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MaRussiya [10]
3 years ago
6

Two electrons are passing 20.0 mm apart. What is the electric repulsive force that they exert on each other

Physics
2 answers:
Pachacha [2.7K]3 years ago
8 0

Answer:

0.5766422350752*10^-24 N

Explanation:

Couloumb's law states that states that there is an electrical force acting on 2 static charges. The magnitude is directly proportional to the product of the 2 charges.

Strength of electrons : q1 = q2 = 1.602 x 10-19. C

Substitute and solve:

F = (9*10^9)(1.602 x 10-19)(1.602 x 10-19) / (0.02)^2

Done.

vladimir1956 [14]3 years ago
4 0
15.0 I’m pretty sure that’s the answer to your question
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Calculate the acceleration of gravity as a function of depth in the earth (assume it is a sphere). You may use an average densit
Ber [7]

Solution :

Acceleration due to gravity of the earth, g $=\frac{GM}{R^2}$

$g=\frac{G(4/3 \pi R^2 \rho)}{R^2}=G(4/3 \pi R \rho)$

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$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-1000) \times 5.5 \times 10^3\right)$

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  $=0.822 \times 10^{-2} \ km/s$

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Acceleration due to gravity at 2000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-2000) \times 5.5 \times 10^3\right)$

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Acceleration due to gravity at 3000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-3000) \times 5.5 \times 10^3\right)$

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Acceleration due to gravity at 4000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-4000) \times 5.5 \times 10^3\right)$

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3 0
3 years ago
At resonance, what is impedance of a series RLC circuit? less than R It depends on many other considerations, such as the values
denis23 [38]

Answer:

at resonance impedence is equal to resistance and quality factor is dependent on R L AND C all

Explanation:

we know that for series RLC circuit impedance is given by

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but we know that at resonance X_L=X_C  

putting  X_L=X_C in impedance formula , impedance will become

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now quality factor of series resonance is given by

Q=\frac{\omega L}{R}=\frac{1}{\omega CR}=\frac{1}{R}\sqrt{\frac{L}{C}}  so from given expression it is clear that quality factor depends on R L and C

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3 years ago
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ycow [4]
The strength of the electric and magnetic fields there is no physical "distance" of oscillation here. nothing is actually moving up and down if you draw light as a sinusoidal wave, the up and down motion is the strength of the EM fields cheers
4 0
3 years ago
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