4 years ago

The answer to the question

1 answer:

4 0

A geostationary orbit can be achieved only at an altitude very close to 35,786 km (22,236 mi) and directly above the equator. This equates to an orbital velocity of 3.07 km/s (1.91 mi/s) and an orbital period of 1,436 minutes, which equates to almost exactly one sidereal day (23.934461223 hours).

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What is the electric potential at a distance of 1.2 m from a 7.5 UC point charge?

grandymaker [24]

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A force of 400-N pushes on a 25-kg box horizontally. The box accelerates at 9 m/s? Find the coefficient of kinetic friction betw

umka2103 [35]

Answer:

Explanation:

Using the Newtons law of motion;

$\sum Fx = ma_x\\Fm - Ff = ma_x$

Fm is the moving force = 400N

Ff is the frictional force = μR

μ is the coefficient of kinetic friction

R is the reaction = mg

m is the mass

a is the acceleration

The equation becomes;

$Fm - \mu R = ma_x\\Fm - \mu mg = ma_x\\400- \mu (25)(9.8) = 25(9)\\400 - 254.8 \mu = 225\\- 254.8 \mu = 225 - 400\\- 254.8 \mu = -175\\ \mu = \frac{-175}{- 254.8} \\\mu = 0.69$

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A car travels 90 meters due north in 15 seconds. Then the car

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I got you kid It’s A- 2.5m/sb

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