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finlep [7]
3 years ago
13

I need to know the answer to number 4

Physics
1 answer:
Alona [7]3 years ago
8 0
I learned this one in Middle School too

<em /><u>B</u><u />aby <u>G</u><u />odzilla <u>E</u>ats <u>H</u>otdogs, <u>N</u><u />ot <u>C</u><u />occa <u>P</u><u />uffs

<u>B</u>elize <u>G</u><u />uatemala <u>E</u><u />l <u /><u />Salvador <u>H</u><u />onduras <u>N</u><u />icaragua <u>C</u><u />osta R<u />ica <u>P</u><u />anama

I hope I helped

ΩΩΩΩΩΩΩΩΩΩ
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Elements with low specific heat values have which characteristic?
scZoUnD [109]
I believe the answer is High thermal energy.
If a substance has a low specific heat, it means it will heat or cool quickly than a high heat compound. 
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4 years ago
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In my trigonometry class, we were assigned a problem on Angular and Linear Velocity.
Rzqust [24]

1) 0.0011 rad/s

2) 7667 m/s

Explanation:

1)

The angular velocity of an object in circular motion is equal to the rate of change of its angular position. Mathematically:

\omega=\frac{\theta}{t}

where

\theta is the angular displacement of the object

t is the time elapsed

\omega is the angular velocity

In this problem, the Hubble telescope completes an entire orbit in 95 minutes. The angle covered in one entire orbit is

\theta=2\pi rad

And the time taken is

t=95 min \cdot 60 =5700 s

Therefore, the angular velocity of the telescope is

\omega=\frac{2\pi}{5700}=0.0011 rad/s

2)

For an object in circular motion, the relationship between angular velocity and linear velocity is given by the equation

v=\omega r

where

v is the linear velocity

\omega is the angular velocity

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In this problem:

\omega=0.0011 rad/s is the angular velocity of the Hubble telescope

The telescope is at an altitude of

h = 600 km

over the Earth's surface, which has a radius of

R = 6370 km

So the actual radius of the Hubble's orbit is

r=R+h=6370+600=6970 km = 6.97\cdot 10^6 m

Therefore, the linear velocity of the telescope is:

v=\omega r=(0.0011)(6.97\cdot 10^6)=7667 m/s

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3 years ago
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Novay_Z [31]

Answer:

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Explanation:

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4 years ago
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Bad White [126]

Answer:

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3 years ago
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Compared to a soft sound, a loud sound at the same pitch....(check all true statements)
Sladkaya [172]

Answer:

One might think of a plucked guitar string - the sound would depend on the original amplitude of the disturbance -

Speed and velocity would still be the same

(b) is correct because the energy transfer depends on the original energy applied.

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