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fiasKO [112]
2 years ago
5

In order to keep an automobile operating, it is necessary to keep adding fuel as it is used up. Explain why this doesn't contrad

ict the law of conservation of energy.
Engineering
1 answer:
Semmy [17]2 years ago
3 0

Answer:

There is conversion of energy from one form to another and this justifies the law of conservation of energy, which states that energy can neither be created nor destroyed but can be converted from one form to another.

Explanation:

The fuel added to automobile is a chemical energy, which provides thermal energy of the combustion engine of the automobile, which is then converted to mechanical energy of the moving parts of the automobile. Thus, there is conversion of energy from one form to another. This justifies the law of conservation of energy, which states that energy can neither be created nor destroyed but can be converted from one form to another.

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the AADT for a section of suburban freeway is 150000 veh/day. Assuming this is an urban radial facility, what range of direction
igomit [66]

Answer:

design hour volumes will be 4000 to 6000

Explanation:

given data

AADT  = 150000 veh/day

solution

we get here design hour volumes that is express as

design hour volumes  = AADT × k × D    ..............1

here k is factor and its  range is 8 to 12 % for urban

and D is directional distribution i.e traffic equal divided by the direction

so here design hour volumes will be 4000 to 6000

7 0
3 years ago
A smooth ceramic sphere (SG 5 2.6) is immersed in a fl ow of water at 208C and 25 cm/s. What is the sphere diameter if it is enc
Aleks [24]

Answer:

a. 4\mu m

b. 1 m

Explanation:

According to the question, the data is as follows

The Density of water at 20 degrees celcius is 1000 kg/m^3

Viscosity is 0.001kg/m/.s

Velocity V = 25 cm/s

V = 0.25 m/s

Now

a. The creeping motion is

As we know that

Reynold Number = (Density of water × V × d) ÷ (Viscosity)

1 = (1,000 × 0.25 × d) ÷ 0.0001

d = (1 × 0.001) ÷ (1,000 × 0.25)

= 4E - 06^m

= 4\mu m

b. Now the sphere diameter is

Reynold Number = (Density of water × V × d) ÷ (Viscosity)

250,000 = (1,000 × 0.25 × d) ÷ 0.0001

d = (250,000 × 0.001) ÷ (1,000 × 0.25)

= 1 m

6 0
3 years ago
(a) A non-cold-worked 1040 steel cylindrical rod has an initial length of 100 mm and initial diameter of 7.50 mm. is to be defor
serg [7]

Answer:

A) 1040 steel is not a possible candidate for this application

B) 35.94%

Explanation:

Initial length = 100 mm =  0.1 m

Initial diameter ( d ) = 7.5 mm = 0.0075 m

Tensile load ( p ) = 18,000 N

Condition : The 1040 steel must not experience plastic deformation or a diameter reduction of more than 1.5 * 10^-5 m

<u>A) would the 1040 steel be a possible candidate for this application</u>

<em>Yield strength of 1040 steel < stress  ( in order to be a possible candidate )</em>

stress = p / A0 = ( 18000 ) / ( \frac{\pi }{4} ) * 0.0075^2

                      = 18,000 / (4.418 * 10^-5 )   =  407.424 MPa

Yield strength of 1040 steel = 450 MPa

stress = 407.424 MPa

∴ Yield strength ( 450 MPa ) > stress ( 407.424 MPa )  

Therefore 1040 steel is not a possible candidate for this application

<u>B) Determine How much cold work would be required to reduce the diameter of the steel to 6.0 mm</u>

Area1 = ( \frac{\pi }{4} ) ( 0.006 )^2 = 2.83 * 10^-5 m^2

therefore % of cold work done = ( A0 - A1 ) / A0  * 100 = 35.94%

6 0
3 years ago
Water at a pressure of 3 bars enters a short horizontal convergent channel at 3.5 m/s. The upstream and downstream diameters of
earnstyle [38]

Answer:

The pressure reduces to 2.588 bars.

Explanation:

According to Bernoulli's theorem for ideal flow we have

\frac{P}{\gamma _{w}}+\frac{V^{2}}{2g}+z=constant

Since the losses are neglected thus applying this theorm between upper and lower porion we have

\frac{P_{u}}{\gamma _{w}}+\frac{V-{u}^{2}}{2g}+z_{u}=\frac{P_{L}}{\gamma _{w}}+\frac{V{L}^{2}}{2g}+z_{L}

Now by continuity equation we have

A_{u}v_{u}=A_{L}v_{L}\\\\\therefore v_{L}=\frac{A_{u}}{A_{L}}\times v_{u}\\\\v_{L}=\frac{d^{2}_{u}}{d^{2}_{L}}\times v_{u}\\\\\therefore v_{L}=\frac{2500}{900}\times 3.5\\\\\therefore v_{L}=9.72m/s

Applying the values in the Bernoulli's equation we get

\frac{P_{L}}{\gamma _{w}}=\frac{300000}{\gamma _{w}}+\frac{3.5^{2}}{2g}-\frac{9.72^{2}}{2g}(\because z_{L}=z_{u})\\\\\frac{P_{L}}{\gamma _{w}}=26.38m\\\\\therefore P_{L}=258885.8Pa\\\\\therefore P_{L}=2.588bars

6 0
3 years ago
A water supply agency is planning to add two reservoirs to its system. Water will flow from Reservoir A to Reservoir B via a 10,
NikAS [45]

Attached is the solution to the above question.

3 0
3 years ago
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