PHYSICS COLLEGE

Singing that is off-pitch by more than about 1% sounds bad. How fast would a singer have to be moving relative to the rest of a band to make this much of a change in pitch due to the Doppler effect

Answers

Answer 1

Answer:

Answer:

-3.396 m/s or 3.465 m/s

Explanation:

v = Speed of sound in air = 343 m/s

$v_s$ = Relative speed of the singer

f = Observed frequency

f' = Actual frequency

1% change can mean $f=1.01f'$

From the Doppler effect equation we have

$f=f'(v)/(v+v_s)\n\Rightarrow 1.01f'=f'(v)/(v+v_s)\n\Rightarrow 1.01=(343)/(343+v_s)\n\Rightarrow v_s=(343)/(1.01)-343\n\Rightarrow v_s=-3.396\ m/s$

The velocity is -3.396 m/s

when $f=0.99f'$

$f=f'(v)/(v+v_s)\n\Rightarrow 0.99f'=f'(v)/(v+v_s)\n\Rightarrow 0.99=(343)/(343+v_s)\n\Rightarrow v_s=(343)/(0.99)-343\n\Rightarrow v_s=3.46464646465\ m/s$

The velocity is 3.465 m/s

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A bowling ball of mass 3 kg is dropped from the top of a tall building. It safely lands on the ground 3.5 seconds later. Neglecting air friction, what is the height of the building in meters? (Give the answer without a unit and round it to the nearest whole number)

Answers

The height of the building is 60 m.

calculation of building height:

The velocity of the ball should be provided by

v = u + gt

here,

u is the initial velocity of the ball = 0

v = 0 + 9.8 x 3.5

v = 34.3 m/s

Now

When the ball hits the ground, energy is conserved;

mgh = ¹/₂mv²

gh = ¹/₂v²

h = (0.5 v²) / g

h = (0.5 x 34.3²) / (9.8)

h = 60.025 m

h = 60 m

Learn more about friction here: brainly.com/question/14455351

Answer:

The height of the building is 60 m.

Explanation:

Given;

mass of the mass of the ball, m = 3 kg

time of motion, t = 3.5 s

The velocity of the ball is given by;

v = u + gt

where;

u is the initial velocity of the ball = 0

v = 0 + 9.8 x 3.5

v = 34.3 m/s

When the ball hits the ground, energy is conserved;

mgh = ¹/₂mv²

gh = ¹/₂v²

h = (0.5 v²) / g

h = (0.5 x 34.3²) / (9.8)

h = 60.025 m

h = 60 m

Therefore, the height of the building is 60 m.

As the captain of the scientific team sent to Planet Physics, one of your tasks is to measure g. You have a long, thin wire labeled 1.73 g/m and a 1.30 kg weight. You have your accurate space cadet chronometer but, unfortunately, you seem to have forgotten a meter stick. Undeterred, you first find the midpoint of the wire by folding it in half. You then attach one end of the wire to the wall of your laboratory, stretch it horizontally to pass over a pulley at the midpoint of the wire, then tie the 1.30 kg weight to the end hanging over the pulley. By vibrating the wire, and measuring time with your chronometer, you find that the wire's second harmonic frequency is 200 Hz . Next, with the 1.30 kg weight still tied to one end of the wire, you attach the other end to the ceiling to make a pendulum. You find that the pendulum requires 313 s to complete 200 oscillations. Pulling out your trusty calculator, you get to work.What value of g will you report back to headquarters?

Answers

Answer:

The value of g is $g =76.2 m/s^2$

Explanation:

From the question we are told that

The mass of the weight is $m = 1.30 kg$

The spring constant $k = 1.73 g/m = 1.73 *10^(-3) \ kg/m$

The second harmonic frequency is $f = 100 \ Hz$

The number of oscillation is $N = 200$

The time taken is $t = 315 \ s$

Generally the frequency is mathematically represented as

$f = (v)/(\lambda)$

At second harmonic frequency the length of the string vibrating is equal to the wavelength of the wave generated

$l = \lambda$

Noe from the question the vibrating string is just half of the length of the main string so

Let assume the length of the main string is $L$

So $l = (L)/(2)$

The velocity of the vibrating string is mathematically represented as

$v = \sqrt{(T)/(\mu) }$

Where T is the tension on the string which can be mathematically represented as

$T = mg$

$v = \sqrt{(mg)/(k) }$

Then

$f = (v)/((L)/(2) )$

=> $v = (fL )/(2)$

=> $\sqrt{(mg)/(k) } = (fL)/(2)$

=> $g = (f^2 L^2 \mu)/(4m)$

substituting values

$g = ((100) * (1.73 *10^(-3) ))/((4 * 1.30)) L^2$

$g = 3.326 m^(-1) s^(-2) L^2$

Generally the period of oscillation is mathematically represented as

$T_p = 2 \pi \sqrt{(L)/(g) }$

=> $L = (T^2 g)/(4 \pi ^2)$

The period can be mathematically evaluated as

$T_p = (t)/(N)$

substituting values

$T_p = (315)/(200)$

$T_p = 1.575 \ s$

Therefore

$L = (1.575^2 * g )/(4 \pi ^2)$

$L = 0.0628 ^2 g$

$g = 3.326 m^(-1) s^(-2) L^2$

substituting for L

$g = 3.326 ((0.0628) g)^2$

=> $g = (1)/((3.326)* (0.0628)^2)$

$g =76.2 m/s^2$

A spectroscope breaks light up into its colors, allowing scientists to analyze light from the solar system and universe. By studying the spectral line patterns, widths, strengths and positions, scientists can determine the speed, position, and _____ of celestial bodies.A) age
B) origin
C) rotation
D) temperature

