            q = \frac{222 (3.822 \ 10^6)^2}{ 9 \ 10^9}

            q = 3.6 10⁵  C

4 0
3 years ago
The term necrosis means
aksik [14]

Necrosis - the death of most or all of the cells in an organ or tissue due to disease, injury, or failure of the blood supply.

Hope this helps, please mark me brainliest

6 0
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Can someone pls help me?!!!! i need it asap!! i’m very stressed
Mariulka [41]

Answer:

Acceleration

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Its speed or velocity change

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3 years ago
Objects with masses of 165 kg and a 465 kg are separated by 0.340 m. (a) Find the net gravitational force exerted by these objec
Dmitrij [34]

Answer:

(a) F_{net} = 4.19 x 10^{-5} N, and its direction is towards m_{2}.

(b) It must be placed inside a hollow shell.

Explanation:

Let, m_{1} = 165 kg, m_{2} = 465 kg, m_{3} = 60 kg, and the distance between m_{1} and m_{2} is 0.340 m.

(a) Since m_{3} is placed midway between m_{1} and m_{2}, then its distance to both masses is 0.170 m.

From the Newton's law of universal gravitation,

F = \frac{Gm_{1}m_{2}  }{r^{2} }

Where all variables have their usual meaning.

Then,

a. F_{net} = F_{23} - F_{13}

F_{13} = \frac{6.67*10^{-11}*165*60 }{(0.17)^{2} }

     = 2.25 x 10^{-5} N

F_{23} = \frac{6.67*10^{-11}*465*60 }{(0.17)^{2} }

     = 6.44 x 10^{-5} N

∴ F_{net} =  = 6.44 x 10^{-5} - 2.25 x 10^{-5}

              = 4.19 x 10^{-5} N

The net force exerted by the two masses on the 60 kg object is 4.19 x 10^{-5}  N.

(ii) /F_{net}/ = /F_{23}/ - /F_{13}/

              = 6.44 x 10^{-5} - 2.25 x 10^{-5}

              = 4.19 x 10^{-5} N

(iii) The direction of the net force is to the right i.e towards m_{2}.

(b) For the net force experienced by the 60 kg object to be zero, it must be placed inside a hollow shell.

7 0
3 years ago
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