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Vera_Pavlovna [14]
3 years ago
11

While driving down the highway, your car must overcome the forces of friction and air resistance in order to keep the car moving

. Experiments have indicated that a force of 560N is necessary to keep a car moving at a constant speed. While driving the car a distance of 100,000m , the car burns up 2.8gal of gasoline. If gasoline contains 100,000,000 Joules of potential energy per gallon, what is the efficiency of the car's engine?
Engineering
1 answer:
ElenaW [278]3 years ago
5 0

Answer:

Efficiency of the engine equals 20%

Explanation:

We know that when the car moves it must do work against the resisting forces to keep moving and this work is spend as energy by the engine to keep the car moving.

we know that

Work=Force\times Displacement

Thus to keep the car moving for 100,000 meters the theoretical work that requires to be done equals  

Work=560N\times 100,000m=56\times 10^{6}Joules

Now the actual energy spend by the car equals the energy spend by burning 2.8 gallons of gasoline.

Thus the energy produced by burning 2.8 gallons of gasoline equals

2.8\times 100\times 10^{6}=280\times 10^{6}Joules

Thus the efficiency is calculated as

\eta =\frac{56\times 10^{6}}{280\times 10^{6}}\times 100\\\\\therefore \eta =20percent

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What is the mechanical advantage of a pulley with 3 support ropes?
snow_tiger [21]

Answer:

The mechanical advantage is 3 to 1

Explanation:

A frictionless pulley with three support ropes carries equal tension on each of the ropes thus;

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Gray cast iron, with an ultimate tensile strength of 31 ksi and an ultimate compressive strength of 109 ksi, has the following s
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Using an appropriate failure theory, find the factor of safety in each case. State the name of the theory that you are using the theory is max stress theory.

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In an orthogonal cutting operation, the tool has a rake angle = 12°. The chip thickness before the cut = 0.32 mm and the cut yie
Snezhnost [94]

Answer:

The shear plane angle and shear strain are 28.21° and 2.155 respectively.

Explanation:

(a)

Orthogonal cutting is the cutting process in which cutting direction or cutting velocity is perpendicular to the cutting edge of the part surface.  

Given:  

Rake angle is 12°.  

Chip thickness before cut is 0.32 mm.

Chip thickness is 0.65 mm.  

Calculation:  

Step1  

Chip reduction ratio is calculated as follows:  

r=\frac{t}{t_{c}}

r=\frac{0.32}{0.65}

r = 0.4923

Step2  

Shear angle is calculated as follows:  

tan\phi=\frac{rcos\alpha}{1-rsin\alpha}

Here, \phi is shear plane angle, r is chip reduction ratio and \alpha is rake angle.  

Substitute all the values in the above equation as follows:  

tan\phi=\frac{rcos\alpha}{1-rsin\alpha}

tan\phi=\frac{0.4923cos12^{\circ}}{1-0.4923sin12^{\circ}}

tan\phi=\frac{0.48155}{0.8976}

\phi=28.21^{\circ}

Thus, the shear plane angle is 28.21°.

(b)

Step3

Shears train is calculated as follows:

\gamma=cot\phi+tan(\phi-\alpha)

\gamma=cot28.21^{\circ}+tan(28.21^{\circ}-12^{\circ})\gamma = 2.155.

Thus, the shear strain rate is 2.155.

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