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KIM [24]
3 years ago
7

A fire nozzle is coupled to the end of a hose with inside diameter D = 77 mm. The nozzle is smoothly contoured and its outlet di

ameter is d = 33 mm. The nozzle is designed to operate at an inlet water pressure of 744 kPa (gage). (a) Determine the design flow rate of the nozzle. (Express your answer in L/s.) (b) Evaluate the axial force required to hold the nozzle in place. (c) Indicate whether the hose coupling is in tension or compression.

Engineering
1 answer:
SashulF [63]3 years ago
4 0

Answer:

Explanation:

Axial Force is normally referred to as the Force acting along the path of the module or assembly. For instance, let’s consider a cylindrical Square column or Building column. Now the column is a structural part of the whole structure and it is planned and designed to take Axial Compression force.

the diagram and step by step explanation in solving this question is in the attached images below

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A communication systems

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What is the primary function of NCEES?
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Answer:

It is a non profit organization that dedicates to licensing professional engineers and surveyors

Explanation:

6 0
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I = 48 mA, R = 1125 2. What is Vs ?
german
54 volts

Ohms law. E= I x R
6 0
4 years ago
Find the number of Btu conducted through a wall in 8 hours. The wall is 8 feet high by 24 feet long and has a total R-value of 1
dedylja [7]

Answer:

ΔQ = 4930.37 BTu

Explanation:

given data

height h = 8ft

Δt = 8  hours

length L = 24 feet

R value = 16.2 hr⋅°F⋅ft² /Btu

inside temperature t1 = 68°F

outside temperature t2 = 16°F

to find out

number of Btu conducted

solution

we get here number of Btu conducted by this expression that s

\frac{\Delta Q}{\Delta t} =\frac{-A}{R} (t2 -t1)     ......................1

here A is area that is = h × L = 8 × 24 = 1492 ft²

put here value we get

\frac{\Delta Q}{8} =\frac{-192}{16.2} (16-68)

solve it we get

ΔQ = 4930.37 BTu

7 0
3 years ago
soccer is also called association football" A soccer ball is a sphere, with circumference of 70 centimeters. in developing a new
timama [110]

Answer: Weight on Mars = 0.02593N

Explanation:

Given; Circumference C of Sphere = 70cm = 0.7m,

Specific Gravity S. G. of material = 1.21,

acceleration due to gravity in the Mars gm = 3.7m/s^2

We know that Weight W = mass m × acceleration due to gravity.

Let the Weight in on the Mars be Wm.

Wm = m × gm

Since we are given gm, we need to calculate for m. (Note that mass m is the same everywhere)

But mass = specific gravity × volume

Since we know the specific gravity, let's go ahead to calculate for the volume of the ball.

We know that Volume of a Sphere V = (4/3)πr^3

To get r, we know that C = 2πr

Therefore, r = C/(2π) = 0.7/(2π) = (7/10)/2π = 7/20π (in meters)

V = (4/3)*π×(7/20π)^3 = 343/6000π^2 (in meter^3)

m = 343/6000π^2 × 1.21 = 7.01×10^(-3)kg

Wm = 7.01×10^(-3) × 3.7 = 0.02593N

8 0
3 years ago
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