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3 years ago

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An 8.00 kg point mass and a 15.0 kg point mass are held in place 50.0 cm apart. A particle of mass m is released from a point be

tween the two masses 20.0 cm from the 8.00 kg mass along the line connecting the two fixed masses.
Find the magnitude and direction of the acceleration of the particle. Answer from book: 2.2 x 10^-9 m/s^2 - toward the 8.00 kg mass....

1 answer:

Verdich [7]3 years ago

3 0

Answer: $a=2.23\times 10^{-9} m/s^2$

Explanation:

Given

Mass of first Point mass $m_1=8 kg$

Mass of second Point $m_2=15 kg$

distance between them $d=50 cm$

third point mass $m_3=m$

Distance between $m\ and\ m_1 is\ 20 cm$

Distance between $m\ and\ m_2 is 30 cm$

Force Due to $m_1\ and\ m F_1=\frac{Gmm_1}{d_1^2}$

$F_1=\frac{8Gm}{(0.2)^2}$

$F_1=200 mG$

$F_2=m\frac{Gmm_2}{d_2^2}$

$F_1=m\frac{15Gm}{(0.3)^2}$

$F_2=166.67 mG$

Net Force

$F_{net}=F_1-F_2$

$=200 mG-166.67 mG$

$=33.33 mG$

$F_{net}=222.33\times 10^{-11} N$

$F_{net}=2.23m\times 10^{-9} N$

acceleration $a=\frac{2.23m\times 10^{-9}}{m}$

$a=2.23\times 10^{-9} m/s^2$

towards 8 kg mass

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Answer:

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Distance = 13.98 * 10^12 m

Explanation:

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work done = ΔK.E ( change in Kinetic energy ) ---- ( 1 )

<em>where :</em>

work done = p * t

= 15 * 10^6 watts * ( 1 year ) = 473040000 * 10^6 J

( note : convert 1 year to seconds )

and ΔK.E = 1/2 mVf^2 given ; m = 1200 kg and initial V = 0

<u>back to equation 1 </u>

473040000 * 10^6 = 1/2 mv^2

Vf^2 = 2(473040000 * 10^6 ) / 1200

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<u>i) Determine how fast the rocket is ( acceleration of the rocket )</u>

a = Vf / t

= 887918.92 / ( 1 year )

= 0.0282 m/s^2

<u>ii) determine distance travelled by rocket </u>

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3 years ago

)If a force of 5.00 N is needed to open a 90.0 cm wide door when applied at the edge opposite the hinges, what force must be app

masha68 [24]

Answer:

A force of 12.857 newtons must be applied to open the door.

Explanation:

In this case, a force is exerted on the door, a moment is performed and the door is opened. If moment remains constant, the force is inversely proportional to distance respect to axis of rotation passing through hinges. That is:

$F \propto \frac{1}{r}$

$F = \frac{k}{r}$ (Eq. 1)

Where:

$F$ - Force, measured in newtons.

$k$ - Proportionality ratio, measured in newton-meters.

$r$ - Distance respect to axis of rotation passing through hinges, measured in meters.

From (Eq. 1) we get the following relationship and clear the final force within:

$F_{A}\cdot r_{A} = F_{B}\cdot r_{B}$

$F_{B}=\left(\frac{r_{A}}{r_{B}} \right)\cdot F_{A}$ (Eq. 2)

Where:

$F_{A}$ , $F_{B}$ - Initial and final forces, measured in newtons.

$r_{A}$ , $r_{B}$ - Initial and final distances, measured in meters.

If we know that $F_{A} = 5\,N$ , $r_{A} = 0.9\,m$ and $r_{B} = 0.35\,m$ , then final force is:

$F_{B}= \left(\frac{0.9\,m}{0.35\,m} \right)\cdot (5\,N)$

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3 years ago

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Answer:

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Answer:

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a.

Using first law of thermodynamics:

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b.

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ANSWER:

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