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2 years ago

Wegut.

Physics

1 answer:

AVprozaik [17]2 years ago

6 0

We/Wm = ge/gm = 120N/1.2N

gm = ge/100 = 0.1 m/s^2

density = mass/volume = 3M/(4pir^3)

Re-arranging this equation, we get

M/r^2 = (4/3)×pi×(density)×r

From Newton's universal law of gravitation, the acceleration due to gravity on the moon gm is

gm = G(M/r^2) = G×(4/3)×pi×(density)×r

Solving for density, we get the expression

density = 3gm/(4×pi×G×r)

= 3(0.1)/(4×3.14×6.67×10^-11×2.74×10^6)

= 130.6 kg/m^3

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A capacitor is charged until it holds 5.0 j of energy. it is then connected across a 10-kω resistor. in 13.6 ms , the resistor d

prohojiy [21]

Answer:

The capacite is C=5.32 uF using the equations of voltage and energy in capacitance

Explanation:

The energy holds is 5 J and the resistor dissipates 2J so the energy total is 3J

Using:

$V_{t}= V_{o}e^{\frac{-t}{R*C} }$

Voltage in this case is the energy dissipated so

$E_{t}= E_{o}e^{\frac{-t}{R*C} }$

$\frac{\sqrt{E_t} }{\sqrt{E_o} } = e^{\frac{-t}{R*C} }$

$\frac{\sqrt{3 J} }{\sqrt{5J} } = e^{\frac{-13.6ms}{10kw*C} }$

Using the equation to find capacitance

$ln 0.775= e^{\frac{-13.6 x10^{3} }{10x10^{3}*C }} \\ln(0.775)= ln * e^{\frac{-13.6 x10^{3s} }{10x10^{3}*C }} \\\\ln(0.775)= {\frac{-13.6 x10^{3} }{10x10^{3}*C }} \\C= \frac{-13.6 x10^{-3} }{10x10x^{3}*ln(0.775) }$

$C= 5.32x10^{-6}$ F

C= 5.32 uF because u is the symbol for micro that is equal to $10^{-6}$

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Answer:

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algol13

Answer:

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Explanation:

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