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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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A 2.98 nF parallel-plate capacitor is charged to an initial potential difference of 49 V and then isolated. The dielectric mater

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3 years ago

Read 2 more answers

A block of ice(m = 14.0 kg) with an attached rope is at rest on a frictionless surface. You pull the block with a horizontal for

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Answer:

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Explanation:

a) We need to calculate the weight, normal force from the ground to the block and the pull force. By 3rd Newton's Law we know that normal force is the reaction of the weight of the block of ice on a horizontal.

The weight of the block ( $W$ ), measured in newtons, is:

$W = m\cdot g$ (1)

Where:

$m$ - Mass of the block of ice, measured in kilograms.

$g$ - Gravitational acceleration, measured in meters per square second.

If we know that $m = 14\,kg$ and $g = 9.807\,\frac{m}{s^{2}}$ , the magnitudes of the weight and normal force of the block of ice are, respectively:

$N = W = (14\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)$

$N = W = 137.298\,N$

And the pull force is:

$F_{pull} = 98\,N$

The weight and the normal force of the block has a magnitude of 137.298 newtons and the pull force exerted on the block has a magnitude of 98 newtons.

b) Since the block of ice is on a frictionless surface and pull force is parallel to the direction of motion and uniform in time, we can apply the Impact Theorem, which states that:

$m\cdot v_{o} +\Sigma F \cdot \Delta t = m\cdot v_{f}$ (2)

Where:

$v_{o}$ , $v_{f}$ - Initial and final speeds of the block, measured in meters per second.

$\Sigma F$ - Horizontal net force, measured in newtons.

$\Delta t$ - Impact time, measured in seconds.

Now we clear the final speed in (2):

$v_{f} = v_{o}+\frac{\Sigma F\cdot \Delta t}{m}$

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$v_{f} = 0\,\frac{m}{s}+\frac{(98\,N)\cdot (1.40\,s)}{14\,kg}$

$v_{f} = 9.8\,\frac{m}{s}$

The final speed of the block of ice is 9.8 meters per second.

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