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djverab [1.8K]
4 years ago
15

When a person has the ability to move a vehicle, they have ____________________.

Engineering
1 answer:
sergiy2304 [10]4 years ago
4 0

(Energy ) i think .....

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A rigid, sealed cylinder initially contains 100 lbm of water at 70 °F and atmospheric pressure. Determine: a) the volume of the
soldier1979 [14.2K]

Answer:

Determine A) The Volume Of The Tank (ft^3) Later A Pump Is Used To Extract ... A rigid, sealed cylinder initially contains 100 lbm of water at 70 degrees F and atmospheric pressure. ... Later a pump is used to extract 10 lbm of water from the cylinder. The water remaining in the cylinder eventually reaches thermal equilibrium ...

3 0
3 years ago
Legal metrology would protect consumers from businesses that do not take measurements according to defined measuring regulations
sergij07 [2.7K]

From what my research concluded, true

5 0
4 years ago
In a wind-turbine, the generator in the nacelle is rated at 690 V and 2.3 MW. It operates at a power factor of 0.85 (lagging) at
Juli2301 [7.4K]

To solve this problem we will apply the concepts related to real power in 3 phases, which is defined as the product between the phase voltage, the phase current and the power factor (Specifically given by the cosine of the phase angle). First we will find the phase voltage from the given voltage and proceed to find the current by clearing it from the previously mentioned formula. Our values are

V = 690V

P_{real} = 2.3MW

Real power in 3 phase

P_{real} = 3V_{ph}I_{ph} Cos\theta

Now the Phase Voltage is,

V_{ph} = \frac{V}{\sqrt{3}}

V_{ph} = \frac{690}{\sqrt{3}}

V_{ph} = 398.37V

The current phase would be,

P_{real} = 3V_{ph}I_{ph} Cos\theta

Rearranging,

I_{ph}=\frac{P_{real}}{3V_{ph}Cos\theta}

Replacing,

I_{ph}=\frac{2.3MW}{3( 398.37V)(0.85)}

I_{ph}= 2.26kA/phase

Therefore the current per phase is 2.26kA

6 0
3 years ago
If the dry-bulb temperature is 95°F and the wet-bulb temperature is also 78°F, what is the relative humid- ity? What is the dew
Sidana [21]

Answer:

Relative humidity 48%.

Dew point 74°F

humidity ratio 118 g of moisture/pound of dry air

enthalpy 41,8 BTU per pound of dry air

Explanation:

You can get this information from a Psychrometric chart for water, like the one attached.

You enter the chart with dry-bulb and wet-bulb temperatures (red point in the attachment) and following the relative humidity curves you get approximately 48%.

To get the dew point you need to follow the horizontal lines to the left scale (marked with blue): 74°F

for the humidity ratio you need to follow the horizontal lines but to the rigth scale (marked with green): 118 g of moisture/pound of dry air

For enthalpy follow the diagonal lines to the far left scale (marked with yellow): 41,8 BTU per pound of dry air

5 0
3 years ago
While discussing run-flat tires: Technician A says that some are self-sealing tires and are designed to quickly and permanently
musickatia [10]

Answer:

The correct option is d ( Neither A nor B)

Explanation:

Technician A made 2 mistakes in his statement.Firstly the tire is self supporting not self sealing.

Secondly, this tire does not provide permanent sealing of punctured area option a is incorrect.

This self-supporting tire after being affected with complete air leakage can temporarily bear the load of the car and avoid rolling over a distance of 80 km at a maximum speed of 55 mph. Here is what technician B suggested incorrectly as the tire after being.Here the technician B suggested incorrectly as the tire after being affected with puncture can not travel at any speed so option B is wrong

Since option a and b are incorrect and c is invalid.

4 0
3 years ago
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