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Cerrena [4.2K]
3 years ago
12

What part of a standing sound wave does a musician seek in playing a musical note of a specific pitch?

Physics
2 answers:
marusya05 [52]3 years ago
6 0
The correct answer to the question above is that the magician is seeking the wavelength of the standing wave. The part of a standing sound wave, which is its wavelength, the magician is seeking when playing a musical note of a specific pitch. 
Anna35 [415]3 years ago
6 0

That is false. I just took the test and it is not the wavelength. The correct answer is antinode NOT wavelength.

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The ball has 7.35 joules of potential energy at position B. At position A, all of the energy changes to kinetic energy. The velo
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I assume that the ball is stationary (v=0) at point B, so its total energy is just potential energy, and it is equal to 7.35 J. 
At point A, all this energy has converted into kinetic energy, which is:
K= \frac{1}{2}mv^2
And since K=7.35 J, we can find the velocity, v:
v= \sqrt{ \frac{2K}{m} }= \sqrt{ \frac{2 \cdot 7.35 J}{1.5 kg} }=3.1 m/s
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3 years ago
When an astronomer sees certain stars and galaxies that look much redder than expected, what conclusion might the astronomer dra
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It all comes to the doppler effect, the red shift means that the galaxy is moving away from us. The redshift is a result from the doppler effect, so as the galaxy moves away the wavelength expands, increasing the wavelength which responds to the red light.

8 0
3 years ago
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Can anyone help me with these questions? TIA!<br> (Don’t actually answer please! :) )
nataly862011 [7]
<h2>Hey There!</h2><h2>_____________________________________</h2><h2>Question 7: </h2>

 \huge\text{Graphs:}

The graph of  

• The I-V for Ohmic Metal wire conductor at constant temperature  always shows a straight line between the Current(I) plotted at Y axis and Voltage(V) plotted at X axis. Picture 1

• The I-V graph for Diode shows that first the current is zero but as we increase the potential difference(voltage), it results in the increase in the current. Picture 2

<h2>_____________________________________ </h2><h2>Question 8: </h2>

\Large\textbf{Diode:}  

A diode is a device that allows current to flow in only one direction.

\Large\textbf{Forward and Reverse Biasing:}  

Forward Bias, When a diode is forward bias (a voltage in the "forward" direction) then the P-side of the diode is attached to the positive terminal and N-side is fixed to the negative side of the battery which is connected, current flows freely through the device. The forward bias decreases the thickness of potential barrier(The potential barrier barrier in which the charge requires additional force for crossing the region)

Reverse Bias, When a diode is Reverse bias(a voltage in the "backward direction) then the P-side of the diode is connected to the negative terminal and N-side is connected to the positive terminal of the battery which is connected. The reverse bias increases the thickness of the potential barrier resulting in the flow of no current.

 \Large\textbf{Answer to the Question "Resistance"}

The Forward bias decreases the resistance of the diode whereas the reversed bias increases the resistance of the diode. As in forward biasing the current is easily flowing through the circuit whereas reverse bias does not allow the current to flow through it.

<h2>_____________________________________ </h2><h2>Best Regards, </h2><h2>'Borz' </h2>

8 0
3 years ago
Suppose you could fit 100 dimes, end to end, between your card with the pinhole and your dime-sized sunball. how many suns could
Naddika [18.5K]

Answer: 100 suns

Explanation:

We can solve this with the following relation:

\frac{d}{x_{sunball-pinhole}}=\frac{D}{x_{sun-pinhole}}

Where:

d=17.91 mm =17.91(10)^{-3}  m is the diameter of a dime

D is the diameter of the Sun

x_{sun-pinhole}=150,000,000 km=1.5(10)^{11}  m is the distance between the Sun and the pinhole

x_{sunball-pinhole}=100 d=1.791 m is the amount of dimes that fit in a distance between the sunball and the pinhole

Finding D:

D=\frac{d}{x_{sunball-pinhole}}x_{sun-pinhole}

D=\frac{17.91(10)^{-3}  m}{1.791 m} 1.5(10)^{11}  m

D=1.5(10)^{9}  m This is roughly the diameter of the Sun

Now, the distance between the Earth and the Sun is one astronomical unit (1 AU), which is equal to:

1 AU=149,597,870,700 m

So, we have to divide this distance between D in order to find how many suns could it fit in this distance:

\frac{149,597,870,700 m}{1.5(10)^{9}  m}=99.73 suns \approx 100 suns

8 0
3 years ago
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