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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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Answer: The students will determine the two fixed points of the thermometer:

Lower fixed point = 0 degree Celsius

Upper fixed point = 100 degree Celsius

Then divide the thermometer with equal intervals

The room temperature will be the point at which the themometric substance remains constant when rising from ice point.

Explanation:

Apparatus available:

Unmarked thermometer

250 cm3 glass beaker

crushed ice 

water

heatproof mat 

clamp, boss and stand

meter rule

Added apparatus

Bunsen burner

Stirrer

Method

The students will determine the two fixed points of the thermometer:

Lower fixed point = 0 degree Celsius

Upper fixed point = 100 degree Celsius

Then divide the thermometer with equal intervals

Procedures

Set up the apparatus of illustrated in the attached figure.

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