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inessss [21]
3 years ago
15

Explain why the failure of a garden hose occurred near its end and why the tear occurred along its length. Use numerical values

to explain your result. Assume the water pressure is 30 psi.
Engineering
1 answer:
Nataliya [291]3 years ago
7 0

Answer:

Most hydraulic systems develops pressure surges that may surpass settings valve. by exposing the hose surge to pressure above the maximum operating pressure will shorten the hose life.

Explanation:

Solution

Almost all hydraulic systems creates pressure surges that may exceed relief valve settings. exposing the hose surge to pressure above the maximum operating  pressure shortens the hose life.

In systems where pressure peaks are severe, select or pick a hose with higher maximum operating  pressure or choose a spiral reinforced hose specifically designed for severe pulsing applications.

Generally, hoses are designed or created to accommodate pressure surges and have operating pressures that is equal to 25% of the hose minimum pressure burst.

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What steel type and ASTM designation is preferred for W-shapes?
Ulleksa [173]

Answer:

The preferred steel type for W-shapes is structural steel and the its preferred ASTM designation is ASTM A992.

Explanation:

The ASTM A992 is a structural steel and it's the most available for w-shapes; besides its availabilty, its ductility improvements makes it the preferred choice; other common designations for this shapes are ASTM A572 Grade 50,0r ASTM A36, but this designations aren't as available as ASTM A992, and it has to be confirmed prior to their specification.

3 0
4 years ago
Find the magnitude of the two pulling forces P and Q when their resultant is 50 N at 20° with Q. P 20°​
morpeh [17]
Two forces P and Q whose resultant is 10Newton are at right angles to each other. If P makes 30 degrees with resultant.Show me the workings of the magnitude of Q in Newton and the diagram of the vectors.
7 0
3 years ago
A 12-ft high retaining wall has backfill of granular soil with an internal angle of friction of 30 and unit weight of 125 pef. W
irakobra [83]

Answer:

P_p = 27000 psf

Explanation:

given,

height of the retaining wall = h = 12 ft

internal angle of friction (∅)= 30°

unit weight = 125 pcf

Rankine passive earth pressure = ?

k_p is the coefficient of passive earth pressure

k_p = \dfrac{1 + sin\phi}{1 - sin\phi}

k_p = \dfrac{1 + sin30^0}{1 - sin30^0}

k_p = 3

Passive earth pressure

P_p = \dfrac{1}{2}k_p \gamma H^2

P_p = \dfrac{1}{2}\times 3\times 125 \times 12^2

      P_p = 27000 psf

Rankine passive earth pressure on the wall is equal to P_p = 27000 psf

7 0
3 years ago
What are two factors that determine the thermal energy of a substance
Sav [38]
Mass and chemical composition
6 0
3 years ago
A soil weighs 2,520 lbs/CY in its in situ condition, 1,970 lb/CY in its loose condition after excavation, and 3,025 lbs/CY in an
azamat

Answer:

load factor = 0.782

Shrink Factor = 0.833

no of truck is  62500

Explanation:

given data

soil weighs in situ condition  =  2,520 lbs/CY

soil weighs in loose condition  =  1,970 lb/CY

soil weighs in embanked state = 3,025 lbs/CY

average volume = 16 LCY

soil from a borrow pit = 1 million CCY

solution

first we get here Load Factor that is express as

load factor = \frac{1,970}{2550}

load factor = 0.782

and Shrink Factor will be as

Shrink Factor = \frac{2520}{3025}

Shrink Factor = 0.833

and

no of truck will be

no of truck = \frac{1000000}{16}

no of truck is  62500

6 0
4 years ago
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