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patriot [66]
3 years ago
5

Fluids at rest possess no flow energy. a)- True b)- False

Engineering
1 answer:
gtnhenbr [62]3 years ago
8 0

Answer:

True.

Explanation:

According the engineering flow they don not possess flow energy when they are in rest.

When they are in motion they show a translation energy.

The features if fluids may be different according the variables of pressure and temperature.

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Which statement explains what causes the balloon to accelerate
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The air leaving through the balloon's mouth pulls the balloon in the same direction as the exiting air, so the balloon experiences a net force. All air surrounding the balloon pushes the balloon forward.
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The use of seatbelts in a car has significantly reduced the number of crash fatalities. Which statement best explains how societ
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as the public need for preventing injuries and deaths from car crashes became known,laws were enacted to mandate the inclusion of seatbelts in almost all passengers vehicles

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3 years ago
An uncovered swimming pool loses 1.0 inch of water off its 1,000 ft^2 surface each week due to evaporation. The heat of vaporiza
soldi70 [24.7K]

Answer:

The affirmation is true, the cover will be worth buying

Explanation:

The equation necessary to use is

E = m*cv,

Where

cv: the heat of vaporization.  

Finding the rate at which the water evaporates (m^3/week).  

The swimming pool loses water at 1 inch/week off its 1,000 ft^2

Than,

1000 ft² * 1 in/wk * 1 ft/12 in = 83.33 ft³/week

To obtains the rate of mass loss it is necessary to multiply it for the density of water

83.33 ft³/week * 62.4 lb/ft³ = 5200 lb/week

Knowing the vaporization heat it is possible to find the rate of heat which is leaving the swimming pool  

5200 lb/week * 1050 BTU/lb = 5460000 btu/week

Over a 15-week period, the pool loses 81.9 million BTU.  

Knowing the cost of energy to heat the pool is $10.00 per million btu

The price = $819

This way, the affirmation is true, the cover will be worth buying

3 0
3 years ago
The heat flux through a 1-mm thick layer of skin is 1.05 x 104 W/m2. The temperature at the inside surface is 37°C and the tempe
miss Akunina [59]

Answer:

a) Thermal conductivity of skin: k_{skin}=1.5W/mK

b) Temperature of interface: T_{interface}=35.6\°C

Heat flux through skin: \frac{Q}{A}=2100W/m^2

Explanation:

a)

k=\frac{QL}{A(T_{2}-T_{1})}

Where: k is thermal conductivity of a material, \frac{Q}{A} is heat flux through a material, L is the thickness of the material, T_{1} is the temperature on the first side and T_{2} is the temperature on the second side

k_{skin}=\frac{QL}{A(T_{2}-T_{1})}

k_{skin}=\frac{Q}{A}*\frac{L}{(T_{2}-T_{1})}

k_{skin}=1.05*10^{4}*\frac{1*10^{-3}}{(37-30)}

k_{skin}=1.5W/mK

b)

k_{insulation}=\frac{k_{skin}}{2}

k_{insulation}=\frac{1.5}{2}

k_{insulation}=0.75W/mK

The heat flux between both surfaces is constant, assuming the temperature is maintained at each surface.

\frac{Q}{A}=\frac{k(T_{2}-T_{1})}{L}

\frac{k_{skin}(T_{skin}-T_{interface})}{L_{skin}}=\frac{k_{insulation}(T_{interface}-T_{insulation})}{L_{insulation}}

\frac{1.5*(37-T_{interface})}{0.001}=\frac{0.75*(T_{interface}-30)}{0.002}

55500-1500T_{interface}=375T_{interface}-11250

1875T_{interface}=66750

T_{interface}=35.6\°C

\frac{Q}{A}=\frac{k_{skin}(T_{skin}-T_{interface})}{L_{skin}}

\frac{Q}{A}=\frac{1.5*(37-35.6)}{0.001}

\frac{Q}{A}=2100W/m^2

3 0
3 years ago
An automobile having a mass of 1100 kg initially moves along a level highway at 120 km/h relative to the highway. It then climbs
soldier1979 [14.2K]

Answer:

-6111.11\ \text{kJ}

863.28\ \text{kJ}

Explanation:

m = Mass of automobile = 1100 kg

v = Velocity of car = 120 km/h = \dfrac{120}{3.6}\ \text{m/s}

h = Height of hill = 80 m

g = Acceleration due to gravity = 9.81\ \text{m/s}^2

Change in kinetic energy

KE=\dfrac{1}{2}m(u^2-v^2)\\\Rightarrow KE=\dfrac{1}{2}\times 1100\times (0-(\dfrac{120}{3.6})^2)\\\Rightarrow KE=-611111.11\ \text{J}

Change in kinetic energy is -6111.11\ \text{kJ}

Change in potential energy is given by

PE=mgh\\\Rightarrow PE=1100\times 9.81\times 80\\\Rightarrow PE=863280\ \text{J}

The change in potential energy is 863.28\ \text{kJ}.

8 0
3 years ago
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