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3 years ago

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Two laptop computers each have a mass of 2 kg. If they are 0.5 m apart on a desk, what is the gravitational force between them?

G = 6.67 × 10-11 N⋅(m/kg)2

1 answer:

Rudik [331]3 years ago

4 0

Answer:

$F = 1.067 \times 10^{-9} N$

Explanation:

As we know that the gravitational force between two masses is given as

$F = \frac{Gm_1m_2}{r^2}$

here we know that

$m_1 = m_2 = 2 kg$

also we know that

$r = 0.5 m$

$G = 6.67 \times 10^{-11} Nm^2/kg^2$

so the force between two masses is given as

$F = \frac{(6.67 \times 10^{-11})(2)(2)}{0.5^2}$

$F = 1.067 \times 10^{-9} N$

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Answer:

A) since the density of the buoy is half the density of sea water when the buoy is at rest on an ocean half of the buoy will be submerged in water

B )angular frequency ( w ) = $\sqrt{\frac{3g}{h} }$

c ) h = 1.34 m ( 4 $\pi ^{2}$ / 3g )

Explanation:

A) since the density of the buoy is half the density of sea water when the buoy is at rest on an ocean half of the buoy will be submerged in water this is because the substances with lesser density floats when placed in a substance with a higher density

B ) when the Buoy is at rest ( t = 0 ) and is then pushed down calculate angular frequency ( w ) of small oscillations in terms of the given variables

volume of cone = hA /3.

h = height of cone, A = Base Area.

therefore the total volume of the Buoy above water level ( at rest )

= ( $\pi r^{2} * \frac{h}{3}$ ) * 2 = $\frac{\pi r^{2} h }{6}$

The part of the buoy immersed in water = x

then the net upward force will be

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note force = Mass * Acceleration

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therefore a = $\frac{3g}{h} * x$

angular frequency ( w ) = $\sqrt{\frac{3g}{h} }$

C ) height of the each cone

$\frac{2\pi }{w} = 1$ therefore w = 2 $\pi$

back to the angular frequency : $\frac{3g}{h} = 4\pi ^{2}$

therefore h = 1.34 m ( 4 $\pi ^{2}$ / 3g )

4 0

3 years ago

On a hot summer day in the state of Washington while kayaking, I saw several swimmers jump from a railroad bridge into the Snoho

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Answer: part a: 19.62m

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part a: 2.83 secs

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$s=ut+0.5at^2$

s=h

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Part b

$v=u+at\\v=0+9.81*2\\v=19.62m/s$

Part c

now we have h=2*19.62=39.24

$39.24=0+0.5*9.81*t^2\\t^2=8\\t=2.83 secs$

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An astronaut finds herself in a predicament in which she has become untethered from her shuttle. She figures that she could get

Blizzard [7]

In order to solve the problem, it is necessary to apply the concepts related to the conservation of momentum, especially when there is an impact or the throwing of an object.

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where,

m=Total mass

$m_O =$ Mass of Object

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$V_f =$ Final Velocity

$V_O =$ Velocity of Object

Our values are:

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Solving to find the final speed, after throwing the object we have

$V_f=\frac{mV_0+m_TV_O}{m-m_O}$

We have three objects. For each object a launch is made so the final mass (denominator) will begin to be subtracted successively. In addition, during each new launch the initial speed will be given for each object thrown again.

That way during each section the equations should be modified depending on the previous one, let's start:

A) $5.3Kg\rightarrow 15m/s$

$V_{f1}=\frac{mV_0+m_TV_O}{m-m_O}$

$V_{f1}=\frac{(98.7)*0+5.3*15}{98.7-5.3}$

$V_{f1}=0.8511m/s$

B) $7.9Kg\rightarrow 11.2m/s$

$V_{f2}=\frac{mV_{f1}+m_TV_O}{m-m_O}$

$V_{f2}=\frac{(98.7)(0.8511)+(7.9)(11.2)}{98.7-5.3-7.9}$

$V_{f2} = 2.0173m/s$

C) $10.5Kg\rightarrow 7m/s$

$V_{f3}=\frac{mV_{f2}+m_TV_O}{m-m_O}$

$V_{f3}=\frac{(98.7)(2.0173)+(10.5)(7)}{98.7-5.3-7.9-10.5}$

$V_{f3} = 3.63478m/s$

Therefore the final velocity of astronaut is 3.63m/s

7 0

3 years ago