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expeople1 [14]
3 years ago
13

The Hubble Space Telescope (HST) orbits 569 km above Earth’s surface. Given that Earth’s mass is 5.97 × 1024 kg and its radius i

s 6.38 × 106 m, what is HST’s tangential speed?
Physics
2 answers:
Sergio039 [100]3 years ago
6 0
<span>We can use an equation to find the gravitational force exerted on the HST. F = GMm / r^2 G is the gravitational constant M is the mass of the Earth m is the mass of the HST r is the distance to the center of the Earth This force F provides the centripetal force for the HST to move in a circle. The equation we use for circular motion is: F = mv^2 / r m is the mass of the HST v is the tangential speed r is the distance to the center of the Earth Now we can equate these two equations to find v. mv^2 / r = GMm / r^2 v^2 = GM / r v = sqrt{GM / r } v = sqrt{(6.67 x 10^{-11})(5.97 x 10^{24}) / 6,949,000 m} v = 7570 m/s which is equal to 7.570 km/s HST's tangential speed is 7570 m/s or 7.570 km/s</span>
xenn [34]3 years ago
6 0

Answer:

Just type 7,570

Explanation:

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<h2>a) Average velocity in first 4 seconds is 64 ft/s upward</h2><h2>b) Average velocity in second 4 seconds is 63.5 ft/s downward</h2>

Explanation:

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    Time = 4 - 0 = 4 s

    \texttt{Velocity = }\frac{256}{4}=64ft/s

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a) Given S(t) = 76 + 128t − 16t²

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    s(8) = 76 + 128 x 8 − 16 x 8² = 78 ft

    Displacement in 4 seconds = 78 - 332 = -254 ft

    Time = 4 - 0 = 4 s

    \texttt{Velocity = }\frac{-254}{4}=-63.5ft/s

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a ball is thrown striaght up in the air and then falls back to earth. if the downward fall takes 2.2s, how fast is the ball trav
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The velocity of the ball when it strikes the ground, given the data is 21.56 m/s

<h3>Data obtained from the question</h3>

From the question given above, the following data were obtained:

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The velocity of the ball when it strikes the ground can be obtained as illustrated below:

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brainly.com/question/22719691

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