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Alexeev081 [22]
3 years ago
8

In an astronomical sense, which is not considered an ice?

Physics
2 answers:
Novay_Z [31]3 years ago
8 0
In an astronomical sense, Hydrogen is not considered ice. Astronomy is the study of celestial bodies (e.g. sun, moon, stars, planets, etc.). In astronomical terms, Hydrogen is an element consisting of one electron and one proton. Hydrogen is the lightest of the elements and is the building block of the universe. Stars<span> form from massive clouds of </span>hydrogen<span> gas.</span>
sveticcg [70]3 years ago
6 0

Answer:

hydrogen

Explanation:

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5.879e+12

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Identify the energy transformations in the image below:
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3 years ago
A cat rides a merry-go-round while turning with uniform circular motion. At time t1 = 2.00 s, the cat's velocity is v with arrow
goldenfox [79]

Answer:

Part a)

a_c = 2.07 m/s^2

Part b)

a_{avg} = 1.32 m/s^2

Explanation:

As we know that it makes half revolution in given time interval

so we have

\frac{T}{2} = t_2 - t_1

\frac{T}{2} = 9 - 2

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now the angular speed is given as

\omega = \frac{2\pi}{T}

\omega = \frac{2\pi}{14}

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now linear speed is given as

v = \sqrt{2.30^2 + 4.00^2}

v = 4.61 m/s

now we have

v = R \omega

4.61 = R(0.448)

R = 10.3 m

Now centripetal acceleration is given as

a_c = \omega^2 R

a_c = 0.448^2 \times 10.3

a_c = 2.07 m/s^2

Part b)

Average acceleration of the cat is given as

a_{avg} = \frac{v_2 - v_1}{\Delta t}

a_{avg} = \frac{2v}{\Delta t}

a_{avg} = \frac{2(4.61)}{9 - 2}

a_{avg} = 1.32 m/s^2

7 0
4 years ago
A student has a displacement of 304 m north in 180 s. What was the student's average velocity?
DanielleElmas [232]

Answer:

v = 1.69 m/s

Explanation:

Given that,

Displacement of the student is 304 m due North and it takes 180 s.

We need to find the student's average velocity. Using formula of velocity.

Velocity = displacement/time

v=\dfrac{304\ m}{180\ s}\\\\v=1.69\ m/s

Hence, the student's average velocity is 1.69 m/s.

6 0
4 years ago
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