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lilavasa [31]
3 years ago
6

(a) What is the escape speed on a spherical asteroid whose radius is 301 km and whose gravitational acceleration at the surface

is 0.412 m/s2
Physics
1 answer:
Allushta [10]3 years ago
7 0

Answer:

V.E=498.02m/s^2

Explanation:

From the question we are told that:

Radius r=301Km

Gravitational acceleration g=0.412 m/s^2

Generally the equation for Escape velocity is mathematically given by

 V.E^2=2gr

 V.E^2=2*0.412m/s^2*301000

 V.E^2=248024

 V.E=\sqrt{248024}

 V.E=498.02m/s^2

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A simple hydraulic lift is made by fitting a piston attached to a handle into a 3.0-cm diameter cylinder. The cylinder is connec
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Answer:

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By Pascal's Law, the pressure on the two pistons should be the same. Pressure is the size of normal force per unit area:

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For the pressures on the two pistons to match:

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F_1, A_1, and A_2 have all been found. The question is asking for F_2. Rearrange this equation to obtain:

\displaystyle F_2 = \frac{F_1}{A_1} \cdot A_2 = F_1 \cdot \frac{A_2}{A_1}.

Evaluate this expression to obtain the value of F_2, which represents the force on the piston with the larger diameter:

\begin{aligned}F_2 &= F_1 \cdot \frac{A_2}{A_1} \\ &= 495\; \rm N \times \frac{2.25\, \pi\; \rm cm^2}{144\, \pi \; \rm cm^2} \approx 3.1 \times 10^4\; \rm N\end{aligned}.

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3 years ago
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We can solve the problem by using Ohm's law.

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