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nignag [31]
2 years ago
15

Two girls are estimating each other's power. One runs up some step

Physics
1 answer:
Sveta_85 [38]2 years ago
8 0

Answer:

A. 450 N

B. 3240 J

C. 0.77 KW

Explanation:

From the question given above, the following data were obtained:

Height of one step = 20 cm

Number of steps = 36

Mass of runner = 45 kg

Time taken = 4.2 s

Next, we shall convert 20 cm to metre (m). This can be obtained as follow:

100 cm = 1 m

Therefore,

20 cm = 20 cm × 1 m /100 cm

20 cm = 0.2 m

Next, we shall determine the total height. This can be obtained as follow:

Height of one step = 0.2 m

Number of steps = 36

Total height =?

Total height = 36 × 0.2

Total height = 7.2 m

A. Determination of the runner's weight.

Mass of runner (m) = 45 kg

Acceleration due to gravity (g) = 10 m/s²

Weight (W) =?

W = m × g

W = 45 × 10

W = 450 N

B. Determination of the increase in the potential energy.

At the ground level, the potential energy (PE₁) is 0 J.

Next, we shall determine the potential energy at 7.2 m. This can be obtained as follow:

Mass of runner (m) = 45 kg

Acceleration due to gravity (g) = 10 m/s²

Total height (h) = 7.2 m

Potential energy at height 7.2 m (PE₂) = ?

PE₂ = mgh

PE₂ = 45 × 10 × 7.2

PE₂ = 3240 J

Final, we shall determine the increase in potential energy. This can be obtained as follow:

Potential energy at ground (PE₁) = 0 J

Potential energy at height 7.2 m (PE₂) = 3240 J

Increase in potential energy =?

Increase in potential energy = PE₂ – PE₁

Increase in potential energy = 3240 – 0

Increase in potential energy = 3240 J

C. Determination of the power.

Energy (E) = 3240 J

Time (t) = 4.2 s

Power (P) =?

P = E/t

P = 3240 / 4.2

P = 771.43 W

Finally, we shall convert 771.43 W to kilowatt (KW). This can be obtained as follow:

1000 W = 1 KW

Therefore,

771.43 W = 771.43 W × 1 KW / 1000 W

771.43 W = 0.77 KW

Therefore, her power is 0.77 KW

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Answer:

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Explanation:

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3 years ago
In which of the two situations described is more energy transferred?
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Answer:

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Explanation:

Each of the situations are analyzed as follows;

Situation A

The temperature of the cup of hot chocolate = 40 °C

The temperature of the interior of the freezer in which the chocolate is placed = -20 °C

We note that at 0°C, the water in the chocolate freezes

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E₁ = m×c₁×ΔT₁

Where;

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ΔT₁ = The change in temperature from 40 °C to 0°C

Therefore, we have;

E₁ = m×4.184×(40 - 0) = 167.360·m kJ

The heat the coffee gives to turn to ice is given as follows;

E₂ = m·H_f

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H_f = The latent heat of fusion = 334 kJ/kg

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The heat required to cool the frozen ice to -20 °C is given as follows;

E₃ = m·c₂·ΔT₂

Where;

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Therefore, we have;

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Situation B

The temperature of the cup of hot chocolate = 90 °C

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The heat transferred by the hot cup of coffee, E, is given as follows;

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Therefore, the total heat transferred in situation A is approximately twice the heat transferred in situation B and is therefore more than the heat transferred in situation B

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Energy transferred in situation A ≈ 2 × Energy transferred in situation B

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