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yan [13]
4 years ago
7

A if two such generic humans each carried 2.5 coulomb of excess charge, one positive and one negative, how far apart would they

have to be for the electric attraction between them to equal their 700 n weight?
Physics
1 answer:
dolphi86 [110]4 years ago
3 0

Using Coulomb's Law <span>

F = {k(Q1)(Q2)}/r² 

Where F =force in Newtons, k = Coulomb constant, Q1 = - Q2= charge of the objects in Coulumbs, and r is the distance between the objects. 

Solving for r: 

r = sqrt[{k(Q1)(Q2)}/F] 

where k = 8.897 x 10^9 N•m²/C², Q1 = 2.5 C, Q2 = -2.5 C, and F = -700 N 

Note: Negative magnitude of Force indicates attraction. 

<span>r = 8,912.77 m</span></span>

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5 0
3 years ago
(a) Calculate the magnitude of the gravitational force exerted on a 445-kg satellite that is a distance of 1.77 earth radii from
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Answer:

a)1396.52 N

b)1396.52 N

c)a_{satellite}= 3.13 m/sec^2

d)a_{earth}=2.32\times10^{-22} m/s^2

Explanation:

The force experienced by the satelite is giveb by

F= \frac{Gm_{satellite}m_{earth}}{r^2} \\

m_{satellite}= 445 Kg

m_{earth}= 6×10^24 Kg

radius r= 1.77Re= 1.77×6.38×10^6 m

now putting values we get

F= \frac{6.67\times10^{-11}(445)(6\times10^24)}{(1.77\times6.38\times10^6)^2}

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a_{satellite}= \frac{F}{m_{satellite}}

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a_{earth}= \frac{F}{m_{earth}}

a_{earth}= \frac{1396.52}{(6\times10^24)}

a_{earth}=2.32\times10^{-22} m/s^2

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