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Anna007 [38]
3 years ago
3

A river barge, whose cross section is approximately rectangular, carries a load of grain. The barge is 27 ft wide and 93 ft long

. When unloaded its draft (depth of submergence) is 5 ft, and with the load of grain the draft is 7 ft. Determine: (a) the unloaded weight of the barge, and (b) the weight of the grain.

Engineering
1 answer:
Monica [59]3 years ago
8 0

Answer:

a) for equilibrium

sum of F vertical is zero

F_{vertical} =0

so that

W_{b} =F_{B}=r_{h2o} *(submerged volume)\\=(62.4lb/ft^3)(5ft*27ft*93ft)\\=783432 lb

b)

F_{vertical} =0\\W_{b}+ W_{mg} =r_{h2o} *(submerged volume)\\W_{g} =(62.4lb/ft^3)*(7ft*27ft*93ft)-783432\\W_{g} =313372lb

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Norma-Jean [14]

Answer:

An estimated amount of $8,200 to $12,000

Explanation:

~~It all really depends on what materials you have purchased for the replacement.

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(The answer is if you're most likely to purchase asphalt shingles)

Hope this Helps!

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2 years ago
You don't know which insert you have, and the inserts are different sizes, meaning the amount needed for a 1:3 ratio is differen
VashaNatasha [74]

Answer:

Explanation:

For ligation process the 1:3 vector to insert ratio is the good to utilize . By considering that we can take 1 ratio of vector and 3 ratio of insert ( consider different insert size ) and take 10 different vials of ligation ( each calculated using different insert size from low to high ) and plot a graph for transformation efficiency and using optimum transformation efficiency we can find out the insert size.

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3 years ago
In a surface grinding operation, the wheel diameter = 8.0 in, wheel width = 1.0 in, wheel speed = 6000 ft/min, work speed = 40 f
Katen [24]

Answer: the answer will be d because it is the right one to be

Explanation:

7 0
3 years ago
Two common methods of improving fuel efficiency of a vehicle are to reduce the drag coefficient and the frontal area of the vehi
qaws [65]

Answer:

\Delta V = 209.151\,L, \Delta C = 217.517\,USD

Explanation:

The drag force is equal to:

F_{D} = C_{D}\cdot \frac{1}{2}\cdot \rho_{air}\cdot v^{2}\cdot A

Where C_{D} is the drag coefficient and A is the frontal area, respectively. The work loss due to drag forces is:

W = F_{D}\cdot \Delta s

The reduction on amount of fuel is associated with the reduction in work loss:

\Delta W = (F_{D,1} - F_{D,2})\cdot \Delta s

Where F_{D,1} and F_{D,2} are the original and the reduced frontal areas, respectively.

\Delta W = C_{D}\cdot \frac{1}{2}\cdot \rho_{air}\cdot v^{2}\cdot (A_{1}-A_{2})\cdot \Delta s

The change is work loss in a year is:

\Delta W = (0.3)\cdot \left(\frac{1}{2}\right)\cdot (1.20\,\frac{kg}{m^{3}})\cdot (27.778\,\frac{m}{s})^{2}\cdot [(1.85\,m)\cdot (1.75\,m) - (1.50\,m)\cdot (1.75\,m)]\cdot (25\times 10^{6}\,m)

\Delta W = 2.043\times 10^{9}\,J

\Delta W = 2.043\times 10^{6}\,kJ

The change in chemical energy from gasoline is:

\Delta E = \frac{\Delta W}{\eta}

\Delta E = \frac{2.043\times 10^{6}\,kJ}{0.3}

\Delta E = 6.81\times 10^{6}\,kJ

The changes in gasoline consumption is:

\Delta m = \frac{\Delta E}{L_{c}}

\Delta m = \frac{6.81\times 10^{6}\,kJ}{44000\,\frac{kJ}{kg} }

\Delta m = 154.772\,kg

\Delta V = \frac{154.772\,kg}{0.74\,\frac{kg}{L} }

\Delta V = 209.151\,L

Lastly, the money saved is:

\Delta C = \left(\frac{154.772\,kg}{0.74\,\frac{kg}{L} }\right)\cdot (1.04\,\frac{USD}{L} )

\Delta C = 217.517\,USD

4 0
3 years ago
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alex41 [277]

Answer:

Gross building area

Explanation:

The Gross building area refers to the entire area of a building covering all the floors. The measurement is expressed in square feet. The Gross building area also includes basements, penthouses, and mezzanines. It is calculated by estimating the exterior dimension of the building. Storage rooms, laundries, staircases are also a part of the gross building area.

6 0
3 years ago
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