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3 years ago

Use the following assumptions for problems 1 and 2:

Engineering

1 answer:

Salsk061 [2.6K]3 years ago

6 0

Answer:

The text file attached has the detailed solution of all the parts individually.

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A laboratory in the Y building keep a vacuum pressure of 0.1 kPa abs. What is the net force acting on the door considering the a

seropon [69]

Answer:

net force acting on the floor is 100 kN

Explanation:

Given data:

$P_{vaccum} = 0.1 kPa$

$P_{atm} = 101.325 kPa$

dimension of floor = 2 m \times 0.5 m

we know that

Net force can be calculated as follow

$f_{net} = P_{vaccum} \times area$

$f_{net} = 0.1\times 10^3 \times 2\times 0.5$

$f_{net} = 0.1\times 10^3 \times 1$

$f_{net} = 100 kN$

Therefore net force acting on the floor is 100 kN

7 0

4 years ago

Given a reservoir watershed of 22 square miles, and assuming a

saveliy_v [14]

Answer:

308 acre-ft of water

Explanation:

Given:

Area of the watershed = 22 square miles

Depth of Rainfall = 0.75 in = $\frac{\textup{0.75}}{\textup{12}}$ =0.0625 ft

Percentage rainfall falling in reservoir as runoff = 35%

Now,

1 square mile = 640 acre

Thus,

22 square miles = 22 × 640 = 14,080 acres

Thus,

The total volume of rainfall = Area of watershed × Depth of the rainfall

The total volume of rainfall = 14,080 acres × 0.0625 ft = 880 acre-ft

also,

only 35% of the total rainfall is contributing as runoff

thus,

Runoff = 0.35 × 880 acre-ft = 308 acre-ft of water

8 0

3 years ago

Discuss typical advantages and disadvantages of an irrigation system?

trapecia [35]

Advantages include low costs and minimal labor.Water stays in the root zone, and foliage stays dry. Drawbacks to surface irrigation include potential overwatering and wasteful runoff.

4 0

3 years ago

Khái niệm về môi trường nhiệt nóng và môi trường nhiện lạnh ?

I am Lyosha [343]

Explanation:

उह्ह्नमजज्ल्ह्ह्बनुतनकुहक्जो

7 0

3 years ago

A steel rod, which is free to move, has a length of 200 mm and a diameter of 20 mm at a temperature of 15oC. If the rod is heate

kherson [118]

Explanation:

thermal expansion ∝L = (δL/δT)÷L ----(1)

δL = L∝L + δT ----(2)

we have δL = 12.5x10⁻⁶

length l = 200mm

δT = 115°c - 15°c = 100°c

putting these values into equation 1, we have

δL = 200*12.5X10⁻⁶x100

= 0.25 MM

L₂ = L + δ L

= 200 + 0.25

L₂ = 200.25mm

12.5X10⁻⁶ *115-15 * 20

= 0.025

20 +0.025

D₂ = 20.025

as this rod undergoes free expansion at 115°c, the stress on this rod would be = 0

3 0

3 years ago