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RideAnS [48]
3 years ago
9

Given that the speed of sound in air is 330m/s ,calculate the frequency of a note which setup a wavelength of 5cm in air.Is this

frequency within the limits of audibility of the human ear?
​
Physics
2 answers:
frosja888 [35]3 years ago
6 0

v = wavelength x frequency

330 = 5 . 10-² m x f

f = 6600 Hz

the frequency human is capable to hear- 20 Hz - 20. K Hz

so human can easily interpret note just like we interpret light

Pie3 years ago
3 0

Explanation:

v = wavelength x frequency

330 = 5 . 10-² m x f

f = 6600 Hz

the frequency that human can hear is about 20 Hz - 20000 Hz

so human can hear the note.

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When the input voltages of a difference amplifier are 5.1 v and 6.4 v, the output voltage is 64.7 v. The inputs are changed to 4
daser333 [38]

Answer:

a) Differential mode gain = 48

b) Common mode gain = 0.4

c) CMRR = 120

Explanation:

The output of a difference amplifier is related to the input by the equation:

V_{0} = A_{1} V_{1} + A_{2} V_{2} \\

When V₁ = 6.4 V, V₂ = 5.1 V and V₀ = 64.7 V, the equation becomes

6.4 A₁ + 5.1 A₂ = 64.7.....................(1)

When V₁ = 5.6 V, V₂ = 4.9 V and V₀ = 35.7 V, the equation becomes

5.6 A₁ + 4.9 A₂ = 35.7.....................(2)

Multiply equation (1) by 5.6  and (2) by 6.4

35.84 A₁ + 28.56A₂ = 362.32.....................(3)

35.84 A₁ + 31.36 A₂ = 228.48....................................(4)

Subtract equation (3) from (4)

2.8 A₂ = -133.84

A₂ = -133.84/2.8

A₂ = -47.8

Put the value of  A₂ into equation (1)

6.4 A₁ + 5.1 (-47.8) = 64.7

6.4 A₁ = 64.7 + 243.78

A₁ = 308.48/6.4

A₁ = 48.2

a) Common mode gain = A₁ + A₂ = 48.2 + (-47.8)

Common mode gain = 0.4

b) Differential mode gain = (A₁ -A₂)/2

Differential mode gain = (48.2 - (-47.8))/2

Differential mode gain = 96/2

Differential mode gain = 48

c) Common Mode Rejection Ratio (CMRR)

CMRR = |\frac{Differential Mode Gain}{Common Mode Gain} |

CMRR = |\frac{48}{0.4} |\\CMRR = 120

4 0
3 years ago
Which data set has the largest range?
MrRissso [65]

Answer:

Option C

Explanation:

We have to check range of all options first

For A:

Largest Value: 5

Smallest Value: 1

So range = Largest value - smallest value

5-1 = 4

For B:

Largest Value: 6

Smallest Value: 4

Range = 6-4 = 2

For C:

Largest Value: 9

Smallest Value: 1

Range = 9-1 = 8

For D:

Largest Value = 9

Smallest Value = 3

Range = 9-3=6

So, the data set in option C has the largest range

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Answer:

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Explanation:

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8 0
2 years ago
Imagine a small child whose legs are half as long as her parent’s legs. If her parent can walk at maximum speed V, at what maxim
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Answer:

\boxed{v=\frac {V}{\sqrt {2}}}

Explanation:

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We know that frequency of simple pendulum follows that f=\frac {1}{2\pi} \sqrt {\frac {g}{l}}

Now, the speed of the father will be V=Lf= L\times (\frac {1}{2\pi} \sqrt {\frac {g}{l}}) while for the child the speed will be v=\frac {L}{2}\times (\frac {1}{2\pi} \sqrt {\frac {g}{0.5l}})

The ratio of the father’s speed to the child’s speed will be

\frac {V}{v}=\frac {\frac {L}{2}\times (\frac {1}{2\pi} \sqrt {\frac {g}{0.5l}})}{ L\times (\frac {1}{2\pi} \sqrt {\frac {g}{l}})}\\\frac {V}{v}=\frac {\sqrt {2}}{2}\\\boxed{v=\frac {V}{\sqrt {2}}}

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ahrayia [7]
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