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2 years ago

Ignore swell and shrinkage for this problem.

Engineering

1 answer:

Komok [63]2 years ago

6 0

Answer:

(a) 5.3

(b) 6

Explanation:

First we calculate time required for excavator to fill 1 truck:

Time to Fill 1 Truck = (No. of Cycles To Fill)(Cycle Time)

Time to Fill 1 Truck = (3)(33 sec)

Time to Fill 1 Truck = 99 sec

Now, we find the time taken by a truck to come back to excavator, after dumping:

Time of Return = Hauling Time + Time to Dump + Time to Come Back

Time of Return = 5 min + 45 sec + 3 min

Time of Return = 300 sec + 45 sec + 180 sec

Time of Return = 525 sec

(a)

In order to balance the output of excavator with number of trucks, following condition must be satisfied:

Time of Return = (No. of Trucks)(Time to Fill 1 Truck)

525 sec = (No. of Trucks)(99 sec)

No. of Trucks = 525 sec/99 sec

No. of Trucks = 5.3

(b)

If we round up the number of trucks, we get:

No. of Trucks = 6

(c)

If we round down the number of trucks, we get:

No. of Trucks = 5

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2 years ago

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2 years ago

Swing arm restraints are intended to prevent a vehicle from falling off a lift.

castortr0y [4]

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2 years ago

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2 years ago

How much cornfield area would be required if you were to replace all the oil consumed in the United States with ethanol from cor

zaharov [31]

Answer:

2377.35 km

Explanation:

Given the following;

1. A cornfield is 1.5% efficient at converting radiant energy into stored chemical potential energy;

2. The conversion from corn to ethanol is 17% efficient;

3. A 1.2:1 ratio for farm equipment to energy production

4. A 50% growing season and,

5. 200 W/m2 solar insolation.

As per our assumptions,1.2/1 is the ratio for farm equipment to energy production,

So USA need around 45.45% (1/(1+1.2) replacement of fuel energy production which is nearly about = 0.4545*10^{20} J/year = \frac{0.4545*10^{20}}{365*24*3600}=1.44121*10^{12} J/sec

Growing season is only part of year ( Given = 50%),

Net efficiency = 1.5%*17%*50%=0.015*0.17*0.5=0.001275 = 0.1275%

Hence , Actual Energy replacement (Efficiency),

=\frac{1.44121*10^{12}}{0.001275} = 1.13*10^{15} J/sec=1.13*10^{15} W

As per assumption (5),

\because 200 W/m2 solar insolation arequired,

So USA required corn field area = 1.13*10^{15}/200 = 5.65*10^{12} m^{2}

Hence, length of each side of a square,

= (5.65*10^{12} )^{0.5} = 2377.35 km

4 0

3 years ago