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3 years ago

10

A coat rack weighs 65.0 lbs when it is filled with winter coats and 40.0 lbs when it is empty. The base of the coat rack has an

area of 452.4 in2. How much more pressure, in psi (pounds per square inch), is exerted by the coat rack on the floor when it is filled with winter coats than when it is empty

1 answer:

Whitepunk [10]3 years ago

8 0

Answer:

0.056 psi more pressure is exerted by filled coat rack than an empty coat rack.

Explanation:

First we find the pressure exerted by the rack without coat. So, for that purpose, we use formula:

P₁ = F/A

where,

P₁ = Pressure exerted by empty rack = ?

F = Force exerted by empty rack = Weight of Empty Rack = 40 lb

A = Base Area = 452.4 in²

Therefore,

P₁ = 40 lb/452.4 in²

P₁ = 0.088 psi

Now, we calculate the pressure exerted by the rack along with the coat.

P₂ = F/A

where,

P₂ = Pressure exerted by rack filled with coats= ?

F = Force exerted by filled rack = Weight of Filled Rack = 65 lb

A = Base Area = 452.4 in²

Therefore,

P₂ = 65 lb/452.4 in²

P₂ = 0.144 psi

Now, the difference between both pressures is:

ΔP = P₂ - P₁

ΔP = 0.144 psi - 0.088 psi

<u>ΔP = 0.056 psi</u>

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Answer:

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3 years ago

Steam at 100°C is condensed into a 38.0 g aluminum calorimeter cup containing 280 g of water at 25.0°C. Determine the amount of

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Answer:

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Explanation:

From the expression of latent heat of steam, we have

Heat supplied by steam = Heat gain water + Heat gain by calorimeter

mathematically,

$m_{s}c_{w} \alpha _{w}$ + $m_{s}l$ = $m_{w}c_{w} \alpha _{w}$ + $m_{c}c_{c} \alpha c_{c}$

L=specific latent heat of water(steam)=2268J/g

$c_{w}$ =specific heat capacity=4.2J/gK

$c_{c}$ =specific heat capacity of calorimeter =0.9J/gk

$m_{w}$ =280g

$m_{c}$ =38g

α=change in temperature

$\alpha _{c}$ =(40-25)=15

$\alpha _{w}$ =(40-25)=15

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Note: the temperature of the calorimeter is the temperature of it content.

From the equation, we can make $m_{s}$ the subject of formula

$m_{s}=\frac{m_{w}c_{w} \alpha +m_{c}c_{c}\alpha}{c_{w}\alpha +l}$

Hence

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3 years ago

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Minchanka [31]

Answer:

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Explanation:

We can calculate the acceleration experimented by the passenger using the formula $v_f^2=v_i^2+2ad$ , taking the initial direction of movement as the positive direction and considering it comes to a rest:

$a=\frac{v_f^2-v_i^2}{2d}=\frac{-v_i^2}{2d}$

Then we use Newton's 2nd Law to calculate the force the passenger of mass m experimented to have this acceleration:

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Which for our values is:

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