You walk 12.0 m West and then 4.00 m south. What is your displacement?

This is theDopplereffect. Sup-pose that, at a particular moment, you are in a train traveling at 34 m/s and acceleratingat 1.2m/

Ilia_Sergeevich [38]

Complete Question

If a sound with frequency fs is produced by a source traveling along a line with speed vs and an observer is traveling with speed vo along the same line from the opposite direction toward the source, then the frequency of the sound heard by the observer is

f_o = [(c+v_o)/(c-v_s)] f_s

where c is the speed of sound, about 332 m/s. (This is the Doppler effect). Suppose that, at a particular moment, you are in a train traveling at 34 m/s and accelerating at 1.2 m/s^2. A train is approaching you from the opposite direction on the other track at 40 m/s, accelerating at 1.4 m/s^2, and sounds its whistle, which has a frequency of 460Hz. At that instant, what is the perceived frequency that you hear and how fast is it changing?

Answer:

The frequency the person hears is $f_o = 557 Hz$

The speed at which it is changing is $\frac{df_o}{dt} = 4.655 Hz/s$

Explanation:

Form the question we are told that

The frequency of the sound produced by source is $f_s$

The speed of the source is $v_s$

The speed of the observer

The frequency of sound heard by observer $f_o =[ \frac{c + v_o }{c - v_s} ] * f_s$

The speed of sound is c with value $c = 332 m/s$

Looking the question we can deduce that the person in the first train is the observer so the

$v_o = 34 m/s$

and the acceleration is $\frac{dv_o}{dt} = 1.2 m/s^2$

The train the travelling in the opposite direction the blew the whistle

is the source

So $v_s = 40 m/s$

and $f_s = 460 Hz$

and the acceleration is $\frac{dv_s}{dt} = 1.4 m/s^2$

We are told that

$f_o =[ \frac{c + v_o }{c - v_s} ] * f_s$

Substituting values we have that

$f_o =[ \frac{332 + 34 }{332 - v40} ] * 460$

$f_o = 557 Hz$

Differentiating $f_o$ using chain rule we have that

$\frac{d f_o}{dt} = \frac{df_o}{dt } * \frac{dv_o}{dt} + \frac{d f_o}{dv_s} * \frac{dv_s}{dt}$

Now

$\frac{df_o}{dt } = \frac{f_s}{c- v_s}$

$\frac{df_o}{dv_s} = \frac{c+ v_o}{c-v_s} f_s$

Substituting this into the equation

$\frac{df_o}{dt} = \frac{f_s}{c-v_s} * \frac{d v_o}{dt} + \frac{c+v_o}{(c-v_s)^2} f_s * \frac{dv_s}{dt}$

Now substituting values

$\frac{df_o}{dt} = \frac{460}{332 - 40} * (1.2) + \frac{332+ 34}{(332- 40)^2} 460 * 1.4$

$\frac{df_o}{dt} = 4.655 Hz/s$

6 0

3 years ago

Why does a liquid need a container and a solid doesn't?

mamaluj [8]

It is mainly because a solid has a much stronger atom bonds compared to liquid
When liquid is poured, it will always tried to fill up its container because the weakness in its atom bond allow it to do so.
Meanwhile, the form of a solid will stay exactly the same no matter where we put it because the strength of their bond will retain their form.

7 0

3 years ago

If element X has 34 protons, how many electrons does it have?

nikitadnepr [17]

It depends on the atom. Assuming the atom is neutral (no extra proton or electrons), the number of electrons is equal to the number of protons.

So if element X has 34 protons, then the number of electrons is 34 too.

hope this helps

5 0

3 years ago

Read 2 more answers

At a distance of 0.220 cmcm from the center of a charged conducting sphere with radius 0.100cm0.100cm, the electric field is 430

liberstina [14]

Answer:

The electric field at a distance of 0.584 cm from the center of the sphere is 61.02 N/C.

Explanation:

Given that,

Initial distance, $d_1=0.22\ cm$

Initial electric field, $E_1=430\ N/C$

We need to find the electric field 0.584 cm from the center of the sphere, $d_2=0.584\ cm$ . The electric field at a distance d is the given by :

$E=\dfrac{kq}{r^2}$

$\dfrac{E_1}{E_2}=\dfrac{r_2^2}{r_1^2}$ (as electric field is inversely proportional to the distance)

$E_2$ is the electric field 0.584 cm from the center of the sphere

$E_2=\dfrac{r_1^2}{r_2^2}E_1$

$E_2=\dfrac{0.22^2}{0.584^2}\times 430$

$E_2=61.02\ N/C$

So, the electric field at a distance of 0.584 cm from the center of the sphere is 61.02 N/C. Hence, this is the required solution.

8 0

3 years ago

A parallel plate capacitor fully charged to voltage V is connected to the battery (the voltage on the plates remains fixed). If

Papessa [141]

Charge will decreases.

A parallel plate capacitor when it is fully charged to voltage V is given as:

C = Q/V

The capacitance of parallel plate capacitor with two plates of Area A separated by distance d and no dielectric material between plates is

C = ε₀ A /d

since from above equation it shows C is proportional to Q and also C is inversely proportional to distance d.

So, ATQ when d increases C will decrease which in result decreases charge on the capacitor.

Thus, Charge will decrease.

Learn more about capacitance here:

brainly.com/question/17115454

#SPJ4

7 0

2 years ago

You walk 12.0 m West and then 4.00 m south. What is your displacement?​

You walk 12.0 m West and then 4.00 m south. What is your displacement?