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salantis [7]
3 years ago
10

What is the difference between a bound orbit and an unbound orbit around the sun?

Physics
1 answer:
Anit [1.1K]3 years ago
8 0

Answer:

Explanation:

The difference between a bound orbit and an unbound orbit around the sun is that:

An object on a bound orbit pursues the same way around the Sun again and again, while an object on an unbound orbit moves toward the Sun only a single time and afterward stays away forever &  never returns.

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What are two things that turbines have been used for?
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2 thing it is used for are commonly like wind turbines and water turbines.
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The process of converting a liquid to a gas is known as _____.
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This is called vaporization.




hope this helps :)

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A toy rocket, launched from the ground, rises vertically with an acceleration of 28 m/s 2 for 9.7 s until its motor stops. Disre
vredina [299]

Answer:

5080.86m

Explanation:

We will divide the problem in parts 1 and 2, and write the equation of accelerated motion with those numbers, taking the upwards direction as positive. For the first part, we have:

y_1=y_{01}+v_{01}t+\frac{a_1t^2}{2}

v_1=v_{01}+a_1t

We must consider that it's launched from the ground (y_{01}=0m) and from rest (v_{01}=0m/s), with an upwards acceleration a_{1}=28m/s^2 that lasts a time t=9.7s.

We calculate then the height achieved in part 1:

y_1=(0m)+(0m/s)t+\frac{(28m/s^2)(9.7s)^2}{2}=1317.26m

And the velocity achieved in part 1:

v_1=(0m/s)+(28m/s^2)(9.7s)=271.6m/s

We do the same for part 2, but now we must consider that the initial height is the one achieved in part 1 (y_{02}=1317.26m) and its initial velocity is the one achieved in part 1 (v_{02}=271.6m/s), now in free fall, which means with a downwards acceleration a_{2}=-9,8m/s^2. For the data we have it's faster to use the formula v_f^2=v_0^2+2ad, where d will be the displacement, or difference between maximum height and starting height of part 2, and the final velocity at maximum height we know must be 0m/s, so we have:

v_{02}^2+2a_2(y_2-y_{02})=v_2^2=0m/s

Then, to get y_2, we do:

2a_2(y_2-y_{02})=-v_{02}^2

y_2-y_{02}=-\frac{v_{02}^2}{2a_2}

y_2=y_{02}-\frac{v_{02}^2}{2a_2}

And we substitute the values:

y_2=y_{02}-\frac{v_{02}^2}{2a_2}=(1317.26m)-\frac{(271.6m/s)^2}{2(-9.8m/s^2)}=5080.86m

3 0
3 years ago
Which tem decorbes the perceived frequency of a sound wave?
JulsSmile [24]

Pitch

Explanation:

The pitch of a sound wave is the perceived frequency of a sound wave.

The quality of sound that makes it discernible to the ear is the pitch.

  • Sound wave is a longitudinal wave that is transmitted by series of rarefaction and compression.
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learn more:

Pitch brainly.com/question/9772227

Sound wave brainly.com/question/2845448

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An example of constant velocity
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Some examples of constant velocity (or at least almost- constant velocity) motion include (among many others): • A car traveling at constant speed without changing direction. A hockey puck sliding across ice. A space probe that is drifting through interstellar space.
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