3 years ago

It is common practice to tin plate electrical terminals to facilitate soldering. Why is it tin that is used?

Engineering

1 answer:

Zielflug [23.3K]3 years ago

4 0

Answer: Its materials are good reactors with electrical terminals and most times we go the easy way

Explanation:

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Shows a closed tank holding air and oil to which is connected a U-tube mercury manometer and a pressure gage. Determine the read

damaskus [11]

Answer:

$P_2-P_1=27209h$

Explanation:

For pressure gage we can determine this by saying:

The closed tank with oil and air has a pressure of P₁ and the pressure of oil at a certain height in the U-tube on mercury is p₁gh₁. The pressure of mercury on the air in pressure gauge is p₂gh₂. The pressure of the gage is P₂.

$P_1+p_1gh_1=p_2_gh_2+P_2$

We want to work out P₁-P₂: Heights aren't given so we can solve it in terms of height: assuming h₁=h₂=h

$P_1-P_2=p_1gh_1-p_2gh_2=(55)\cdot{32.2}h-845\cdot{32.2}h$

$P_2-P_1=27209h$

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3 years ago

Read two numbers from user input. Then, print the sum of those numbers. Hint -- Copy/paste the following code, then just type co

Softa [21]

Answer:

I am Providing Answer in C Language Program.

Explanation:

Please find attachment regarding code of taking two numbers input and adding them.

I would like to recommend you please use software which supports C language.

#include <stdio.h>

int main () {

int a, b, sum;

printf ("\ nEnter two no:");

scanf ("% d% d", & d, & e);

sum1 = d + e;

printf ("Sum:% d", sum1);

return (0);

}

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3 years ago

Indicates the design of the building and<br> adjoining areas

Lelu [443]

Answer:

Architectural Plan

Explanation:

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A PLL is set up so that its VCO free-runs at 8.9 MHz. The VCO does not change frequency unless its input is within plus or minus

pickupchik [31]

It’s A 75khz because it’s plus or minus so if u add it would be too much

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2 years ago

A digital filter is given by the following difference equationy[n] = x[n] − x[n − 2] −1/4y[n − 2].(a) Find the transfer function

slega [8]

Answer:

$y(z) = x(z) - x(z) {z}^{ - 2} - \frac{1}{4} y(z) {z}^{ - 2} \\ y(z) + \frac{1}{4} y(z) {z}^{ - 2} = x(z) - x(z) {z}^{ - 2} \\ y(z) (1 + \frac{1}{4}{z}^{ - 2}) = x(z)(1 - {z}^{ - 2}) \\ h(z) = \frac{y(z)}{x(z)} = \frac{(1 + \frac{1}{4}{z}^{ - 2})}{(1 - {z}^{ - 2})}$

The rest is straightforward...

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2 years ago