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liraira [26]
3 years ago
11

I WILL GIVE 20 POINTS!!

Engineering
2 answers:
Ludmilka [50]3 years ago
6 0
Use a resume header
Alex777 [14]3 years ago
5 0

Answer:

Use a resume header

Explanation:

Create a Summary

Research industry, employer keywords

there are some hints okay

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A glass tube is inserted into a flowing stream of water with one opening directed upstream and the other end vertical. If the wa
Furkat [3]

Answer:

h=0.46m

Explanation:

From the question we are told that:

Velocity of water V=3m/s

Height=?

Generally, the equation for Water Velocity is mathematically given by

V=\sqrt{2gh}

Therefore Height h is given as

h=\frac{v}{2g}

h=\frac{3^2}{2*9.81}

h=0.46m

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3 years ago
Free brainlist because im new and i just want to but you have t friend me first
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3 years ago
A) For Well A, provide a cross-section sketch that shows (i) ground elevation, (ii) casing height, (iii) depth to
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3 years ago
A medium-sized jet has a 3.8-mm-diameter fuselage and a loaded mass of 85,000 kg. The drag on an airplane is primarily due to th
SCORPION-xisa [38]

Answer:

F_{thrust} ≅ 111 KN

Explanation:

Given that;

A medium-sized jet has a 3.8-mm-diameter i.e diameter (d) = 3.8

mass = 85,000 kg

drag co-efficient (C) = 0.37

(velocity (v)= 230 m/s

density (ρ) = 1.0 kg/m³

To calculate the thrust; we need to determine the relation of the drag force; which is given as:

F_{drag} = \frac{1}{2} × CρAv²

where;

ρ = density of air wind.

C = drag co-efficient

A = Area of the jet

v = velocity of the jet

From the question, we can deduce that the jet is in motion with a constant speed; as such: the net force acting on the jet in the air = 0

SO, F_{drag}-F_{thrust} = 0

We can as well say:

F_{drag}= F_{thrust}

We can now replace F_{thrust} with F_{drag} in the above equation.

Therefore, F_{thrust} = \frac{1}{2} × CρAv²

The A which stands as the area of the jet is given by the formula:

A=\frac{\pi d^2}{4}

We can now have a new equation after substituting our A into the previous equation as:

F_{thrust} = \frac{1}{2} × Cρ (\frac{\pi d^2}{4})v^2

Substituting our data from above; we have:

F_{thrust} = \frac{1}{2} × (0.37)(1.0kg/m^3)(\frac{\pi(3.8m)^2 }{4})(230m/s)^2

F_{thrust} = \frac{1}{8}   (0.37)(1.0kg/m^3)({\pi(3.8m)^2 })(230m/s)^2

F_{thrust} = 110,990N

F_{thrust}  in N (newton) to KN (kilo-newton) will be:

F_{thrust} = (110,990N)*\frac{1KN}{1,000N}

F_{thrust} = 110.990 KN

F_{thrust} ≅ 111 KN

In conclusion, the jet engine needed to provide 111 KN thrust in order to cruise at 230 m/s at an altitude where the air density is 1.0 kg/m³.

5 0
3 years ago
How to calculate tension.
Evgen [1.6K]

Answer:

Tension can be easily explained in the case of bodies hung from chain, cable, string

Explanation

uniform speed, tension; T = W.

T=m(g±a)

3 0
2 years ago
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