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Anton [14]
3 years ago
6

An air compressor compresses 6 L of air at 120 kPa and 22°C to 1000 kPa and 400°C. Determine the flow work, in kJ/kg, required b

y the compressor. The gas constant of air is R = 0.287 kPa·m3/kg·K. The flow work required by the compressor is
Physics
1 answer:
Mariana [72]3 years ago
6 0

Answer:

The work flow required by the compressor = 100.67Kj/kg

Explanation:

The solution to this question is obtained from the energy balance where the initial and final specific internal energies and enthalpies are taken from A-17 table from the given temperatures using interpolation .

The work flow can be determined using the equation:

M1h1 + W = Mh2

U1 + P1alph1 + ◇U + Workflow = U2 + P2alpha2

Workflow = P2alpha2 - P1alpha1

Workflow = (h2 -U2) - (h1 - U1)

Workflow = ( 684.344 - 491.153) - ( 322.483 - 229.964)

Workflow = ( 193.191 - 92.519)Kj/kg

Workflow = 100.672Kj/kg

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The total distance a group of movers move a refrigerator is 22 m. If they complete this task in 15 minutes, their efforts would
lianna [129]

Answer:

Their efforts would be expressed in units of Joules per second

Explanation:

The unit of their efforts can be derived from the formula of power which is given by the product of mass, acceleration and distance (the product is energy with unit joules) divided by time taken to complete the task (unit is seconds)

Therefore, the unit of their efforts would be joules per second

4 0
3 years ago
Two dump trucks each have a mass of 1,500 kg. The distance of the dump truck
xxTIMURxx [149]

Answer:

6.00 x 10⁻⁸N

Explanation:

Given parameters:

Mass of each dump trucks  = 1500kg

Distance between them  = 50m

Unknown:

New gravitational force between them = ?

Solution:

From Newton's law of universal gravitation,

        F = \frac{G m1 m2}{r^{2} }  

F is the gravitational force

G is the universal gravitation constant

m is the mass

r is the distance

           F  = \frac{6.67 x 10^{-11} x 1500  x 1500}{50^{2} }    = 6.00 x 10⁻⁸N

4 0
2 years ago
A rabbit is hopping along at an approximately constant speed of 3.9 m/s. The rabbit passes a crouched cat ready to chase the rab
Firlakuza [10]

Answer:

t  = 7,8 s

Explanation:

From the instant, the rabbit passes the cat. The cat star running acceleration of 0,5 m/s² .

When the cat arrives at the speed of 3,9 m/s the cat catches the rabbit

Then for the cat arrives at 3,9 m/s nedds

v = vo + a*t     vo  = 0  then   v = a*t

3,9 ( m/s) = 0,5 ( m/s² ) * t

t  = 7,8 s

v  =  3,9 m/s =

4 0
2 years ago
At its closest point, Mercury is approximately 46 million kilometers from the sun. What is this distance in AU?
alina1380 [7]
1 astronomical unit 1 AU = 1.4960 * 10^11 meters
it is the average distance between earth and sun
mercury to sun distance is = 46,000,000 * 1000 meters
               = 4.6 * 10^9 meters = 4.6 * 10^9 / 1.4960 * 10^11  AU
               =  3.0.74 / 100  = 0.0374 AU

5 0
3 years ago
Read 2 more answers
Estimate the volume of a typical house (2050 feet squared in size and 10 feet tall) answer in units of meters squared
Aloiza [94]

Answer:

Volume = 6248.48 m^{3}

Explanation:

Given:

The area of the house A = 2050\ ft^{2}

The height of the house h=10\ ft

We need to find the volume of a typical house.

Solution:

We find the volume of the house by multiplying the area of the house and height of the house.

Volume = Area\times height

Volume = A\times h

Area and height of the house are known, so we substitute these value in above equation.

Volume = 2050\times 10

Volume = 20500\ ft^{3}

Now we convert the unit from feet to meter.

Divide the volume by 3.2808 for m^{3}

Volume = \frac{20500}{3.2808}

Volume = 6248.48\ m^{3}

Therefore, the volume of the house is 6248.48 m^{3}

8 0
3 years ago
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