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blondinia [14]
3 years ago
13

How could you increase the sleds acceleration

Physics
1 answer:
gladu [14]3 years ago
6 0
By putting this special transportation plastic on the bottom of the sled, because the transportation plastic is slick. that is what the transport bins slide around on. (p.s) the plastic is really expensive!
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The string of a bow and arrow is pulled tight.
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Answer:

elastic

Explanation:

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What is the formula for Ba+2 and F-?
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BaF2 here is the chemical compound

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As mass increases what happens to the kinetic energy
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How does a resistor affect the speed of the current that passes through it?
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A small block with mass 0.0400 kg slides in a vertical circle of radius 0.600 m on the inside of a circular track. During one of
Maurinko [17]

Answer:

Explanation:

Given that

The mass of the body is 0.04kg

M=0.04kg

The radius of the paths is 0.6m

r=0.6m

The normal force exerted at A is 3.9N

Fa=3.9N

The normal force exerted at B is 0.69N

Fb=0.69N

Then work done by friction from point A to B will be the change in K.E

W=∆K.E+P.E

So we need to know the velocity at both point A and B

Then since the centripetal force is given as

Ft=mv²/r

Then,

For point A

Fa=mv²/r

3.9=0.04v²/0.6

3.9=0.0667v²

v²=3.9/0.0667

v²=58.5

v=√58.5

v=7.65m/s

Va=7.65m/s

Now at point B

Fb=mv²/r

0.69=0.04v²/0.6

0.69=0.0667v²

v²=0.69/0.0667

v²=10.35

v=√10.35

v=3.22m/s

Vb=3.22m/s

Then, the work done is

W=∆K.E+P.E

P.E is given as mgh

The height will be 2R =1.2m

P.E=mgh

P.E=0.04×9.81×1.2

P.E=0.471J

Final kinetic energy at B minus initial kinetic energy at A

W=K.Eb-K.Ea

K.E is given as 1/2mv²

W=1/2m(Vb²-Va²) +P.E

W=0.5×0.04(3.22²-7.65²) +0.471

W=0.5×0.04×(-48.1541) +0.471

W=-0.96+0.471

W=-0.49J

work was done on the block by friction during the motion of the block from point A to point B is 0.49J.

Friction opposes motions and that is why the work done is negative

3 0
3 years ago
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