Answers

It is D - temperature

Final answer:

A spectroscope analyses light to determine various parameters of celestial bodies. The missing parameter in this context is the 'temperature' of the celestial body (option D). The spectral lines, based on their pattern and strengths helps in determining this.

Explanation:

A spectroscope decomposes or breaks white light into its spectrum of colors, allowing scientists to study them and understand various aspects of celestial bodies. When scientists analyze the spectral line patterns, widths, strengths, and positions, they can discern essential parameters. These parameters include the speed and position of the celestial body, and more importantly, the correct answer to your question, its temperature (option D). This is because every element when heated, absorbs or emits light at characteristic wavelengths, that give us the 'spectral lines'. By studying these we can determine the temperature of the celestial body.

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1. What is the frequency of light waves with wavelength of 5 x 10-⁷ m?

Answers

Taking into account the definition of wavelength, frecuency and propagation speed, the frequency of light waves with wavelength of 5×10⁻⁷ m is 6×10¹⁴ Hz.

Definition of wavelength

First of all, wavelength is the minimum distance between two successive points on the wave that are in the same state of vibration. It is expressed in units of length (m).

Definition of frequency

On the other side, frequency is the number of vibrations that occur in a unit of time. Its unit is s⁻¹ or hertz (Hz).

Definition of propagation speed

Finally, the propagation speed is the speed with which the wave propagates in the medium, that is, it is the magnitude that measures the speed at which the wave disturbance propagates along its displacement.

The propagation speed relate the wavelength (λ) and the frequency (f) inversely proportional using the following equation:

v = f× λ

All electromagnetic waves propagate in a vacuum at a constant speed of 3×10⁸ m/s, the speed of light.

Frequency of light waves with wavelength of 5×10⁻⁷ m

In this case, you know:

v= 3×10⁸ m/s
f= ?
λ= 5×10⁻⁷ m

Replacing in the definition of propagation speed:

3×10⁸ m/s = f× 5×10⁻⁷ m

Solving:

3×10⁸ m/s ÷ 5×10⁻⁷ m= f

f= 6×10¹⁴ Hz

In summary, the frequency of light waves with wavelength of 5×10⁻⁷ m is 6×10¹⁴ Hz.

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Answer:

Speed of light =m/s

wavelength = m

frequency = ?

we have

Speed = frequency × wavelength

$3* 10^8$ = frequency × $5 * 10^(-7)$

Frequency = $(3*10^8)/(5*10^(-7))=6*10^(14)$ hz

On December 26, 2004, a great earthquake occurred off the coast of Sumatra and triggered immense waves (tsunami) that killed some 200000 people. Satellites observing these waves from space measured 800 from one wave crest to the next and a period between waves of 1.0 hour.Part AWhat was the speed of these waves in m/s?Express your answer using two significant figures.=Part BWhat was the speed of these waves in km/h ?Express your answer using two significant figures.=

Answers

Answers:

a) 222.22 m/s

b) 800.00 km/h

Explanation:

The speed of a wave is given by the following equation:

$v=f \lambda$

Where:

$v$ is the speed

$f=(1)/(T)$ is the frequency, which has an inverse relation with the period $T=1 h$

$\lambda=800 km$ is the wavelength

Solving with the given units:

$v=(1)/(T)\lambda$

$v=(1)/(1 h)800 km$

$v=800.00 km/h$ This is the speed of the wave in km/h

Transforming this speed to m/s:

$v=800.00 (km)/(h) (1 h)/(3600 s) (1000 m)/(1 km)$

$v=222.22 m/s$ This is the speed of the wave in m/s

We often refer to the electricity at a typical household outlet as being 120 V. In fact, the voltage of this AC source varies; the 120 V is __________. We often refer to the electricity at a typical household outlet as being 120 V. In fact, the voltage of this AC source varies; the 120 V is __________. the minimum value of the voltage the peak value of the voltage the average value of the voltage the rms value of the voltage

Answers

Explanation:

We often refer to the electricity at a typical household outlet as being 120 V. In fact, the voltage of this AC source varies; the 120 V is "the rms value of the voltage".

The rms value of voltage is given by :

$V_(rms)=(V_(pk))/(√(2))$

Where

$v_(pk)$ is the peak value of voltage

So, the correct option is (d). " rms value of voltage".

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