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dalvyx [7]
3 years ago
15

A house is maintained at 1 atm and 24°C. Outdoor air at 5.5°C infiltrates into the house through the cracks and warm air inside

the house is forced to leave the house at a rate of 90 m3 /hr. The gas constant of air is R = 0.287 kPa·m3 /kg·K. The constant pressure specific heat of air at room temperature is cp = 1.005 kJ/kg:°C.
a. Create a schematic representation of the system, indicating all the mass and energy interactions across the boundary.
b. Determine the mass flow rate of the warm air exiting the house (in kg/s).
c. Determine the mass flow rate of the cold air entering the house (in kg/s).
d. Determine the rate of net energy loss of the house due to mass transfer (in kW).
e. List the assumptions and approximations that were needed to solve this problem.
Physics
1 answer:
Lunna [17]3 years ago
3 0

Answer:

5454gjb

Explanation:

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Friction in machines is commonly reduced by using what?
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I am not sure if this is what your looking for but most machines use oil 
8 0
3 years ago
(11%) Problem 5: A submarine is stranded on the bottom of the ocean with its hatch 25 m below the surface. In this problem, assu
V125BC [204]

Answer:

F = 1.24*10^4 N

Explanation:

Given

Depth of the ship, h = 25 m

Density of water, ρ = 1.03*10^3 kg/m³

Diameter of the hatch, d = 0.25 m

Pressure of air, P(air) = 1 atm

Pressure of water =

P(w) = ρgh

P(w) = 1.03*10^3 * 9.8 * 25

P(w) = 2.52*10^5 N/m²

P(net) = P(w) + P(air) - P(air)

P(net) = P(w)

P(net) = 2.52*10^5 N/m²

Remember,

Pressure = Force / Area, so

Force = Area * Pressure

Area = πr² = πd²/4

Area = 3.142 * 0.25²/4

Area = 3.142 * 0.015625

Area = 0.0491 m²

Force = 0.0491 * 2.52*10^5

F = 12373 N

F = 1.24*10^4 N

5 0
3 years ago
Read 2 more answers
A car accelerates at a constant rate from 0 to 50 mph in three fourths min. How far does the car travel during that​ time?
Helen [10]

Answer:

the car have travelled 0.31 mile during that​ time

Explanation:

Applying the Equation of motion;

s = 0.5(u+v)t

Where;

s = distance travelled

u = initial speed = 0 mph

v = Final speed = 50 mph

t = time taken = 3/4 min = 3/4 ÷ 60 hours = 1/80 hour

Substituting the given values into the equation;

s = 0.5(0+50)×(1/80)

s = 0.3125 miles

s ~= 0.31 mile

the car have travelled 0.31 mile during that​ time

8 0
3 years ago
Can I PLEASE get some help? I REALLY need it!
soldi70 [24.7K]
The answer is C. Hope this helps.
7 0
3 years ago
For a standard production car, the highest road-tested acceleration ever reported occurred in 1993, when a Ford RS200 Evolution
Ann [662]

Answer:

a = 8.06 m/s²

Explanation:

The acceleration of this car can be found using the first equation of motion:

v_f = v_i + at\\\\a = \frac{v_f-v_i}{t}

where,

a = acceleration = ?

vf = final speed = 26.8 m/s

vi = initial speed = 0 m/s

t = time = 3.323 s

Therefore,

a = \frac{26.8\ m/s-0\ m/s}{3.323\ s}

<u>a = 8.06 m/s²</u>

3 0
3 years ago
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