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inessss [21]
3 years ago
7

Several different loads are going to be used with the voltage divider from Part A. If the load resistances are 300 kΩkΩ , 200 kΩ

kΩ , and 100 kΩkΩ , what is the output voltage that is the most different from the design output voltage vovov_o = 6 VV ? Express your answer in V to three significant figures.

Engineering
1 answer:
harina [27]3 years ago
7 0

Answer:

attached below

Explanation:

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Consider a very long rectangular fin attached to a flat surface such that the temperature at the end of the fin is essentially t
zepelin [54]

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Explanation:

4 0
3 years ago
A 50-kN hydraulic press performs pressing and clamping actions. The clamping cylinder force is 4 kN. The pressing cylinder strok
galben [10]

Answer:

The attached figure shows the hydraulic circuit using one sequence valve to control two simultaneous operations performed in proper sequence in one direction only. In the other direction, both the operations are simultaneous.

When we keep the 4/2 DCV in crossed arrow position, oil under pressure is supplied to the inlet port of the sequence valve. It directly flows to Head end port-1. Hence Cylinder 'C1' extends first.

By the end of the extension of cylinder 'C1', pressure in the line increases and hence poppet of sequence valve is lifted off from its seat and allows oil to flow to port-2 and hence, Cylinder 'C2 extends completing the pressing operation.

In the straight-arrow position of 4/2 DCV the oil under pressure reaches the rod end of both the cylinders C1 and C2 simultaneously through port-3. This causes both the cylinders to retract simultaneously.

Also, a Flow control valve is provided tho control the velocity of clamping

Explanation:

find attached the figure

4 0
3 years ago
What is the metal removal rate when a 2 in-diameter hole 3.5 in deep is drilled in 1020 steel at cutting speed of 120 fpm with a
Studentka2010 [4]

Answer:

a) the metal removal rate is 14.4 in³/min

b) the cutting time is 0.98 min

Explanation:

Given the data from the question

first we find the rpm for the spindle of the drilling tool, using the equation

Ns = 12V/πD

V is the cutting speed(120 fpm) and D is the diameter of the hole( 2 in)

so we substitute

Ns = 12 × 120 / π2

Ns = 1440 / 6.2831

Ns = 229.18 rmp

Now we find the metal removal rate using the equation

MRR = (πD²/4) Fr × Ns

Fr is the feed rate( 0.02 ipr ),

so we substitute

MRR = ((π × 2²)/4) × 0.02 × 229.18

MRR = 14.3998 ≈ 14.4 in³/min

Therefore the metal removal rate is 14.4 in³/min

Next we find the allowance for approach of the tip of the drill

A = D/2

A = 2/2

= 1 in

now find the time required to drill the hole

Tm = (L + A) / (Fr × Ns)

Lis the the depth of the hole( 3.5 in)

so we substitute our values

Tm = (3.5 + 1) / (0.02 × 229.18  )

Tm = 4.5 / 4.5836

Tm = 0.98 min

Therefore the cutting time is 0.98 min

8 0
2 years ago
2. The device that varies incoming line pressure using centrif-
Kryger [21]

Answer:

I believe  reverse boost valve.

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3 years ago
PLZ ASAP WILL GIVE BRAINLIST
gavmur [86]

Answer: A, B, C & F (interacting w computers, making decisions & solving problems, evaluating information & getting information).

Explanation: Those are the correct & verified answers.

7 0
2 years ago
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