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Ivanshal [37]
4 years ago
6

The hot-wire anemometer is an instrument used for measuring velocities or temperatures. If, during its calibration, the output s

ignal E is measured as 0, 1.7, 3.3, and 5.6 V at velocities V of 0, 1, 1.5, and 2 m/s, obtain the gradient dE/dV at V = 0 m/s, using the polynomial representation of the function E(V).

Engineering
1 answer:
Alexus [3.1K]4 years ago
6 0

Answer:

Explanation:

An expression was assumed for E(V) = av^2 + bv + c

each values for substituted for corresponding values of the velocity and a three unknown equation was formed, which was solved simultaneously to get the three unknowns (a, b and c).

The unknowns was substiutted back ito the original equation.

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Answer:

class TriangleNumbers

{

public static void main (String[] args)

{

 for (int number = 1; number <= 10; ++number) {

  int sum = 1;

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  for (int summed = 2; summed <= number; ++summed) {

   sum += summed;

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  System.out.print(" = " + Integer.toString(sum) + '\n');

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}

}

Explanation:

We need to run the code for each of the 10 lines. Each time we sum  numbers from 1 to n. We start with 1, then add numbers from 2 to n (and print the operation). At the end, we always print the equals sign, the sum and a newline character.

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What is the purpose of a presentation model? to fabricate a prototype to test the efficiency of the manufacturing process to tes
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Presentation Model is a pattern that pulls presentation behavior from a view. As such it's an alternative to Supervising Controller and Passive View.

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Is there a picture or something?
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A ball A is thrown vertically upward from the top of a 24-m-high building with an initial velocity of 7 m/s . At the same instan
solniwko [45]

Answer:

The height from ground at which they pass each other is 22.656 m

The time at which they pass each other is 1.6 sec.

Explanation:

For ball A, we have:

height = (h)a

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g = - 9.8 m/s² (for upward motion)

Using 2nd eqn. of motion:

h = Vi t + (1/2)gt²

(h)a = 7t + (1/2)(-9.8)t²

(h)a = 7t - 4.9t²    

this is the height of ball A with reference as the building. Taking ground as reference, we have to add the height of building, that is, 24 m.

(h)a = 7t - 4.9t² + 24  ______ eqn (1)

For ball B, we have:

height = (h)b

Initial Velocity = Vi = 22 m/s

g = - 9.8 m/s² (for upward motion)

Using 2nd eqn. of motion:

h = Vi t + (1/2)gt²

(h)b = 22t + (1/2)(-9.8)t²

(h)b = 22t - 4.9t²    ______ eqn (2)

Now, when the too balls pass each other, there height must be same.

Therefore,

(h)a = (h)b

using eqn (1) and eqn (2):

7t - 4.9t² + 24 = 22t - 4.9t²

22t - 7t = 24

t = 24/15

<u>t = 1.6 sec</u>

Now, for the height, at which they pass each other put t = 6sec in eqn (2)

Therefore,

h = (22)(1.6) - (4.9)(1.6)²

h = 35.2 - 12.544

<u>h = 22.656 m</u>

<u></u>

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