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earnstyle [38]
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
Physics
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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Leto [7]

Using the kinematic equation below we can determine the distance traveled if t=2, a=7.4m/s^2.  First we must determine the final velocity:

v_{final}=v_{initial}+\frac{1}{2}at\\\\v_{final}=0+(7.4m/s^2)(2s)=34.8m/s

Now we will determine the distance traveled:

v_{final}^2=v_{initial}^2+2a \Delta x\\\\\Delta x = \frac{v_{final}^2}{2a} =\frac{(34.8)^2}{(2)(7.4)}=81.83 m

Therefore, the drag racer traveled 81.83 meters in 2 seconds.

3 0
3 years ago
Three identical 6.0-kg cubes are placed on a horizontal frictionless surface in contact with one another. The cubes are lined up
Westkost [7]

Answer:

24 N

Explanation:

m = mass of the cube = 6.0 kg

Consider the three cubes together as one.

M = mass of the three cubes together = 3 m = 3 (6.0) = 18 kg

a = acceleration of the combination = 2 ms⁻²

F = Force applied on the combination

Using Newton's second law

F = ma = (18) (2) = 36 N

F_{L} = Force by the left cube on the middle cube

Consider the forces acting on left cube, from the force diagram, we have

F - F_{L} = ma \\36 - F_{L} = (6) (2)\\F_{L} = 24 N

4 0
2 years ago
What does the term Hubble time mean in cosmology, and what is the current best calculation for the Hubble time?
balu736 [363]

Hubble time in cosmology means the estimated age of the universe and the best calculation for it is T=1/H, where H is the Hubble constant
4 0
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
KonstantinChe [14]

Answer:

7.2g

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

\alpha _{s}=(100-40)=60

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

m_{s}=\frac{(280*4.2*15) +(38*0.9*15)}{(4.2*60) +2268} \\m_{s}=\frac{18153}{2520}\\ m_{s}=7.2g

Hence the amount of steam needed is 7.2g

6 0
3 years ago
A car traveling at 27.4 m/s hits a bridge abutment. A passenger in the car, who has a mass of 65.0 kg, moves forward a distance
Minchanka [31]

Answer:

F=43570.9N

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:

F=ma=\frac{-mv_i^2}{2d}

Which for our values is:

F=\frac{-(65kg)(27.4m/s)^2}{2(0.56m)}=43570.9N

6 0
3 years ago
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