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Sauron [17]
3 years ago
9

14. The load across a 50.0-V battery consists of a

Engineering
1 answer:
Paladinen [302]3 years ago
7 0
The answer you’re looking for is
A.
- I hope this helps, I got it right. Enjoy the rest of your day/night, please make me brainliest!
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For powder compaction using a single-action punch, derive an expression for the distribution of axial pressure within a die of r
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a) 2∪p/lb (l+b)dH

b) po exp( 4∪x/l)

Explanation:

please check the attachment for proper explanation and proper sign notations thanks.

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Which of the following is a valid reason why a scientist might reject a scientific theory
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3 years ago
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How many trips would one rubber-tired Herrywampus have to make to backfill a space with a geometrical volume of 5400 cubic yard?
nikklg [1K]

Answer:

If analyzed by volume capacity, more trips are needed to fill the space, thus the required trips are 288

Explanation:

a) By volume.

The shrinkage factor is:

\frac{5400cu-yd}{1-0.25} =7200cu-yd

The volume at loose is:

V_{loose} =V_{bank} (1+swell-factor)=7200(1+0.2)=8640cu-yd

If the Herrywampus has a capacity of 30 cubic yard:

\frac{8640cu-yd}{30cu-yd/trip} =288trip

b) By weight

The swell factor in terms of percent swell is equal to:

pounds-per-cubic-yard-loose=\frac{pounds-per-cubic-yard-bank}{\frac{percent-swell}{100}+1 }

pounds-per-cubic-yard-loose=\frac{3000}{\frac{20}{100} +1} =2500lb/cu-yd

The weight of backfill is:

8640cu-yd*2500\frac{lb}{cu-yd} *\frac{1ton}{2000lb} =10800ton

The Herrywampus has a capacity of 40 ton:

\frac{10800}{40ton/trip} =270trip

If analyzed by volume capacity, more trips are needed to fill the space, thus the required trips are 288

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Javier is well versed in computer-aided design and is adept at spatial reasoning. What other skill would be
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