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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:

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Explanation:

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Advocard [28]

Answer:

<em>No, a rigid body cannot experience any acceleration when the resultant force acting on the body is zero.</em>

Explanation:

If the net force on a body is zero, then it means that all the forces acting on the body are balanced and cancel out one another. This sate of equilibrium can be static equilibrium (like that of a rigid body), or dynamic equilibrium (that of a car moving with constant velocity)

For a body under this type of equilibrium,

ΣF = 0   ...1

where ΣF is the resultant force (total effective force due to all the forces acting on the body)

For a body to accelerate, there must be a force acting on it. The acceleration of a body is proportional to the force applied, for a constant mass of the body. The relationship between the net force and mass is given as

ΣF = ma   ...2

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a is the acceleration of the body

Substituting equation 2 into equation 1, we have

0 = ma

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a = 0

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An atom with more or less neutrons than expected is a what ?
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(a) Calculate the acceleration due to gravity on the surface of the Sun.
Ira Lisetskai [31]
<h2>a)Acceleration due to gravity on the surface of the Sun is 274.21 m/s²</h2><h2>b) Factor of increase in weight is 27.95</h2>

Explanation:

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                      g=\frac{GM}{r^2}

 Here we need to find acceleration due to gravity of Sun,

                G = 6.67259 x 10⁻¹¹ N m²/kg²

    Mass of sun, M = 1.989 × 10³⁰ kg

    Radius of sun, r = 6.957 x 10⁸ m

Substituting,

                g=\frac{6.67259\times 10^{-11}\times 1.989\times 10^{30}}{(6.957\times 10^8)^2}\\\\g=274.21m/s^2

Acceleration due to gravity on the surface of the Sun = 274.21 m/s²

b) Acceleration due to gravity in earth = 9.81 m/s²

   Ratio of gravity = 274.21/9.81 = 27.95

   Weight = mg

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