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2 years ago

How far from the surface of Earth is the magnitude of Earth's gravitational field equal to 7.86 N/kg?

Physics

1 answer:

AfilCa [17]2 years ago

4 0

Answer:

7.48 x 10⁵ m

Explanation:

g = 7.86 N/kg

M = 5.97 x 10²⁴ kg, R = 6.37 x 10⁶ m.

Find height h

g = GM/(R + h)²

(R + h)² = GM/g = 6.67 x 10⁺¹¹ x 5.97 x 10²⁴ /7.86 = 5.066 x 10¹³

R + h = 7.12 x 10⁶ m

h = 7.12 x 10⁶ - 6.37 x 10⁶ = 7.48 x 10⁵ m

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A round pipe of varying diameter carries petroleum from a wellhead to a refinery. At the wellhead, the pipe's diameter is 50.9 c

seraphim [82]

Answer:

Flow rate 2.34 m3/s

Diameter 0.754 m

Explanation:

Assuming steady flow, the volume flow rate along the pipe will always be constant, and equals to the product of flow speed and cross-section area.

The area at the well head is

$A = \pi r_w^2 = \pi (0.509/2)^2 = 0.203 m^2$

So the volume flow rate along the pipe is

$\dot{V} = Av = 0.203 * 11.5 = 2.34 m^3/s$

We can use the similar logic to find the cross-section area at the refinery

$A_r = \dot{V}/v_r = 2.34 / 5.25 = 0.446 m^2$

The radius of the pipe at the refinery is:

$A_r = \pi r^2$

$r^2 =A_r/\pi = 0.446/\pi = 0.141$

$r = \sqrt{0.141} = 0.377m$

So the diameter is twice the radius = 0.38*2 = 0.754m

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Sladkaya [172]

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600 watts may be your answer:)

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sasho [114]

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