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Contact [7]
3 years ago
8

Calculate the value of two equal charges if they repel one another with a force of 10 N having 10 m distance apart in a vacuum.

What would be the size of the charges if they were situated in an insulting liquid whose permittivity was ten times that of a vacuum.
​
Physics
1 answer:
MAVERICK [17]3 years ago
4 0

Explanation:

a) From Coulombs law,

F = (kq1q2)/r²...............1

F = 10 N

r = 10m

k = 9 × 10^9 m/F

q1 = q2

from equation 1, make q the subject

q = (Fr²)/2k

q = (10×10²)/(2×9×10^9)

q = 1000/1.8×10^10

q = 5.556×10^-8 C

b) K = k/ko

k = Kko = 10times permittivity of free-space

Goodluck

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Then  acceleration a = F/m

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An instrument is thrown upward with a speed of 15 m/s on the surface of planet X where the acceleration due to gravity is 2.5 m/
Katen [24]
<h2>Answer: 12 s</h2>

Explanation:

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In this sense, the main movement equation in the Y axis is:

y-y_{o}=V_{o}.t-\frac{1}{2}g.t^{2}    (1)

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y  is the instrument's final position  

y_{o}=0  is the instrument's initial position

V_{o}=15m/s is the instrument's initial velocity

t is the time the parabolic movement lasts

g=2.5\frac{m}{s^{2}}  is the acceleration due to gravity at the surface of planet X.

As we know y_{o}=0  and y=0 when the object hits the ground, equation (1) is rewritten as:

0=V_{o}.t-\frac{1}{2}g.t^{2}    (2)

Finding t:

0=t(V_{o}-\frac{1}{2}g.t^{2})   (3)

t=\frac{2V_{o}}{g}   (4)

t=\frac{2(15m/s)}{2.5\frac{m}{s^{2}}}   (5)

Finally:

t=12s

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