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3 years ago

Introduction for site visit

Engineering

1 answer:

Stella [2.4K]3 years ago

4 0

Answer:

Site visit is when an external evaluation team goes to an institution to evaluate verbal, written and visual evidence. explanatory context. Site visits, which often last several days, are part of the quality evaluation process.

Explanation:

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When an airplane is flying 200 mph at 5000-ft altitude in a standard atmosphere, the air velocity at a certain point on the wing

Alexandra [31]

Answer:

Pressure = $P2=73.13 lbf/ft^2$

Explanation:

The pressure is calculated by Bernoulli's equation.

$P1+\frac{1}{2} dVi^2=P2+\frac{1}{2} dVf^2\\$

Solving it for P2

$P2=P1 +\frac{1}{2} d(Vi^2-Vf^2)$

Now inserting values

P1 = 0 as it is given that initial point is at atmospheric point.

d= density = 2.05 sl/ft3

Vi = 200 mph = 293.33 ft/s

Vf = 273 mph = 400.4 ft/s

Putting these values

$P2 = 0 + 0.5* 2.05(293.33^2-400.4^2)\\P2=-76sl/ft.s^2$

Changing the units to pounds per square foot

$P2=73.13 lbf/ft^2$

7 0

4 years ago

The tank shown in the accompanying figure is being filled by pipes 1 and 2. If the water level is to remain constant, what is th

VMariaS [17]

Answer: 2.93 ft/sec

Explanation: Calculate the volume/sec entering from the two inlets (Pipes 1 and 2), add them, and then calculate the flow in Pipe 3.

The table illustrates the approach. I calculated the volume of each pipe for a 1 foot section with the indicated diameters, divided by 2 for the radius of each using V = πr²h. Units of V are in^3/foot length. Now we can multiply that volume by the flow rate, in ft/sec, to obtain the flow rate in in^3/sec.

Add the two rates from Pipes 1 and 2 (62.14 in^3/sec) to arrive at the flow rate for Pipe 3 necessary to keep the water level constant. Calculate the volume of 1 foot of Pipe 3 (21.21 in^3/foot) and then divide this into the inflow sum of 62.14 in^3/sec to find the flow rate of Pipe 3 (in feet/sec) necessary to keep the water level constant.

That is 2.93 ft/sec.

6 0

3 years ago

Technician A says that the connecting rod and main bearing caps should be marked before removing to ensure that they can be inst

Rama09 [41]

Answer:

Technician A

Explanation:

This is because when the dimacations are made, the surface area and length respectively can be maintained to reduce to idea of force which may lead to damage

3 0

2 years ago

Convert the following pairs of voltage and current waveforms to phasor form. Each pair of waveforms corresponds to an unknown el

Lisa [10]

Answer:

The solutions are attached below:

Explanation:

3 0

3 years ago

Calculate the reluctance of a 4-meter long toroidal coil made of low-carbon steel with an inner radius of 1.75 cm and an outer r

My name is Ann [436]

Answer:

R = 31.9 x 10^(6) At/Wb

So option A is correct

Explanation:

Reluctance is obtained by dividing the length of the magnetic path L by the permeability times the cross-sectional area A

Thus; R = L/μA,

Now from the question,

L = 4m

r_1 = 1.75cm = 0.0175m

r_2 = 2.2cm = 0.022m

So Area will be A_2 - A_1

Thus = π(r_2)² - π(r_1)²

A = π(0.0225)² - π(0.0175)²

A = π[0.0002]

A = 6.28 x 10^(-4) m²

We are given that;

L = 4m

μ_steel = 2 x 10^(-4) Wb/At - m

Thus, reluctance is calculated as;

R = 4/(2 x 10^(-4) x 6.28x 10^(-4))

R = 0.319 x 10^(8) At/Wb

R = 31.9 x 10^(6) At/Wb

8 0

4 years ago

Introduction for site visit​

Introduction for site visit