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2 years ago

How do you sound waves reach our ears?

Physics

1 answer:

iren2701 [21]2 years ago

5 0

Answer:

Sound waves enter the outer ear and travel through a narrow passageway called the ear canal, which leads to the eardrum. The eardrum vibrates from the incoming sound waves and sends these vibrations to three tiny bones in the middle ear. These bones are called the malleus, incus, and stapes.

Explanation:

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You have a resistor of resistance 200 ? and a 6.00-?F capacitor. Suppose you take the resistor and capacitor and make a series c

Ne4ueva [31]

Answer:

eazy

Explanation:

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3 years ago

I shared a picture of the problem. It’s a basic Physics question and an Algebra question.

julia-pushkina [17]

Hence the expression of ω in terms of m and k is

$\omega = \sqrt{\frac{k}{m}$

Given the expressions;

$T_s = 2 \pi \sqrt{\frac{m}{k} } \ and \ T_s = \frac{2 \pi}{\omega}$

Equating both expressions we will have;

$2 \pi \sqrt{\frac{m}{k} } = \frac{2 \pi}{\omega}$

Divide both equations by 2π

$\frac{2 \pi\sqrt{\frac{m}{2 \pi} } }{2 \pi}=\frac{\frac{2 \pi}{\omega} }{2\pi}\\\sqrt{\frac{m}{2 \pi} } = \frac{1}{\omega}\\$

Square both sides

$(\sqrt{\frac{m}{k} } )^2 = (\frac{1}{\omega} )^2\\\frac{m}{k} = \frac{1}{\omega ^2} \\\omega ^2 = \frac{k}{m}$

Take the square root of both sides

$\sqrt{\omega ^2} =\sqrt{\frac{k}{m} } \\\omega = \sqrt{\frac{k}{m}$

Hence the expression of ω in terms of m and k is

$\omega = \sqrt{\frac{k}{m}$

3 0

2 years ago

Use technology and the given confidence level and sample data to find the confidence interval for the population mean muμ. Assum

Ronch [10]

Answer:

a. μ $_{95%} =$ 3 ± 1.8 = [1.2,4.8]

b. The correct answer is option D. No, because the sample size is large enough.

Explanation:

a. The population mean can be determined using a confidence interval which is made up of a point estimate from a given sample and the calculation error margin. Thus:

μ $_{95%} = x_$ ±(t*s)/sqrt(n)

where:

μ $_{95%} =$ = is the 95% confidence interval estimate

x_ = mean of the sample = 3

s = standard deviation of the sample = 5.8

n = size of the sample = 41

t = the t statistic for 95% confidence and 40 (n-1) degrees of freedom = 2.021

substituting all the variable, we have:

μ $_{95%} =$ 3 ± (2.021*5.8)/sqrt(41) = 3 ± 1.8 = [1.2,4.8]

b. The correct answer is option D. No, because the sample size is large enough.

Using the the Central Limit Theorem which states that regardless of the distribution shape of the underlying population, a sampling distribution of size which is ≥ 30 is normally distributed.

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2 years ago

cycling at 13.0 m/s on your new aero carbon fiber bike, how much times would it take a pro cyclist to ride in 120 km? Give your

sp2606 [1]

We have that the time in seconds, minutes, and hours is

$t=1.083*10^{-4}s$