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Feliz [49]
3 years ago
12

If only 10 pounds is required to lift a 500-lb block, how much chain must be played out to lift the engine 3.0 inches?

Physics
1 answer:
adelina 88 [10]3 years ago
3 0

Answer:

150 inches (12.5 ft)

Explanation:

The work done to lift the 500 pound block 3 inches should be the same as that to lift the 10 pond object a given distance.

The following is the equation one needs to solve:

10 \,lb\,* \,x\,=\,500\,lb\,*\,3\,in\\10 \,lb\,* \,x\,=\,1500\,lb\,in\\

therefore solving for the distance "x" gives as the answer (in inches):

10 \,lb\,* \,x\,=\,1500\,lb\,in\\x\,=\,\frac{1500\,lb\,in}{10\,lb} \\\\x\,=150\,in

which can also be given in feet as: 12.5 ft

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The final volume of the gas is 144.25 L

Explanation:

For an ideal gas kept at constant pressure, the work done by the gas on the surroundings is given by

W=p\Delta V = p(V_f - V_i)

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For the gas in the cylinder in this problem,

p = 2.00 atm

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And we also know the work done,

W = 288 J

So we can solve the equation for V_f, the final volume:

V_f = V_i + \frac{W}{p}=0.250 + \frac{288}{2.00}=144.25 L

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A 8.34 × 103-kg lunar landing craft is about to touch down on the surface of the moon, where the acceleration due to gravity is
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Answer:

21870.3156 N

Explanation:

u = Initial velocity

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s = Displacement

a = Acceleration

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v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-18.2^2}{2\times 162}\\\Rightarrow a=-1.02234\ m/s^2

The acceleration of the craft should be 1.02234 m/s²

F=ma\\\Rightarrow F=8.34\times 10^3\times 1.02234\\\Rightarrow F=8526.3156\ N

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W=mg\\\Rightarrow W=8.34\times 10^3\times 1.6\\\Rightarrow W=13344\ N

Thrust

F_t=F+W\\\Rightarrow F_t=8526.3156+13344\\\Rightarrow F_t=21870.3156\ N

The thrust needed to reduce the velocity to zero at the instant when the craft touches the lunar surface is 21870.3156 N

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