Answer:
19063.6051 g
Explanation:
Pressure = Atmospheric pressure + Gauge Pressure
Atmospheric pressure = 97 kPa
Gauge pressure = 500 kPa
Total pressure = 500 + 97 kPa = 597 kPa
Also, P (kPa) = 1/101.325 P(atm)
Pressure = 5.89193 atm
Volume = 2.5 m³ = 2500 L ( As m³ = 1000 L)
Temperature = 28 °C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T₁ = (28.2 + 273.15) K = 301.15 K
Using ideal gas equation as:
PV=nRT
where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value = 0.0821 L.atm/K.mol
Applying the equation as:
5.89193 atm × 2500 L = n × 0.0821 L.atm/K.mol × 301.15 K
⇒n = 595.76 moles
Molar mass of oxygen gas = 31.9988 g/mol
Mass = Moles * Molar mass = 595.76 * 31.9988 g = 19063.6051 g
False! Just saying. You could be under the influence, or just have no clue as to what you're doing.
Answer:
Explanation:
It wouldn't work because the wind energy she would be collecting would actually come from the car engine.
The relative wind velocity observed from a moving vehicle is the sum of the actual wind velocity and the velovity of the vehicle.
u' = u + v
While running a car will generate a rather high wind velocity, and increase the power generated by a wind turbine, the turbine would only be able to convert part of the wind energy into electricity while adding a lot of drag. In the end, it would generate less energy that what the drag casuses the car to waste to move the turbine.
Regenerative braking uses an electric generator connected to the wheel axle to recover part of the kinetic energy eliminated when one brakes the vehicle. Normal brakes dissipate this energy as heat, a regenerative brake uses it to recharge a batttery. Note that is is a fraction of the energy that is recovered, not all of it.
A "regenerative accelerator" makes no sense. Braking is taking kinetic energy out of the vehicle, while accelerating is adding kinetic energy to it. Cars accelerate using the power from their engines.
Answer:
C: Viscosity, the resistance to flow that fluids exhibit
Explanation:
Did it on Edge :)
Friction losses in pipes can be reduced by decreasing the length of the pipes, reducing the surface roughness of the pipes, and increasing the pipe diameter. Thus, options (c),(e), and (f) hold correct answers.
Friction loss is a measure of the amount of energy a piping system loses because flowing fluids meet resistance. As fluids flow through the pipes, they carry energy with them. Unfortunately, whenever there is resistance to the flow rate, it diverts fluids, and energy escapes. These opposing forces result in friction loss in pipes.
Friction loss in pipes can decrease the efficiency of the functions of pipes. These are a few ways by which friction loss in pipes can be reduced and the efficiency of the piping system can be boosted:
- <u><em>Decrease the length of the pipes</em></u>: By decreasing pipe lengths and avoiding the use of sharp turns, fittings, and tees, whenever possible result in a more natural path for fluids to flow.
- <u><em>Reduce the surface roughness of the pipes</em></u>: By reducing the interior surface roughness of pipes, a smooth and clearer path is provided for liquids to flow.
- <u><em>Increase the pipe diameter: </em></u>By widening the diameters of pipes, it is ensured that fluids squeeze through pipes easily.
You can learn more about friction losses at
brainly.com/question/13348561
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