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kipiarov [429]
3 years ago
13

Please help me please help

Physics
1 answer:
Anna [14]3 years ago
5 0

Answer:

2.12x 10 -3

Explanation:

i dont really have explanation but like that's how it works

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A negatively charged balloon has 4 μC of charge. How many excess electrons are on this bal- loon? The elemental charge is 1.6 ×
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Data:

The charge of a body depends on the amount of electrons it gains or loses. Q = n * e, where "Q" is charge, "n" is the number of plus or minus electrons, and "e" is the fundamental charge of an electron 1,6 * 10 ^{-19}C<span>. To know if the body has gained or lost, we look at the signal of its charge, remembering that the electron is negative. The charge of the body is 4 μC (positive), so there is a lack of electrons! 

Q = 4 </span>μC → Q = 4*10^{-6}
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n = ?<span>

We have:
</span>Q = n*e
n =  \frac{Q}{e}
n =  \frac{4*10^{-6}}{1,6 * 10 ^{-19}}
n = 2,5*10^{-6-(-19)}
n = 2,5*10^{-6+19}
\boxed{n = 2,5*10^{13}electrons}
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The speed of sound in ice, water, and steam is shown. What best explains the speed of sound in different states of matter?
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I NEED HELP PLEASE, THANKS! :)
Zina [86]

Answer:

charge C = greatest net force

charge B = the smallest net force

ratio  = 9 : 1

Explanation:

we know that in Electrostatic Forces, when 2 charges are at same sign then they repel each other and if they are different signed charges then they attract each other

so as per Coulomb's formula of Electrostatic Forces

F = \frac{k\ q_1\ q_2}{r^2}     .....................1

and here k is 9 × 10^9 N.m²/c² and we consider each charge at distance d

so two charge force at A to B is

F1 = \frac{k\ q^2}{d^2}

and force between charges at A to C, at 2d distance

F1 = \frac{k\ q^2}{(2d)^2}  =  \frac{k\ q^2}{4d^2}

force between charges at A to D,  3d distance

F1 = \frac{k\ q^2}{(3d)^2}  = \frac{k\ q^2}{9d^2}  

so

Charge a It receives force to the left from b and c and to the right from d

so at a will be

F(a)  = -F1 - F2 + F3             ....................2

put here value

F(a) = -\frac{k\ Q^2}{d^2}-\frac{k\ Q^2}{4d^2}+\frac{k\ Q^2}{9d^2}

solve it

F(a) = \frac{k\ q^2}{d^2}(-1-\frac{1}{4}+\frac{1}{9})  

F(a) = -\frac{41}{36}\ F1   = 1.13 F1  

and

Charge b It  receives force to the right from a and d and to the left from c

F(b) = F1 - F1 + F2            ....................3

F(b)  =  \frac{k\ q^2}{d^2}-\frac{k\ q^2}{d^2}+\frac{k\ q^2}{4d^2}    

F(b)  = \frac{1}{4} \ F1    =  0.25 F1

and

Charge c It receives forces to the right from all charges.

F(c) = F2 + F 1 + F 1      ....................4

F(c) = \frac{k\ q^2}{4d^2}+\frac{k\ q^2}{d^2}+\frac{k\ q^2}{d^2}      

F(c) =  \frac{9}{4} \ F1   = 2.25 F1

and

Charge d It receives forces to the left from all charges

F(d) = - F3 - F2 -F 1      ....................5

F(d) = -\frac{k\ q^2}{9d^2}-\frac{k\ q^2}{4d^2}-\frac{k\ q^2}{d^2}  

so

F(d) = -\frac{49}{36} \ F1    = 1.36 F1

and

now we get here ratio of the greatest to the smallest net force that is

ratio = \frac{2.25}{0.25}

 ratio  = 9 : 1

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