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3 years ago

If two twins (54 kg each) were 0.02 m apart, what is the force of gravity between them?

Physics

1 answer:

Lelechka [254]3 years ago

3 0

Answer:

Force, $F=4.86\times 10^{-4}\ N$

Explanation:

We have,

Masses of two twins are 54 kg each

They are placed at a distance of 0.02 m

It is required to find the force of gravity between them. The formula used to find the gravitational force between masses is given by :

$F=G\dfrac{m_1m_2}{r^2}$

plugging all the known values:

$F=6.67\times 10^{-11}\times \dfrac{54^2}{(0.02)^2}\\\\F=4.86\times 10^{-4}\ N$

So, the force of gravity between them is $4.86\times 10^{-4}\ N$ .

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Explanation:

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3 years ago

A 1-kg collar (located at point (2,2) from the origin) is pulled along a vertical, frictionless bar with a force of 10 N applied

faltersainse [42]

Answer:

The acceleration of the collar is 10 m/s²

Explanation:

Given;

mass of the collar, m = 1 kg

applied force on the bar, F = 10 N

The acceleration of the collar can be calculated by applying Newton's second law of motion;

F = ma

where;

F is the applied force

m is mass of the object

a is the acceleration

a = F / m

a = 10 / 1

a = 10 m/s²

Therefore, the acceleration of the collar is 10 m/s²

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2 years ago

A 66-foot-tall woman walks at 55 ft/s toward a street light that is 2424 ft above the ground. What is the rate of change of th

I am Lyosha [343]

Answer:

a. $\frac{dx}{dt}=20ft/s$

b. $\frac{d(x+L)}{dt}==25ft/s$

Explanation:

Using the triangle theorem both triangle the woman makes between the light so the rate of change of length can use geometry first

$tan(\beta)=\frac{24ft}{L+x}=\frac{6ft}{x}$

Solve to find the rate relation

$x=\frac{24}{6}*L$

$x=4*L$

Now the rate of the change rate

$\frac{dx}{dt}=4*\frac{dL}{dt}$

$\frac{dx}{dt}=4*5ft/s=20ft/s$

Finally the rate of her shadow moving

$\frac{d(x+L)}{dt}=\frac{dx}{dt}+\frac{dL}{dt}$

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2 years ago

Given the picture below, what is the volume of the object?

Valentin [98]

The answer is D, 10 ml

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3 years ago

Ten high-technology batteries are tested for 200 hours each. One failed at 20 hours; another failed at 140 hours; all others com

Bas_tet [7]

Answer:

Failure rate = 20%

MTBF = 880 hours

Explanation:

given data

batteries = 10

tested = 200 hours

one failed = 20 hours

another fail at = 140 hours

solution

we know that Mean Time between Failures is express as = (Total up time) ÷ (number of breakdowns) ....................1

so here Total up time will be

Total up time = 200 × 10

Total up time = 2000

and here

Number of breakdown = 1 at 20 hour and another at 140 hour = 2

so it will be = (Total up time) ÷ (number of breakdowns) .......2

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so here gap between occurrences is

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and

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MTBF = 1000 - 120

MTBF = 880 hours

and

Failure rate (FR) will be

Failure rate (FR) = 1 ÷ MTBF ................3

Failure rate (FR) = R÷T ......................4

as here R is the number of failures and T is total time

so Failure rate (FR) = 20%

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3 years ago