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3 years ago

How many types of engineering specialist are there?

1 answer:

3 0

There are now six major branches of engineering: Mechanical, Chemical, Civil, Electrical, Management, and Geotechnical, and literally hundreds of different subcategories of engineering under each branch.

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Firefighters are holding a nozzle at the end of a hose while trying to extinguish a fire. The nozzle exit diameter is 8 cm, and

ivanzaharov [21]

Question

Determine the average water exit velocity

Answer:

53.05 m/s

Explanation:

Given information

Volume flow rate, $Q=16 m^{3}/min$

Diameter d= 8cm= 0.08 m

Assumptions

The flow is jet flow hence momentum-flux correction factor is unity
Gravitational force is not considered
The flow is steady, frictionless and incompressible
Water is discharged to the atmosphere hence pressure is ignored

We know that Q=AV and making v the subject then

$V=\frac {Q}{A}$ where V is the exit velocity and A is area

Area, $A=\frac {\pi d^{2}{4}$ where d is the diameter

By substitution

$V=\frac {16\times 4}{\pi 0.08^{2}}=3183.098862 m/min$

To convert v to m/s from m/s, we simply divide it by 60 hence

$V=\frac {3183.098862 m/min}{60 s}=53.0516477 m/s\approx 53.05 m/s$

3 0

3 years ago

Gas is kept in a 0.1 m diameter cylinder under the weight of a 100 kg piston that is held down by a spring with a stiffness k =

Artyom0805 [142]

Answer:

The spring is compressed by 0.275 meters.

Explanation:

For equilibrium of the gas and the piston the pressure exerted by the gas on the piston should be equal to the sum of weight of the piston and the force the spring exerts on the piston

Mathematically we can write

$Force_{pressure}=Force_{spring}+Weight_{piston}$

we know that

$Force_{pressure}=Pressure\times Area=300\times 10^{3}\times \frac{\pi \times 0.1^2}{4}=750\pi Newtons$

$Weight_{piston}=mass\times g=100\times 9.81=981Newtons$

Now the force exerted by an spring compressed by a distance 'x' is given by $Force_{spring}=k\cdot x=5\times 10^{3}\times x$

Using the above quatities in the above relation we get

$5\times 10^{3}\times x+981=750\pi \\\\\therefore x=\frac{750\pi -981}{5\times 10^{3}}=0.275meters$

5 0

3 years ago

A cylinder with a 6.0 in. diameter and 12.0 in. length is put under a compres-sive load of 150 kips. The modulus of elasticity f

jeka94

Answer:

Final Length = 11.992 in

Final Diameter = 6.001 in

Explanation:

First we calculate the cross-sectional area:

Area = A = πr² = π(3 in)² = 28.3 in²

Now, we calculate the stress:

Stress = Compressive Load/Area

Stress = - 150 kips/28.3 in²

Stress = -5.3 ksi

Now,

Modulus of Elasticity = Stress/Longitudinal Strain

8000 ksi = -5.3 ksi/Longitudinal Strain

Longitudinal Strain = -6.63 x 10⁻⁴

but,

Longitudinal Strain = (Final Length - Initial Length)/Initial Length

-6.63 x 10⁻⁴ = (Final Length - 12 in)/12 in

Final Length = (-6.63 x 10⁻⁴)(12 in) + 12 in

<u>Final Length = 11.992 in</u>

we know that:

Poisson's Ratio = - Lateral Strain/Longitudinal Strain

0.35 = - Lateral Strain/(- 6.63 x 10⁻⁴)

Lateral Strain = (0.35)(6.63 x 10⁻⁴)

Lateral Strain = 2.32 x 10⁻⁴

but,

Lateral Strain = (Final Diameter - Initial Diameter)/Initial Diameter

2.32 x 10⁻⁴ = (Final Diameter - 6 in)/6 in

Final Diameter = (2.32 x 10⁻⁴)(6 in) + 6 in

<u>Final Diameter = 6.001 in</u>

8 0

3 years ago

I'll mark brainliest plz help

Citrus2011 [14]

Answer:

Explanation:

There are three points in time we need to consider. At point 0, the mango begins to fall from the tree. At point 1, the mango reaches the top of the window. At point 2, the mango reaches the bottom of the window.

We are given the following information:

y₁ = 3 m

y₂ = 3 m − 2.4 m = 0.6 m

t₂ − t₁ = 0.4 s

a = -9.8 m/s²

t₀ = 0 s

v₀ = 0 m/s

We need to find y₀.

Use a constant acceleration equation:

y = y₀ + v₀ t + ½ at²

Evaluated at point 1:

3 = y₀ + (0) t₁ + ½ (-9.8) t₁²