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trasher [3.6K]
2 years ago
5

1. What's the maximum overall length of the part? 2. What material is used to fabricate the part

Engineering
1 answer:
kiruha [24]2 years ago
8 0

Answer:

1. The maximum length overall length of the part is 4 inches

2. The material with which the part is fabricated is 0.25 inch thick Hot Rolled Steel

Explanation:

The information given in the drawing are;

The angle of projection of the drawing = Third angle projection

The thickness of the material = 0.25" plate

The material with which the part is fabricated = 0.25 inch thick HRS

HRS is the acronym for Hot Rolled Steel

Therefore;

<u>The material with which the part is fabricated is 0.25 inch thick Hot Rolled Steel</u>

The scale of the drawing = 1:1 = which is 1 is to 1 drawing

The maximum length overall length of the part, l = 4 units

By visually testing the 1:1 scale drawing measurements of 4 units is comparable 4 inches measurement with an online ruler

Therefore;

<u>The maximum length overall length of the part, l = 4 inches</u>

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castortr0y [4]

Yes, the green LED emits light when we connect the banana plug wires across it.

This indicates that the threshold voltage is lower for the green LED light than the blue.

<u>Explanation</u>:

A banana plug is named for its resemblance to the shape of a banana. They are wider in the middle of the plug, and narrower at the top and bottom. The banana plug can be easily plugged or unplugged into the ports of speaker or receiver.

LED lights emit photons when it is applied with electrical charge. LED lights are more efficient and last longer than incandescent light bulbs. Green light is commonly provides the calming effect. It is generally used in hyper-pigmentation treatment.

4 0
2 years ago
A program is seeded with 30 faults. During testing, 21 faults are detected, 15 of which are seeded faults and 6 of which are ind
Vesna [10]

Answer:

Estimated number of indigenous faults remaining undetected is 6

Explanation:

The maximum likelihood estimate of indigenous faults is given by,

N_F=n_F\times \frac{N_S}{n_S} here,

n_F = the number of unseeded faults = 6

N_S = number of seeded faults = 30

n_s = number of seeded faults found = 15

So NF will be calculated as,

N_F=6\times \frac{30}{15}=12

And the estimate of faults remaining is  N_F-n_F = 12 - 6 = 6

8 0
3 years ago
How will the delay and active power per device change as you increase the doping density of both the N- and the P-MOSFET?
Murljashka [212]

Answer:

hello your question is incomplete attached below is the missing part of the  question

Consider an inverter operating a power supply voltage VDD. Assume that matched condition for this inverter. Make the necessary assumptions to get to an answer for the following questions.

answer : Nd ∝ rt

Explanation:

Determine how the delay and active power per device will change as the doping density of N- and P-MOSFET increases

Pactive ( active power ) = Efs * F

Pactive = \frac{q^2Nd^2*Xn^2}{6Eo} * f

also note that ; Pactive ∝ Nd2 (

tD = K . \frac{Vdd}{(Vdd - Vt )^2}  since K = constant

Hence : Nd ∝ rt

5 0
3 years ago
I want a problems and there solutions of The inception of cavitation?​
Ugo [173]

Answer:

The overview of the given scenario is explained in explanation segment below.

Explanation:

  • The inception of cavitation, that further sets the restriction for high-pressure and high-free operation, has always been the matter of substantial experimental study over the last few generations.
  • Cavitation inception would be expected to vary on the segment where the local "PL" pressure mostly on segment keeps falling to that are below the "Pv" vapor pressure of the fluid and therefore could be anticipated from either the apportionment of the pressure.

    ⇒  A cavitation number is denoted by "σ" .

4 0
3 years ago
A single fixed pulley is used to lift a load of 400N by the application of an effort of 480N in 10s through a vertical height of
Allushta [10]

Answer:

(a) the velocity ratio of the machine (V.R) = 1

(b) The mechanical advantage of the machine (M.A) = 0.833

(c) The efficiency of the machine (E) = 83.3 %

Explanation:

Given;

load lifted by the pulley, L = 400 N

effort applied in lifting the, E = 480 N

distance moved by the effort, d = 5 m

(a) the velocity ratio of the machine (V.R);

since the effort applied moved downwards through a distance of d, the load will also move upwards through an equal distance 'd'.

V.R = distance moved by effort / distance moved by the load

V.R = 5/5 = 1

(b) The mechanical advantage of the machine (M.A);

M.A = L/E

M.A = 400 / 480

M.A = 0.833

(c) The efficiency of the machine (E);

E = \frac{M.A}{V.R} \times 100\%\\\\E = 0.833 \ \times \ 100\%\\\\ E = 83.3 \ \%

4 0
2 years ago
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