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3 years ago

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A twin-sized air mattress used for camping has dimensions of 100 cm by 194 cm by 14 cm when blown up. The weight of the mattress

is 4 kg. How heavy a person (in N) could the air mattress hold if it is placed in freshwater

1 answer:

Paul [167]3 years ago

8 0

Answer:

$2625.156\ \text{N}$

Explanation:

Dimensions of mattress 100 cm by 194 cm by 14 cm

$m_m$ = Mass of mattress = 4 kg

$\rho$ = Density of water = $1000\ \text{kg/m}^3$

g = Acceleration due to gravity = $9.81\ \text{m/s}^2$

Volume of mattress

$V=100\times 194\times 14=271600\ \text{cm}^3=0.2716\ \text{m}^3$

Weight of water displaced is equal to the buoyant force

Mass of water

$m=\rho V\\\Rightarrow m=1000\times 0.2716\\\Rightarrow m=271.6\ \text{kg}$

Mass of person would be

$m_p=m-m_m=271.6-4\\\Rightarrow m_p=267.6\ \text{kg}$

Weight of the person would be

$w=m_pg\\\Rightarrow w=267.6\times 9.81\\\Rightarrow w=2625.156\ \text{N}$

The air mattress could hold a person that weighs up to $2625.156\ \text{N}$ .

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Starting from Newton’s law of universal gravitation, show how to find the speed of the moon in its orbit from the earth-moon dis

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Explanation:

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$m=7.349(10)^{22} kg$ is the mass of the Moon

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Being $V$ the orbital velocity of the moon

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$m.a_{C}=G\frac{Mm}{r^{2}}$ (4)

Simplifying:

$a_{C}=G\frac{M}{r^{2}}$ (5)

Making (5)=(3):

$\frac{V^{2}}{r}=G\frac{M}{r^{2}}$ (6)

Finding $V$ :

$V=\sqrt{\frac{GM}{r}}$ (7)

$V=\sqrt{\frac{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24} kg)}{3.9(10)^{8} m}}$ (8)

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If we've got a normal distribution, then we can convert it to standard normal distribution and its values will give us the z score.

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Learn more about z-score here:

brainly.com/question/21262765

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