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choli [55]
3 years ago
13

Two electrons in a vacuum exert force of F = 3.8E-09 N on each other. They are then moved such that they are separated by x = 8.

2 times their original distance. What is the force that the electrons experience at the new distance, F_n, in newtons? Numeric: A numeric value is expected and not an expression. F_n = ______________________________ How far apart were the electrons originally in meters? Numeric: A numeric value is expected and not an expression. d = _________________________________
Physics
1 answer:
iren [92.7K]3 years ago
8 0

Answer:

F_n = 5.65E-11 N

d =  1.20682E-31 m

Explanation:

F = 3.8E-09 N

where

m = Mass of electron = 9.109E−31 kilograms

G = Gravitational constant = 6.67E-11 m³/kgs²

x = Distance between them

F=G\frac{m^2}{x^2}\\\Rightarrow 3.8E-09=G\frac{m^2}{x^2}

For F_n

F_n=G\frac{m^2}{x^2}\\\Rightarrow F_n=G\frac{m^2}{(8.2x)^2}\\\Rightarrow F_n=G\frac{m^2}{67.24x^2}

Dividing the above equations we get

\frac{F}{F_n}=\frac{G\frac{m^2}{x^2}}{G\frac{m^2}{67.24x^2}}\\\Rightarrow \frac{F}{F_n}=67.24\\\Rightarrow F_n=\frac{F}{67.24}\\\Rightarrow F_n=\frac{3.8E-09}{67.24}\\\Rightarrow F_n=5.65E-11\ N

F_n = 5.65E-11 N

F=G\frac{m^2}{x^2}\\\Rightarrow x=\sqrt{\frac{Gm^2}{F}}\\\Rightarrow x=\sqrt{\frac{G}{F}}m\\\Rightarrow x=\sqrt{\frac{6.67E-11}{3.8E-09}}9.109E-31\\\Rightarrow x=1.20682E-31\ m

d =  1.20682E-31 m

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2 years ago
A 6.00-mH solenoid is connected in series with a 5.0-μF capacitor and an AC source. The solenoid has internal resistance 3.0 Ω w
son4ous [18]

Answer:

5773.50269 Hz

23 A

Explanation:

L = Inductance = 6 mH

C = Capacitance = 5 μF

R = Resistance = 3 Ω

\epsilon = Maximum emf = 69 V

Resonant angular frequency is given by

\omega=\dfrac{1}{\sqrt{LC}}\\\Rightarrow \omega=\dfrac{1}{\sqrt{6\times 10^{-3}\times 5\times 10^{-6}}}\\\Rightarrow \omega=5773.50269\ Hz

The resonant angular frequency is 5773.50269 Hz

Current is given by

I=\dfrac{\epsilon}{R}\\\Rightarrow I=\dfrac{69}{3}\\\Rightarrow I=23\ A

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3 years ago
A cat, walking along the window ledge of a new york apartment, knocks off a flower pot, which falls to the street 280 feet below
Harlamova29_29 [7]
H = 280 ft, the height of the flower pot.
g = 32 ft/s²

Neglect air resistance.
Note that 1 ft/s = 15/22 mi/h

The initial vertical velocity is zero.
Let v =  the velocity with which the flower pot hits the ground.
Then
v² = 2gh
    = 2*(32 ft/s²)*(280 ft)
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v = 133.866 ft/s

Also,
v = (133.866 ft/s)*(15/22 (mi/h)/(ft/s)) = 91.272 mi/h

Answer:  133.9 ft/s or 91.3 mi/h

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3 years ago
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