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ycow [4]
3 years ago
11

When a "superball" is dropped, can it rebound to a hightgreater than its origional hight?

Physics
1 answer:
Serhud [2]3 years ago
5 0

Answer:

The superball cannot rebound to a higher height than its original height because that would go against the laws of energy conservation. For it to bounce higher than its original height, that means the potential energy acting on it is higher than the one it had been dropped with originally, and that energy must be an external one. However, it should be noted that if the superball is thrown downwards with an extra energy, then it can rebound higher than the original height.

Explanation:

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50 kg of water at 75o C is cooled to 25o C. How much heat was given off?
attashe74 [19]

Answer:

b the answer is b

Explanation:

b is the awnser because it cools after the heat on the water witch lets the steam out

8 0
2 years ago
What everyday things exist bc of science?
fredd [130]
Paper..pencils...tables, walls...everything that is man made science played a huge factor in it because the first science was when some hairy dude was like i wonder what happens when i do this...same thing with food and drinks..how do you think milk was made..some dude was like hmmm..i wonder..and i think you can figure out the rest..hope i helped
6 0
3 years ago
A hydrogen atom has its electron in the n = 6 level. the radius of the electron's orbit in the bohr model is 1.905 nm.
gayaneshka [121]
Are there any options or is it not multiple choice.
5 0
3 years ago
Suppose a 65 kg person stands at the edge of a 6.5 m diameter merry-go-round turntable that is mounted on frictionless bearings
Hatshy [7]

Answer:0.316 rad/s

Explanation:

Given

mass of Person m=65 kg

velocity of person v=3.8 m/s

diameter of turntable d=6.5 m

moment of Inertia of the table I_0=1850 kg-m^2

Moment of inertia of Person I

I=mr^2=65\times (\frac{6.5}{2})^2

I=686.56 kg-m^2

initial angular velocity \omega _1=\frac{v}{r}

\omega _1=\frac{3.8}{3.25}=1.17 rad/s

Conserving Angular momentum

I\omega _1=(I+I_0)\omega _2 , where \omega _2=final\ angular\ velocity

686.56\times 1.17=(686.56+1850)\times \omega _2

\omega _2=\frac{686.56}{2536.56}\times 1.17

\omega _2=0.316 rad/s

4 0
3 years ago
The first charged object is exerting a force on the second charged object. Is the second charged object necessarily exerting a f
mario62 [17]

Answer:

Explanation:

Of course because it's Newton's Law that if body A exerts force on body B, then body B will exert equal but opposite force on body A.

HAPPY LEARNING:)

6 0
3 years ago
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