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aivan3 [116]
2 years ago
10

Why did boy (the observer) hear the pitch of the sound change? What does this tell us about the frequency of the sound? Support

your conclusion based on results from the experiment and your knowledge of sound waves.
Physics
1 answer:
timofeeve [1]2 years ago
5 0

Answer:

<u>because of the doppler effect</u>

Explanation:

<em>Remember</em>, the doppler effect refers to the changes in sound (frequency of sound) observed by a person who is in a position relative to the wave source.

In this example, we notice as the train comes closer to the boy, the sound becomes louder also increasing the pitch slightly, the doppler effect sets in when the train passes the boy because the boy notices a decrease in the pitch of the moving train.

We learn from the change in the observed sound of the train that the frequency of the sound is determined by the distance of the observer from the wave source.

In other words, the closer the source of the sound to the observer; the faster it travels to the observer, however, the farther it is; the lesser it is; the greater the sound heard.

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When will the motion of an object be
mafiozo [28]

Answer:

C. The forces are balanced.

Explanation:

Every answer other than C means the object moved.

5 0
2 years ago
A body which has surface area 5cm² and temperature of 727°C radiates 300J of energy in one minute. Calculate it's emissivity giv
cestrela7 [59]
<h2>Answer: 0.17</h2>

Explanation:

The Stefan-Boltzmann law establishes that a black body (an ideal body that absorbs or emits all the radiation that incides on it) "emits thermal radiation with a total hemispheric emissive power proportional to the fourth power of its temperature":  

P=\sigma A T^{4} (1)  

Where:  

P=300J/min=5J/s=5W is the energy radiated by a blackbody radiator per second, per unit area (in Watts). Knowing 1W=\frac{1Joule}{second}=1\frac{J}{s}

\sigma=5.6703(10)^{-8}\frac{W}{m^{2} K^{4}} is the Stefan-Boltzmann's constant.  

A=5cm^{2}=0.0005m^{2} is the Surface area of the body  

T=727\°C=1000.15K is the effective temperature of the body (its surface absolute temperature) in Kelvin.

However, there is no ideal black body (ideal radiator) although the radiation of stars like our Sun is quite close.  So, in the case of this body, we will use the Stefan-Boltzmann law for real radiator bodies:

P=\sigma A \epsilon T^{4} (2)  

Where \epsilon is the body's emissivity

(the value we want to find)

Isolating \epsilon from (2):

\epsilon=\frac{P}{\sigma A T^{4}} (3)  

Solving:

\epsilon=\frac{5W}{(5.6703(10)^{-8}\frac{W}{m^{2} K^{4}})(0.0005m^{2})(1000.15K)^{4}} (4)  

Finally:

\epsilon=0.17 (5)  This is the body's emissivity

3 0
3 years ago
Three objects each with a mass of 10.0 kg are
Vera_Pavlovna [14]

Answer: F net=0

Explanation:

Its in the picture.

7 0
3 years ago
Read 2 more answers
A) A mass spectrometer has a velocity selector that allows ions traveling at only one speed to pass with no deflection through s
KengaRu [80]

Answer:

(A). The speed of the ions is 1.2\times10^{6}\ m/s

(B). The radius of curvature of a singly charged lithium ion is 2.0\times10^{6}\ m

Explanation:

Given that,

Electric field = 60000 N/C

Magnetic field = 0.0500 T

(A). We need to calculate the velocity

For no deflection

F_{E}=F_{B}

Eq=Bqv

v = \dfrac{E}{B}

v=\dfrac{60000}{0.0500}

v=1.2\times10^{6}\ m/s

(B). We need to calculate the radius

Using magnetic force balance by centripetal force

Bqv=\dfrac{mv^2}{r}

r=\dfrac{mv^2}{Bqv}

Put the value into the formula

r=\dfrac{1.16\times10^{-26}\times(1.2\times10^{6})^2}{0.0500\times1.6\times10^{-19}}

r=2.0\times10^{6}\ m

Hence, (A). The speed of the ions is 1.2\times10^{6}\ m/s

(B). The radius of curvature of a singly charged lithium ion is 2.0\times10^{6}\ m

6 0
3 years ago
The current through an inductor of inductance L is given by I(t) = Imax sin(ωt).
sammy [17]

Answer:

(a) emf_L=-LI_{max}\omega cos(\omega t)

(b) neither increasing or decreasing

(c) opposite to the flow of charge carriers

Explanation:

The current through an inductor of inductance L is given by:

I(t)=I_{max}sin(\omega t)   (1)

(a) The induced emf is given by the following formula

emf_L=-L\frac{dI}{dt}    (2)

You derivative the expression (1) in the expression (2):

emf_L=-L\frac{d}{dt}(I_{max}sin(\omega t))\\\\emf_L=-LI_{max}\omega cos(\omega t)

(b) At t=0 the current is zero

(c) At t = 0 the emf is:

emf_L=-\omega LI_{max}

w, L and Imax have positive values, then the emf is negative. Hence, the induced emf is opposite to the flow of the charge carriers.

(d) read the text carefully

6 0
2 years ago
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