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tensa zangetsu [6.8K]
3 years ago
13

How much potential energy does a 40-kg medicine ball gain when it is lifted 5 m? Use the formula: GPE= mass X gravity (9.8 m/s2)

X Height
Question 16 options:

A 20.4 J


B 100 J


C 1960 J


D 200 J
Physics
1 answer:
Ronch [10]3 years ago
4 0
The answer is c: <span>1960 J

</span>Potential Energy :
<span>PE = m x g x h = 40*9.8*5=1960
</span>
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1.3636

Explanation:

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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
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Answer:

Explanation:

Given that

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The normal force exerted at B is 0.69N

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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

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For point A

Fa=mv²/r

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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

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3 0
3 years ago
Some element can be either solid or liquid. At the melting point, the liquid has 8 × 10-22 J more enthalpy per atom than the sol
OlgaM077 [116]

Answer:

481.76 J/mol

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N_A = Avogadro's number = 6.022\times 10^{23}

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\Delta H=8\times 10^{-22}\times 6.022\times 10^{23}\\\Rightarrow \Delta H=481.76\ J/mol

Entropy is given by

\Delta S=6\times 10^{-24}\times 6.022\times 10^{23}\\\Rightarrow \Delta S=3.6132\ J/mol K

Latent heat of fusion is given by

L_f=\Delta H\\\Rightarrow L_f=481.76\ J/mol

The latent heat of fusion is 481.76 J/mol

Melting point is given by

T_m=\dfrac{L_f}{\Delta S}\\\Rightarrow T_m=\dfrac{8\times 10^{-22}\times 6.022\times 10^{23}}{6\times 10^{-24}\times 6.022\times 10^{23}}\\\Rightarrow T_f=133.33\ K

Melting occurs at 133.33 K

3 0
3 years ago
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