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3 years ago

9

Assuming that the density of a person is the same as the density of water, what buoyant force does the atmosphere exert on a per

son whose mass is 70 kg?

1 answer:

Charra [1.4K]3 years ago

4 0

Answer:

the buoyant force F will be 0.84 N

Explanation:

Assuming that a person has the density of water , then its volume could be calculated through

V= m/ρ , where m= mass of the person , ρ= density of water

then

V= m/ρ = 70 kg/ 1000 kg/m³ = 0.07 m³

then the buoyant force will be equal to the mass of displaced air in the volume V.

if we assume that the density of air in the atmosphere is ρa=1.225 kg/m³ then the buoyant force F will be

F= ρa*V*g = 0.07 m³*1.225 kg/m³ *9.8 m/s² =0.84 N

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3 years ago

Asteroid Ida was photographed by the Galileo spacecraft in 1993, and the photograph revealed that the asteroid has a small moon,

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Answer:

The orbital speed of Dactyl is $5.55m/s$

Explanation:

The orbital speed can be determined by the combination of the universal law of gravity and Newton's second law:

$F = G\frac{M \cdot m}{r^{2}}$ (1)

Where G is gravitational constant, M is the mass of the asteroid, m is the mass of the moon and r is the distance between them

In the other hand, Newton's second law can be defined as:

$F = ma$ (2)

Where m is the mass and a is the acceleration

Then, equation 2 can be replaced in equation 1

$m\cdot a = G\frac{M \cdot m}{r^{2}}$ (2)

However, a will be the centripetal acceleration since the moon Dactyl describe a circular motion around the asteroid

$a = \frac{v^{2}}{r}$ (3)

$m\frac{v^{2}}{r} = G\frac{M \cdot m}{r^{2}}$ (4)

Therefore, v can be isolated from equation 4:

$m \cdot v^{2} = G \frac{M \cdot m}{r^{2}}r$

$m \cdot v^{2} = G \frac{M \cdot m}{r}$

$v^{2} = G \frac{M \cdot m}{rm}$

$v^{2} = G \frac{M}{r}$

$v = \sqrt{\frac{G M}{r}}$ (5)

Finally, the orbital speed can be found from equation 5:

Notice, that it is necessary to express r in units of meters.

$r = 95km \cdot \frac{1000m}{1km}$ ⇒ $95000m$

$v = \sqrt{\frac{(6.672x10^{-11}N.m^{2}/kg^{2})(4.4x10^{16}kg)}{95000m}}$

$v = 5.55m/s$

Hence, the orbital speed of Dactyl is $5.55m/s$

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3 years ago

Use the dimensional analysis and check the correctness of given equation:-<br> PV= nRT

Harman [31]

PV=nRT

Here

P=Pressure

V=Volume

n=Molarity

R=universal gas constant

T=Temperature.

LHS

$\\ \tt\bull\leadsto PV$

$\\ \tt\bull\leadsto [ML^2T^{-2}][M^0L^3T^0]$

$\\ \tt\bull\leadsto [ML^5T^{-2}]$

RHS

$\\ \tt\bull\leadsto nRT$

$\\ \tt\bull\leadsto [M^0L^{3}T^0][M^1 L^2 T^{-2}K^{-1}][M^0L^0T^0K^1]$

$\\ \tt\bull\leadsto [ML^5T^{-2}]$

LHS=RHS

hence verified

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3 years ago

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Answer:

False

Explanation:

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