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3 years ago

The gravitational force between two volleyball players is 3.3x10^-7.

Physics

1 answer:

Pepsi [2]3 years ago

5 0

The distance of separation between the two masses is 0.927 m.

<h3>Gravitational force:</h3>

This is the force that exists between two masses in the universe.

To calculate the distance of separation of the masses, we use the formula below.

F = GMm/r².............. Equation 1

Where:

F = Gravitational force
m = First mass
M = Second mass
G = Universal constant
r = distance of seperation.

Make r the subject of the equation.

r = √(GMm/F)................... Equation 2

From the question,

Given:

F = 3.3×10⁻⁷ N
m = 61 kg
M = 75 kg
G = 6.69×10⁻¹¹ Nm²/kg²

Substitute these values into equation 2

r = √(61×75×6.69×10⁻¹¹)/(3.3×10⁻⁷)
r = 0.927 m

Hence, The distance of separation between the two masses is 0.927 m

Learn more about Gravitational force here: brainly.com/question/11359658

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Answer:

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Explanation:

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2 years ago

A silver wire 2.6 mm in diameter transfers a charge of 420 C in 80 min. Silver contains 5.8 x 10- free electrons per cubic meter

kifflom [539]

Answer:

a). 87.5 mA or $87.5 x10^{-3}$ A

b). 1.78 $\frac{m}{s}$

Explanation:

$d=2.6 mm \\Q=420C\\t=80min\\n=5.8x10^{28} \\q=1.6x10^{-19}$

n the number of free electrons is 28 in text reference and if they don't give q is take as the charge of electron.

a).

$I=\frac{Q}{t}\\ I= \frac{420 C}{80 min}*\frac{1min}{60 s} =\frac{420 C}{4800s}\\ I=87.5 x10^{-3}$ A

b).

$I=n*abs (q)*V_{d}*A$

$A= \pi * (\frac{d}{2})^{2} \\A=\pi (*\frac{2.6x10^{-3} m}{2})^{2} \\A=5.309x10^{-6}$

$V_{d} =\frac{I}{n*abs(q)*A} \\V_{d}=\frac{87.5 x10^{-2} }{5.8x^10{28} *1.6x^{-19} *5.3x^{6} }\\V_{d}=1.78 \frac{m}{s}$

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