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krek1111 [17]
1 year ago
10

A rocket lifts a payload upward from the surface of Earth. The radius of Earth is R, and the weight of the payload on the surfac

e of Earth is W. The force of Earth's gravity on the payload is W/2 when the rocket's distance from the center of the Earth is
A. R
B. sqrt(2)R
C. 2R
D. 2sqrt(2)R
E. 4R
Physics
1 answer:
topjm [15]1 year ago
4 0

Answer:

the correct answer is B      r =√2   R

Explanation:

To solve this exercise we will use the force of gravity, at two points when the rocket is on the platform and when it is at the desired height.

Rocket with cargo on the platform

              F_{g} = G \frac{ m M }{R^2} = W

where m is the mass of the rocket and the charge, M the mass of the earth and R the distance from the center of the earth

Rocket loaded at desired height

              F_{g} = G \frac{m M}{r^2 } = \frac{W}{2}

we write our two equations

        W = g m M / R²

        \frac{W}{2} = G m M / r²

we match to solve them

         Gm M / R² = 2 Gm M / r²

         1 / R² = 2 / r²

          r =√2   R

therefore the correct answer is B

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Answer:

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Explanation:

Water molecules are linked by hydrogen bonds that require a lot of heat (energy) to break, which is released when the temperature drops. That energy is called specific heat or thermal capacity (ĉ) when it is enough to change the temperature of 1g of the substance (in this case water) by 1°C. Water ĉ equals 1 cal/(g.°C).

Given that ĉ = Q / (m.ΔT),

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m= mass of the substance (unity: grams or g), and

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A semi is traveling down the highway at a velocity of v = 26 m/s. The driver observes a wreck ahead, locks his brakes, and begin
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Answer:

fcosθ + Fbcosθ  =Wtanθ

Explanation:

Consider the diagram shown in attachment

fx= fcosθ (fx: component of friction force in x-direction ; f: frictional force)

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