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uranmaximum [27]
2 years ago
15

While observing a planet through a telescope, a scientist observes a natural satellite orbiting the planet. What is the best cla

ssification for this satellite?
Physics
2 answers:
olganol [36]2 years ago
6 0
Satellites are objects which orbit the planet. The natural satellite of Earth, in this case, is moon
mina [271]2 years ago
4 0

Answer;

The moon

The best classification for this satellite is the moon.

Explanation;

The moon is the second brightest object in the sky after the Sun.

It moves around the Sun at the same time it moves around the Earth. The Moon orbits the Earth because of the pull of Earth's gravity. The Moon orbits around the Earth once per month.

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Walking across a carpet is an example of charge being transferred by
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static electricity and friction

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A tennis player strikes a tennis ball from underneath with her racket. The ball is sent straight up with an initial velocity of
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So the acceleration of gravity is 9.8 m/s so that’s how quickly it will accelerate downwards. You can use a kinematic equation to determine your answer. We know that initial velocity was 19 m/s, final velocity must be 0 m/s because it’s at the very top, and the acceleration is -9.8 m/s. You can then use this equation:

Vf^2=Vo^2+2ax

Plugging in values:

361=19.6x

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Wave A has an amplitude of 2 and wave B has an amplitude of 2 as shown below. What will happen when the crest of wave A meets th
puteri [66]
Since the two waves have equal amplitudes, if the crest of one wave
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3 years ago
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Why is accuracy important in dodgeball?
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Accuracy and power are the only things that matter in dodge ball. If you are not accurate, then your target does not get hit.
8 0
3 years ago
A bullet is fired straight up from a gun with a
melamori03 [73]

Answer: 815.51 m

Explanation:

This situation is related to projectile motion or parabolic motion, in which the initial velocity of the bullet has only y-component, since it was fired straight up. In addition, we are dealing with constant acceleration (due gravity), therefore the following equations will be useful to solve this problem:

V=V_{o}+gt (1)

V^{2}=V_{o}^{2}+2gy (2)

Where:

V is the final velocity of the bullet

V_{o}=152 m/s is the initial velocity of the bullet

g=-9.8 m/s^{2} is the acceleration due gravity, always directed downwards

t=6.9 s is the time

y is the vertical position of the bullet at t=6.9 s

Let's begin by finding V from (1):

V=152 m/s-9.8 m/s^{2}(6.9 s) (3)

V=84.38 m/s (4)

Now we have to substitute (4) in (2):

(84.38 m/s)^{2}=(152 m/s)^{2}-2(9.8 m/s^{2})y (5)

Isolating y:

y=815.511 m This is the displacement  of the bullet after 6.9 s

8 0
3 years ago
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