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alexgriva [62]
3 years ago
7

A 75-hp compressor in a facility that operates at full load for 2500 h a year is powered by an electric motor that has an effici

ency of 93 percent. If the unit cost of electricity is $0.11/kWh, the annual electricity cost of this compressor is
Physics
1 answer:
stealth61 [152]3 years ago
3 0

Answer:

C = 16544.35\, USD

Explanation:

The annual electricity cost is:

C = (\frac{1}{0.93} )\cdot(75\, h.p.)\cdot(0.746\, kW)\cdot(2500\,h)\cdot(\frac{3600\,s}{1\,h} )\cdot(\frac{1\,kWh}{3600\,kJ} )\cdot (0.11\,\frac{USD}{kWh} )

C = 16544.35\, USD

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qaws [65]

-- We know that the y-component of acceleration is the derivative of the
y-component of velocity.

-- We know that the y-component of velocity is the derivative of the
y-component of position.

-- We're given the y-component of position as a function of time.

So, finding the velocity and acceleration is simply a matter of differentiating
the position function ... twice.

Now, the position function may look big and ugly in the picture.  But with the
exception of  't' , everything else in the formula is constants, so we don't even
need any fancy processes of differentiation.  The toughest part of this is going
to be trying to write it out, given the text-formatting capabilities of the wonderful
envelope-pushing website we're working on here.

From the picture . . . . . y (t) = (1/2) (a₀ - g) t² - (a₀ / 30t₀⁴ ) t⁶

First derivative . . . y' (t) = (a₀ - g) t  -  6 (a₀ / 30t₀⁴ ) t⁵  =  (a₀ - g) t  -  (a₀ / 5t₀⁴ ) t⁵

There's your velocity . . . /\ .

Second derivative . . . y'' (t) = (a₀ - g) -  5 (a₀ / 5t₀⁴ ) t⁴ = (a₀ - g) -  (a₀ /t₀⁴ ) t⁴

and there's your acceleration . . . /\ .
That's the one you're supposed to graph.

a₀ is the acceleration due to the model rocket engine thrust
     combined with the mass of the model rocket
'g' is the acceleration of gravity ... 9.8 m/s² or 32.2 ft/sec²
t₀  is how long the model rocket engine burns

Pick, or look up, some reasonable figures for a₀ and t₀
and you're in business.

The big name in model rocketry is Estes.  Their website will give you
all the real numbers for thrust and burn-time of their engines, if you
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6 0
3 years ago
The atmospheric pressure above a swimming pool changes from 753 to 772 mm of mercury. the bottom of the pool is a rectangle (11
Evgesh-ka [11]
Change in pressure : 772-753 = 19 mm Hg 
Since 1 mm Hg = 133 Pa Hg
 Then, 19 mm Hg = 2527 Pa Hg
 Area of rectangular pool = L* B = 11 x 21 = 231 m^2
 To find the force, we multiply the area by the change in pressure: 2527 (231) = 583737N
5 0
3 years ago
Read 2 more answers
What kinds of natural phenomena does the inverse-square law apply to?
uysha [10]

Answer:

It applies to diverse phenomena

Explanation:

not sure if u need one, just ask if you do :D

4 0
2 years ago
Read 2 more answers
A wheelbarrow full of bricks is lifted to the top of a wall. If the mass of the loaded wheelbarrow is 3000.0 g and the height of
Rina8888 [55]

Answer:

the Gravitational potential energy is 13.23 J

Explanation:

The computation of the GPE is shown below:

GPE stands for Gravitational potential energy

The following formula should be used for the same

= mass × gravity × height

= 3000 g × 9.8m/sec^2 × 0.45 m

= 13.23 J

Hence, the Gravitational potential energy is 13.23 J

We simply applied the above formula so that we can easily determine the GPE

6 0
3 years ago
You walk forward 105 meters, in 15 seconds. What is your overall velocity? I need help immediately please
damaskus [11]
Velocity = displacement/ time
105 meters/ 15 seconds = 7 m/s
8 0
2 years ago
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