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Leto [7]
3 years ago
9

The water level in a large tank is maintained at height H above the surrounding level terrain. A rounded nozzle placed in the si

de of the tank discharges a horizontal jet of water. The nozzle is located a distance h below the free surface. Neglecting friction and surface tension, determine the range xmax where the stream strikes the ground.

Engineering
1 answer:
Margaret [11]3 years ago
8 0

Answer:

Explanation : Please go through the attached documents for the step by step solution.

Please note that, The file with the diagram is the first one and it is placed first.

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Your boss needs to recommend either the in-situ soil or the borrow material. Although you do not have much background on soil be
Andrej [43]

Answer: see attached file for the answer

7 0
3 years ago
What is Maglev Technology?<br> I don't want a copied and pasted answer please.
Levart [38]

Answer:

I do not know much but I looked and read, this I learned I hope this is good

The Maglev is a system of train transportation that uses two sets of magnets

working repel one another when matching poles face each other. Here, both magnetic attraction and repulsion are used to move the train car along the guideway.

Explanation:

Brainliest please

5 0
3 years ago
(3) Calculate the heat flux through a sheet of brass 7.5 mm (0.30 in.) thick if the temperatures at the two faces are 150°Cand 5
bezimeni [28]

Answer:

a.) 1.453MW/m2,  b.)  2,477,933.33 BTU/hr  c.) 22,733.33 BTU/hr  d.) 1,238,966.67 BTU/hr

Explanation:

Heat flux is the rate at which thermal (heat) energy is transferred per unit surface area. It is measured in W/m2

Heat transfer(loss or gain) is unit of energy per unit time. It is measured in W or BTU/hr

1W = 3.41 BTU/hr

Given parameters:

thickness, t = 7.5mm = 7.5/1000 = 0.0075m

Temperatures 150 C = 150 + 273 = 423 K

                        50 C = 50 + 273 = 323 K

Temperature difference, T = 423 - 323 = 100 K

We are assuming steady heat flow;

a.) Heat flux, Q" = kT/t

K= thermal conductivity of the material

The thermal conductivity of brass, k = 109.0 W/m.K

Heat flux, Q" = \frac{109 * 100}{0.0075} = 1,453,333.33 W/m^{2} \\ Heat flux, Q" = 1.453MW/m^{2} \\

b.) Area of sheet, A = 0.5m2

Heat loss, Q = kAT/t

Heat loss, Q = \frac{109*0.5*100}{0.0075} = 726,666.667W

Heat loss, Q = 726,666.667 * 3.41 = 2,477,933.33 BTU/hr

c.) Material is now given as soda lime glass.

Thermal conductivity of soda lime glass, k is approximately 1W/m.K

Heat loss, Q=\frac{1*0.5*100}{0.0075} = 6,666.67W

Heat loss, Q = 6,666.67 * 3.41 = 22,733.33 BTU/hr

d.) Thickness, t is given as 15mm = 15/1000 = 0.015m

Heat loss, Q=\frac{109*0.5*100}{0.015} =363,333.33W

Heat loss, Q = 363,333.33 * 3.41 = 1,238,966.67 BTU/hr

5 0
3 years ago
A circular column is fixed at the base and not supported at the top. If the column needs to be 15ft and hold 10kips, what is the
muminat

Answer:

The required size of column is length = 15 ft and diameter = 4.04 inches

Explanation:

Given;

Length of the column, L = 15 ft

Applied load, P = 10 kips = 10 × 10³ Psi

End condition as fixed at the base and free at the top

thus,

Effective length of the column, \L_e = 2L = 30 ft = 360 inches

now, for aluminium

Elastic modulus, E = 1.0 × 10⁷ Psi

Now, from the Euler's critical load, we have

P =\frac{\pi^2EI}{L_e^2}

where, I is the moment of inertia

on substituting the respective values, we get

10\times10^3 =\frac{\pi^2\times1.0\times10^7\times I}{360^2}

or

I = 13.13 in⁴

also for circular cross-section

I = \frac{\pi}{64}\times d^4

thus,

13.13 = \frac{\pi}{64}\times d^4

or

d = 4.04 inches

The required size of column is length = 15 ft and diameter = 4.04 inches

3 0
3 years ago
Test de evaluare
ivolga24 [154]
C why it’s c bc they I just got it right
7 0
3 years ago
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