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alukav5142 [94]
1 year ago
13

Josie ran at an average speed of 16 m/s. if her mass 10 kg, what was her kinetic energy as she crosses her finish line?

Physics
2 answers:
levacccp [35]1 year ago
4 0

I have to be honest with you . . . It's not possible to determine the answer with only the information given in the question.

Kinetic energy = (1/2) (mass) (speed squared).

Josie's Kinetic energy as she crosses the finish line is

(1/2) (her mass) (her speed as she crosses the finish line)².

That's (5 kg) x (her speed as she crosses the finish line)² .

But read the question again.

WE DON'T know her speed as she crosses the finish line.

The question only tells us her average speed for the whole race.

We can be confident that her <u><em>Average</em></u> Kinetic Energy was

(5 kg) (16 m/s)²  =  1,280 Joules.

But we can't tell what her kinetic energy was as she crossed the finish line, beause we don't know what her speed was at that instant.

Helen [10]1 year ago
3 0

Answer: 1280 J

KE=\frac{1}{2}mv^{2}  \\\\KE= \frac{1}{2} (10 kg)(16 m/s)^2\\=1280 J

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Answer:

New pressure of the gas increases by 26.5% with respect to initial pressure, new volume decreases 27% with respect to initial volume and new temperature decreases 8% with respect to initial volume.

Explanation:

If we assume the gas is a perfect gas we can use the perfect gas equation:

PV=nRT

  • For Isothermal process:

\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}(1)

Where subscripts 1 shows before the isothermal process and 2 after it, because isothermal means constant temperature T1=T2, and pressure increases by 10% means P2=1,1*P1, using these facts on (1) we have:

V_{2}=\frac{V_{1}}{1.1} (2)

  • For Isobaric process:

\frac{P_{2}V_{2}}{T_{2}}=\frac{P_{3}V_{3}}{T_{3}} (3)

Where subscripts 2 shows before the isobaric process and 3 after it, because isobaric means constant pressure P2=P3, and volume decreases by 20% means V3=0.8*V2, using these facts on (3) we have:

T_{3}=0.8T_{2} (4)

  • For Isochoric process:

\frac{P_{3}V_{3}}{T_{3}}=\frac{P_{4}V_{4}}{T_{4}} (5)

Where subscripts 3 shows before the isochoric process and 4 after it, because isochoric means constant volume V3=V4, and temperature increases by 15% means T4=1.15*T3, using these facts on (5) we have:

P_{4}=1.15P_{3} (6)

So now because P4=1.15*P3, P2=P3 and P2=1.1*P1:

P_{4}=1.15*1.1P_{1}=1.265P1

This is, the new pressure of the gas increases by 26.5%  with respect to initial pressure.

Similarly, we have V3=V4, V3=0.8*V2 and V1=1,1*V2:

V_{4}=\frac{0.8}{1.1}V_{1}=0.72V1

so the final volume decreases 27% with respect to initial volume.

T4=1,15*T3, T3=0.8*T2 and T1=T2:

T_{4}=1.15*0.8T_{1}=0.92T1

The new temperature decreases 8% with respect to initial volume.

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The vaule of the g constant (the acceleration of all objects due to gravity)on earth is
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A proton traveling due west in a region that contains only a magnetic field experiences a vertically upward force (away from the
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Explanation:

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If you stood on a planet having a mass four times higher than Earth's mass, and a radius two times 70) lon longer than Earth's r
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CHECK COMPLETE QUESTION BELOW

you stood on a planet having a mass four times that of earth mass and a radius two times of earth radius , you would weigh?

A) four times more than you do on Earth.

B) two times less than you do on Earth.

C) the same as you do on Earth

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Answer:

OPTION C is correct

The same as you do on Earth

Explanation :

According to law of gravitation :

F=GMm/R^2......(a)

F= mg.....(b)

M= mass of earth

m = mass of the person

R = radius of the earth

From law of motion

Put equation b into equation a

mg=GMm/R^2

g=GMm/R^2

g=GM/R^2

We know from question a planet having a mass four times that of earth mass and a radius two times of earth radius if we substitute we have

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g= G4M/4R^2

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g= GM/R^2

Therefore, g remain the same

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