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Nimfa-mama [501]
4 years ago
9

How does centripetal force act on a satellite in orbit? It changes the direction of the satellite. It keeps the satellite spinni

ng. It exerts an unbalanced force on the satellite.
Physics
2 answers:
sladkih [1.3K]4 years ago
6 0

Answer:

The force of gravity in keeping an object in circular motion is an example of centripetal force. Since it acts always perpendicular to the motion, gravity does not do work on the orbiting object if it is in a circular orbit.

Explanation:

enyata [817]4 years ago
3 0

gravity keeps the saellite in place

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A horizontal 52-n force is needed to slide a 50-kg box across a flat surface at a constant velocity of 3.5 m/s. What is the coef
3241004551 [841]

Answer: 0.61

Explanation:

This is calculation based on friction.

Since the box rests on a flat surface, the force that exists between them is known as frictional force.

Since the friction is dynamic (velocity is not zero)

The frictional force = kinetic energy gained by the body.

Ff = 1/2mv^2

coefficient of kinetic friction × normal reaction = 1/2mv^2

Since normal reaction is equal to the weight(force acting along the vertical component)

Normal reaction= mg = 50 × 10 = 500N. Therefore,

coefficient of kinetic friction × 500 = 1/2×50×3.5^2

coefficient of kinetic friction = 50×3.5^2/1000

coefficient of kinetic friction= 0.61

6 0
3 years ago
What would happen if you charged a balloon by
Luda [366]

When you rub a balloon against your hair or clothing, electrons that were previously on the hair/clothing will "jump" onto the balloon. Therefore, the balloon now has a negative charge accumulated on its surface.

When you bring that balloon near another balloon with a neutral charge, they will stick to each other, because the electrons on the surface will be attracted to the positive charges on the other. The positive charges that were previously randomly oriented now line up at the surface. However, after some time, the electrons move around back to their former random positions.

8 0
3 years ago
A 50-cm-long spring is suspended from the ceiling. A 330 g mass is connected to the end and held at rest with the spring unstret
lukranit [14]

Explanation:

Given that,

Length of the spring, l = 50 cm = 0.5 m

Mass connected to the end, m = 330 g = 0.33 kg

The mass is released and falls, stretching the spring by 30 cm before coming to rest at its lowest point. On applying Newton's second law, 10 cm below the release point, x = 15 cm

(a) When the mass is connected, the force of gravity is balanced by the force in spring.

kx=mg\\\\k=\dfrac{mg}{x}\\\\k=\dfrac{0.33\times 9.8}{0.15}\\\\k=21.56\ N/m

(b) The amplitude of the oscillation will be 15 cm as it is half of the total distance travelled.

(c) The frequency of the oscillation is given by :

f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}} \\\\f=\dfrac{1}{2\pi}\sqrt{\dfrac{21.56}{0.33}} \\\\f=1.28\ Hz

Hence, this is the required solution.

7 0
4 years ago
Why does the earth stay in orbit around the sun instead of drifting away from it into space?
pychu [463]

Answer:

b it is b part answer i think so

3 0
2 years ago
A man walks 3.50 mi due east, then turns and walks 2.57 mi due north. How far
BaLLatris [955]

Answer:

4.34 mi at 36.3^{\circ} north of east

Explanation:

The displacement of an object in motion is a vector connecting its initial position to the final position of motion.

In this problem, the man has 2 different motions:

- 3.50 mi due east

- 2.57 mi due north

We can take the east direction as positive x-direction and north as positive y-direction, so these two motions can be written as:

x=+3.50 mi

y=+2.57 mi

Since the two motions are perpendicular to each other, the resultant displacement can be found by using Pythagorean's theorem; therefore:

d=\sqrt{x^2+y^2}=\sqrt{3.50^2+2.57^2}=4.34 mi

We can also find the direction using the equation:

tan \theta = \frac{y}{x}

And therefore,

\theta=tan^{-1}(\frac{y}{x})=tan^{-1}(\frac{2.57}{3.50})=36.3^{\circ}

7 0
4 years ago
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