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wariber [46]
3 years ago
10

Find the change in the force of gravity between two planets when the distance between them is reduced to one-tenth of the origin

al distance.

Physics
1 answer:
Fofino [41]3 years ago
7 0
The equation for gravitational force is provided. If you were to put 1/10 times d (distance) than 1/10 would be squared = 1/100. So the gravitational force would be 100 times more.

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You need to get to class, 300 meters away, and you arrive in 1 minute. What speed were you walking? (calculate in meters per sec
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5 meters per second. 300 / 60(seconds)
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Help
elena55 [62]

Answer:

I took the test, and the correct answer is not Hydrosphere and Geosphere, but instead, its actually Hydrosphere and Atmosphere

Explanation:

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3 years ago
Two children stretch a jump rope between them and send wave pulses back and forth on it. The rope is 2.6 m long, its mass is 0.5
Irina-Kira [14]

Answer:

15.13 m/s

Explanation:

The wave speed of the stretched rope can be calculated using the following formula

v = \sqrt{\frac{F_T}{\mu}}

where F_T = 44N is the tension on the rope and \mu = m/L = 0.5 / 2.6 = 0.1923 kg/m is the density of the rope per unit length

v = \sqrt{\frac{44}{0.1923}} = \sqrt{228.8} = 15.13 m/s

6 0
3 years ago
The distance from the dwarf planet Pluto to the sun is 4.43 × 109 km at perihelion and 7.37 × 109 km at aphelion. Find the eccen
kaheart [24]

Answer:

0.249

Explanation:

Perihelion = 4.43 x 10^9 km

Aphelion = 7.37 x 10^9 km

Let e be the eccentricity and a be the length of semi major axis.

The relation between the semi major axis, perihelion and aphelion s given by

Semi major axis = half of sum of perihelion and aphelion

a = \frac{4.43+7.37}{2}\times10^{9}

a = 5.9 x 10^9 km

The relation between the perihelion, semi major axis and the eccentricity is given by

Perihelion = a (1 - e)

4.43 x 10^9 = 5.9 x 10^9 (1 - e)

0.751 = 1 - e

e = 0.249

3 0
3 years ago
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