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4 years ago

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A train at rest emits a sound at a frequency of 1045 Hz. An observer in a car travels away from the sound source at a speed of 2

2.4 m/s. What is the frequency heard by the observer? Assume the speed of sound in air to be 344 m/s. Answer in units of Hz

1 answer:

11111nata11111 [884]4 years ago

6 0

<h2>Answer:976.95Hz</h2>

Explanation:

Let $c$ be the speed of sound in medium.

Let $v_{s}$ be the speed of the source.

Let $v_{o}$ be the speed of the observer.

Let $f$ be the frequency emitted by the source when the source is in rest.

Let $f_{new}$ be the frequency observed by the observer.

When the observer is moving away and the source is at rest,

$f_{new}=f\times \dfrac{c-v_{o}}{c}$

Given,

$c=344ms^{-1}$

$v_{s}=0ms^{-1}$

$v_{o}=22.4ms^{-1}$

$f=1045Hz$

$f_{new}=1045\times \frac{344-22.4}{344}=1045\times \frac{321.6}{344}=976.95Hz$

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A student measures the diameter of a small cylindrical object and gets the following readings: 4.32, 4.35, 4.31, 4.36, 4.37, 4.3

Zinaida [17]

Answer:

a. $\bar{d}=4.34 cm$

b. $\sigma=0.023 cm$

c. $\rho=(0.0089\pm 0.00058) kg/cm^{3}$

Explanation:

a) The average of this values is the sum each number divided by the total number of values.

$\bar{d}=\frac{\Sigma_{i=1}^(N)x_{i}}{N}$

$x_{i}$ is values of each diameter
N is the total number of values. N=6

$\bar{d}=\frac{4.32+4.35+4.31+4.36+4.37+4.34}{6}$

$\bar{d}=4.34 cm$

b) The standard deviation equations is:

$\sigma=\sqrt{\frac{1}{N}\Sigma^{N}_{i=1}(x_{i}-\bar{d})^{2}}$

If we put all this values in that equation we will get:

$\sigma=0.023 cm$

Then the mean diameter will be:

$\bar{d}=(4.34\pm 0.023)cm$

c) We know that the density is the mass divided by the volume (ρ = m/V)

and we know that the volume of a cylinder is: $V=\pi R^{2}h$

Then:

$\rho=\frac{m}{\pi R^{2}h}$

Using the values that we have, we can calculate the value of density:

$\rho=\frac{1.66}{3.14*(4.34/2)^{2}*12.6}=0.0089 kg/cm^{3}$

We need to use propagation of error to find the error of the density.

$\delta\rho=\sqrt{\left(\frac{\partial\rho}{\partial m}\right)^{2}\delta m^{2}+\left(\frac{\partial\rho}{\partial d}\right)^{2}\delta d^{2}+\left(\frac{\partial\rho}{\partial h}\right)^{2}\delta h^{2}}$

δm is the error of the mass value.
δd is the error of the diameter value.
δh is the error of the length value.

Let's find each partial derivative:

1. $\frac{\partial\rho}{\partial m}=\frac{4m}{\pi d^{2}h}=\frac{4*1.66}{\pi 4.34^{2}*12.6}=0.0089$

2. $\frac{\partial\rho}{\partial d}=-\frac{8m}{\pi d^{3}h}=-\frac{8*1.66}{\pi 4.34^{3}*12.6}=-0.004$

3. $\frac{\partial\rho}{\partial h}=-\frac{4m}{\pi d^{2}h^{2}}=-\frac{4*1.66}{\pi 4.34^{2}*12.6^{2}}=-0.00071$

Therefore:

$\delta\rho=\sqrt{\left(0.0089)^{2}*0.05^{2}+\left(-0.004)^{2}*0.023^{2}+\left(-0.00071)^{2}*0.5^{2}}$

$\delta\rho=0.00058$

So the density is:

$\rho=(0.0089\pm 0.00058) kg/cm^{3}$

I hope it helps you!

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3 years ago

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$\text{Hello there! :)}$

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4 years ago

A girl and a boy pull in opposite directions on strings attached to each end of a spring balance. Each child exerts a force of 2

BabaBlast [244]

Answer:

yo they deleted my answer. The answer is 0N

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so when two forces pull on an object from opposite sides with the same force (in this case its 20N), then the object is in equilibrium at 0N.

So its clear that there is one person on the the opposite side.

SOOO generally<u>: (left or down) would be considered </u><u>negative</u><u> in an equation. And the other person (right or up) would be considered </u><u>positive</u><u>.</u> So if both forces are the same numbers on opposite sides then the answer is 0 (if you add both of them).

<em>0 is the number of equilibrium.</em>

OK, so the equation would be -20N + 20N and then badda bing badda boom viola, the answer: 0N

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3 years ago

Dario, a prep cook at an Italian restaurant, spins a salad spinner and observes that it rotates 20.0 times in 5.00 seconds and t

BabaBlast [244]

Answer:

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Explanation:

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Where for definition,

$\omega_i = \frac{\theta}{t}$

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re-arrange to solve $\alpha$ ,

$\alpha = \frac{-8\pi}{2*12\pi}$

$\alpha = -\frac{8\pi}{3}rad/s^2$

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$\omega_f-\omega_i = \alpha t$

Re-arrange to find t, and considering that $\omega_f=0$

$t= \frac{\omega_i}{\alpha}$

$t=\frac{-8\pi}{-8\pi/3}$

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3 years ago

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ruslelena [56]

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Hope this helps

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4 years ago