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Vlad [161]
2 years ago
9

What is the purpose of a rectifier assembly

Engineering
1 answer:
Elden [556K]2 years ago
6 0

Answer:

Explanation:

I assume you are talking about electronics. A rectifier changes Alternating current into direct current.

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(25%) A well-insulated compressor operating at steady state takes in air at 70 oF and 15 psi, with a volumetric flow rate of 500
lubasha [3.4K]

Answer:

You can look it up

Explanation: if you don't know what it is look it up on .

6 0
3 years ago
6.3.9 A coin was tossed n = 1000 times, and the proportion of heads observed was
4vir4ik [10]

Answer:

No

Explanation:

51 / 100 = 510 / 1000

Chance of getting a head is 1 / 2 of total throws

= 1 / 2 × 1000

= 500 is the probability

and the number of heads was just 10 more the the probability...if the was a greater gap, there would be evidence to say the coin is unfair

4 0
3 years ago
One of the flaws in the engineers' reasoning for galloping gertie's design was that they attributed prior failures of suspension
Jlenok [28]
False i think it would be
3 0
3 years ago
A common procedure for measuring the velocity of an air stream involves insertion of an electrically heated wire (called a hot-w
timurjin [86]

Answer:

V = 6.33 m/s

Explanation:

Given:

- The length of the wire L = 0.02 m

- The diameter of the wire D = 0.0005 m

- The calibration expression V = 0.0000625*h^2

- Environment temperature T_inf = 298 K

- Surface temperature T_s = 348 K

- The voltage drop dV = 5 V

- The electric current I = 0.1 A

Find:

- the velocity of Air

Solution:

- Calculate the surface area of the wire:

                             A = pi*D*L

                             A = pi*(0.0005)*(0.02) = 0.00003142 m^2

- The rate of energy in the wire P:

                             P = I*dV = 0.1*5 = 0.5 W

- Apply Newton's Law of Cooling:

                            P = h*A*(T_s - T_inf)

                            h =  P /A*(T_s - T_inf)

Plug in the values:

                             h= 0.5/ 0.00003142*(348 - 298)

                             h = 318.27 W /m^2K

- Using the calibration relationship given, compute the velocity of air:

                             V = 6.25*10^-5 * h^2

                             V = 6.25*10^-5 * (318.27)^2

                             V = 6.33 m/s

5 0
3 years ago
What measures can be taken to mitigate or prevent the formation of stress corrosion cracks?
sasho [114]

Answer:

a) Selection and control of material

b) Control of stress

c) Control of the environment

Explanation:

The measures that can be taken to control and prevent the formation of strss corrosion cracks are:

a) Selection and control of material

It deals with the selection of the best quality material that can resist the stress corrosion cracks to a high extent.

b) Control of stress

It is to control or reduce the intensity of the major stress that causes the stress corrosion cracks.

c) Control of the environment

It includes the prevention of the material from the corrosive environment by removing the component from the region or controlling the environment condition around the object.

7 0
3 years ago
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