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givi [52]
3 years ago
10

The engine of a 1500-kg automobile has a power rat- ing of 75 kW. Determine the time required to accelerate this car from rest t

o a speed of 100 km/h at full power on a level road. Is your answer realistic
Engineering
1 answer:
creativ13 [48]3 years ago
3 0

Answer:

7.72 seconds

The answer is not realistic as the time is very less to reach to a speed of 100 km/hr

also, we have not taken other factors into consideration like wind drag etc.

Explanation:

Data provided in the question:

Mass of the engine = 1500 kg

Power rating = 75 kW = 75,000 W

Initial speed, v₁ = 0

Final speed = 100 km/hr = 100\times\frac{5}{18} = 27.78 m/s

Now,

Power = Work done ÷ Time

also,

Work done = Final energy - Initial energy

= \frac{1}{2}mv_2^2-\frac{1}{2}mv_1^2

= \frac{1}{2}\times1500\times27.78^2-\frac{1}{2}\times1500\times0^2

= 578703.70 J

thus,

75,000 = 578703.70 ÷ time

or

time = 7.716 s ≈ 7.72 seconds

The answer is not realistic as the time is very less to reach to a speed of 100 km/hr

also, we have not taken other factors into consideration like wind drag etc.

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5 0
3 years ago
Is it better to do blue prints with paper and pencil or a computer program if you’re going to design a house? Why?
Hatshy [7]

Answer:

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Explanation:

I use Revit and its way better to do all that you can see 2D 3D the measurements and its super easy to use hope this helps

6 0
3 years ago
From your cooling load (8890.007 Btu/hr = 2.605kW, determine mass flow rate of refrigerants. Use the following "rule of thumb" e
NeX [460]

Answer:

0.740833917 ton/hr

Explanation:

Given:

Cooling load, 8890.007 Btu/hr = 2.605 kW

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According to the thumb rule

1 ton of refrigerant = 12000Btu

Hence for 8890.007 Btu/hr,

the mass flow rate of the refrigerant is =8890.007 / 12000

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7 0
3 years ago
EJERCICIO 6
Ket [755]

Answer:

mnfokfnfi3or

Explanation:

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8 0
3 years ago
In wet mill ethanol plants, a total energy of 74,488 Btu (British thermal units, a common energy unit) is used to produce 1 gall
V125BC [204]

Answer:

The energy yield for one gallon of ethanol is 2.473 %.

Explanation:

The net energy yield (\% e), expressed in percentage for one gallon of ethanol is the percentage of the ratio of the difference of the provided energy (E_{g}), measured in Btu, and the energy needed to produce the ethanol (E_{p}), measured in Btu, divided by the energy needed to produce the ethanol. That is:

\% e =\frac{E_{g}-E_{p}}{E_{p}} \times 100\,\% (1)

If we know that E_{g} = 76330\,Btu and E_{p} = 74488\,Btu, then the net energy yield of 1 gallon of ethanol:

\%e = \frac{76330\,Btu-74488\,Btu}{74488\,Btu}\times 100\,\%

\%e = 2.473\,\%

The energy yield for one gallon of ethanol is 2.473 %.

4 0
3 years ago
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