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hodyreva [135]
3 years ago
9

The valu

Physics
1 answer:
natulia [17]3 years ago
4 0

Answer:

\% Error = 2.6\%

Explanation:

Given

x: 1.54, 1.53, 1.44, 1.54, 1.56, 1.45

Required

Determine the percentage error

First, we calculate the mean

\bar x = \frac{\sum x}{n}

This gives:

\bar x = \frac{1.54+ 1.53+ 1.44+ 1.54+ 1.56+ 1.45}{6}

\bar x = \frac{9.06}{6}

\bar x = 1.51

Next, calculate the mean absolute error (E)

|E| = \sqrt{\frac{1}{6}\sum(x - \bar x)^2}

This gives:

|E| = \sqrt{\frac{1}{6}*[(1.54 - 1.51)^2 +(1.53- 1.51)^2 +.... +(1.45- 1.51)^2]}

|E| = \sqrt{\frac{1}{6}*0.0132}

|E| = \sqrt{0.0022}

|E| = 0.04

Next, calculate the relative error (R)

R = \frac{|E|}{\bar x}

R = \frac{0.04}{1.51}

R = 0.026

Lastly, the percentage error is calculated as:

\% Error = R * 100\%

\% Error = 0.026 * 100\%

\% Error = 2.6\%

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An organ pipe open at both ends is 1.5 m long. A second organ pipe that is closed at one end and open at the other is 0.75 m lon
telo118 [61]

Answer:

A. 110Hz,220Hz, 330 Hz

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for organ open at open both ends;

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L = λ /4 + λ /4 = λ /2

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Wave equation is given by;

V = Fλ

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F = V/ λ

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F₀  is the fundamental frequency

F₀ = 330 / 2(1.5)

F₀ = 330 / 3

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the length of the organ for the first overtone, L = A---->N + N----->A + A----->N +  N----->A

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F₁ = 220 Hz

Thus, F₁ = 2F₀

For open organ at one end

the length of the organ for the fundamental frequency, L = N------A

L = λ /4

λ = 4L

F₀ = V/4L

F₀ = 330 / (4 x 0.75)

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the length of the organ for the first overtone, L = N-----N + N-----A

L = λ/2 + λ / 4

L = 3λ /4

F₁ = 3F₀

F₁ = 3 x 110

F₁ = 330 Hz

Thus the fundamental frequency for both organs is 110 Hz,

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The first overtone for the organ open at one end is 330 Hz

The correct option is "A. 110Hz,220Hz, 330 Hz"

6 0
3 years ago
An airplane is flying at a speed of 45 m/s when it drops a 40 kg food package to the Polar exploration team. If the plane drops
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3 0
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PLEASE HELP ASAP
suter [353]
Match:
A-2
B-1
C-3
D-4

:)




6 0
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