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2 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]2 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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Answer:

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(b)

If friction present,

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(a) $7.32\cdot 10^7 Hz$

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$f=\frac{c}{\lambda}$

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Substituting into the equation, we find

$f=\frac{3.0\cdot 10^8 m/s}{4.1 m}=7.32\cdot 10^7 Hz$

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The angular wave number of a wave is given by

$k=\frac{2\pi}{\lambda}$

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$\lambda$ is the wavelength of the wave

For this wave, we have

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$k=\frac{2\pi}{4.1 m}=1.53 m^{-1}$

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For an electromagnetic wave,

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E is the magnitude of the electric field component

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For this wave,

E = 310 V/m

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(e) z-axis

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(f) 127.5 W/m^2

The time-averaged rate of energy flow of an electromagnetic wave is given by:

$I=\frac{E^2}{2\mu_0 c}$

where we have

E = 310 V/m is the amplitude of the electric field

$\mu_0$ is the vacuum permeability

c is the speed of light

Substituting into the formula,

$I=\frac{(310 V/m)^2}{2(4\pi\cdot 10^{-7} H/m) (3\cdot 10^8 m/s)}=127.5 W/m^2$

(g) $1.53\cdot 10^{-8} kg m/s$

For a surface that totally absorbs the wave, the rate at which momentum is transferred to the surface given by

$\frac{dp}{dt}=\frac{A}{c}$

where the <S> is the magnitude of the Poynting vector, given by

$=\frac{EB}{\mu_0}=\frac{(310 V/m)(1.03\cdot 10^{-6} T)}{4\pi \cdot 10^{-7}H/m}=254.2 W/m^2$

and where the surface is

A = 1.8 m^2

Substituting, we find

$\frac{dp}{dt}=\frac{(254.2 W/m^2)(1.8 m^2)}{3\cdot 10^8 m/s}=1.53\cdot 10^{-8} kg m/s$

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$p=\frac{}{c}$

where we have

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Substituting, we find

$p=\frac{254.2 W/m^2}{3\cdot 10^8 m/s}=8.47\cdot 10^{-7} N/m^2$

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