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4 years ago

Estimate the volume of a piece of molecular cloud that has the same amount of water as your body.

Physics

1 answer:

noname [10]4 years ago

6 0

Question:

The water molecules now in your body were once part of a molecular cloud. Only about onemillionth of the mass of a molecular cloud is in the form of water molecules, and the mass density of such a cloud is roughly 2.0×10−21 g/cm^3.

Estimate the volume of a piece of molecular cloud that has the same amount of water as your body.

Answer:

The volume of cloud that has the same density as the amount of water in our body is 1.4×10²⁵ cm³

Explanation:

Here, we have mass density of cloud = 2.0×10⁻²¹ g/cm^3

Density = Mass/Volume

Volume = Mass/Density = If the mass is 40 kg and the body is made up of 70% by mass of water, we have

28 kg water = 28000 g

Therefore the Volume = 28 kg/ 2.0×10⁻²¹ g/cm^3 = 1.4×10¹⁹ m³ = 1.4×10²⁵ cm³.

Therefore, the volume of cloud that has the same density as the amount of water in our body = 1.4×10²⁵ cm³.

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3 years ago

A force of 9 pounds stretches a spring 1 foot. A mass weighing 6.4 pounds is attached to the spring, and the system is then imme

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Answer:

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Explanation:

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3 years ago

Part A (4 pts) Consider light of wavelength λ = 670nm traveling in air. The light is incident at normal incidence upon a thin fi

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Answer:

Explanation:

On both sides of the film , the mediums have lower refractive index.

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2μt = λ / 2

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4 years ago

A hydraulic lift has two connected pistons with cross-sectional areas 15 cm2 and 670 cm2. It is filled with oil of density 510 k

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Answer:

Explanation:

Given

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$A_1=15\ cm^2$

$A_2=670\ cm^2$

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mass of car place on Large area $M=1100\ kg$

Suppose a mass of m kg is placed on smaller area

According to pascal law's intensity of pressure is same at every point on Liquid

$P_1=P_2$

$\frac{F_1}{A_1}=\frac{F_2}{A_2}$

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$m=1100\times \frac{15}{670}$

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3 years ago

For exercise, an athlete lifts a barbell that weighs 400 N from the ground to a height of 2.0 m in a time of 1.6 s. Assume the e

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Answer:

Explanation:

(ΔK + ΔUg + ΔUs + ΔEch + ΔEth = W)

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ΔK = 0

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ΔUg = positive

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b )

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c )

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3200 J

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= 2000 W .

d )

1 ) Since he is doing same amount of work , his metabolic energy output is same as that in earlier case .

2 ) Since he is doing the same exercise in less time so his power is increased . Hence in the second day his power is more .

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4 years ago