Physics, 14.09.2019 10:30, heidiheffron574

An opera singer in a convertible sings a note at 550 hz while cruising down the highway at 92 km/h . the speed of sound in the air is 343 m/s. a) what is the frequency heard by a person standing beside the road in front of the car?
b) what is the frequency heard by a person standing beside the road behind the car?

Answers: 2

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answered: drewje12

A) 594 Hz

B) 512 Hz

Explanation:

Howdy!

This is a typical Doppler effect problem, whose formula is:

$f = \frac{v+v_{r} }{v+v_{s}} f_{0}$

Where :

$v_{r}$

is the velocity of the receiver (in our case is equal to zero)

$v_{s}$

is the velocity of the source (in our case is 92 km/h) which is positive when the source is moving away (second case) and negative otherwise.

$v_{s}$

Is the velocity of the wave in the medium.

Before we start calculating we need to have the velocity of the source in the same units as the velocity of the waves:

92km/h = 92*(1000)/3600 m/s = 25.5 m/s

When the source is moving towards the receiver the sign of the velocity is negative, so:

$f = \frac{343}{343-25.5} 550 Hz$

f = 594.1 Hz

B) Now the velocity of the source must change sign:

$f = \frac{343}{343+25.5} 550 Hz$

f = 511.9 Hz

As a sanity check we know that when the source is moving towards the source the frequency is higher and when the source is moving away from the source the frequency is lower.

answered: Guest

Answer: a charge of -1.0 × 10⁻⁴ c must be placed at x = 26 m.the sign of the charge will be the same sign as the -4.0 µc charge.======electricstatic force between two point charges is given by $f = \dfrac{kq_1q_2}{r^2}$ where k is coulomb's constant, 8.99×10⁹ n·m²/c², q₁ and q₂ are the point charges, and r is the distance between the two point charges.assign positive to q₁let q₁ = 6.0 µc, q₂ = -4.0 µc, q₀ represent the charge at the origin, and q₃ represent this new charge to place.the charge at the origin q₀ would experience the following force f₀₁ from q₁, with r = 6.0 m (note that 1 µc = 10⁻⁶ c) $f_{01} = \dfrac{kq_0q_1}{(\text{6.0 m})^2}$ the force from q₂ $f_{02} = \dfrac{kq_0q_2}{(\text{15 m})^2}$ the force from q₃ $f_{03} = \dfrac{kq_0q_1}{(\text{26 m})^2}$ the net force on the origin charge, q₀, has to be zero for no electrostatic forcehaving an equation where k and q₀ can be divided out of the equation will allow us to solve for q₃ $\begin{aligned} f_{net} & = 0\\ f_{01} + f_{02} + f_{03} & = 0 \\ \dfrac{kq_0q_1}{(\text{6.0 m})^2} + \dfrac{kq_0q_2}{(\text{15 m})^2} + \dfrac{kq_0q_3}{(\text{26 m})^2} & = 0 \\ kq_0 \left( \dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2} + \dfrac{q_3}{(\text{26 m})^2}\right) & = 0 \\ \dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2} + \dfrac{q_3}{(\text{26 m})^2} & = 0 \end{aligned}$ isolating: $\begin{aligned} \dfrac{q_3}{(\text{26 m})^2} & = -\dfrac{q_1}{(\text{6.0 m})^2} - \dfrac{q_2}{(\text{15 m})^2} \\ q_3 & = -(\text{26 m})^2\left(\dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2}\right) \\ q_3 & = -(\text{26 m})^2\left(\dfrac{6.0\times10^{-6} \text{ c}}{(\text{6.0 m})^2} + \dfrac{-4.0\times10^{-6} \text{ c}}{(\text{15 m})^2}\right) \\ q_3 & = -1.0064888 \times 10^{-4}\text{ c} \\ q_3 & = -1.0\times 10^{-4}\text{ c} \end{aligned}$ a charge of -1.0 × 10⁻⁴ c must be placed at x = 26 m.the sign of the charge will be the same as the -4.0 µc charge

answered: Guest

Simple, don't do it. just don't go to that dock