1). The equation is: (speed) = (frequency) x (wavelength)
Speed = (256 Hz) x (1.3 m) = 332.8 meters per second
2). If the instrument is played louder, the amplitude of the waves increases.
On the oscilloscope, they would appear larger from top to bottom, but the
horizontal size of each wave doesn't change.
If the instrument is played at a higher pitch, then the waves become shorter,
because 'pitch' is directly related to the frequency of the waves, and higher
pitch means higher frequency and more waves in any period of time.
If the instrument plays louder and at higher pitch, the waves on the scope
become taller and there are more of them across the screen.
3). The equation is: Frequency = (speed) / (wavelength)
(Notice that this is exactly the same as the equation up above in question #1,
only with each side of that one divided by 'wavelength'.)
Frequency = 300,000,000 meters per second / 1,500 meters = 200,000 per second.
That's ' 200 k Hz ' .
Note:
I didn't think anybody broadcasts at 200 kHz, so I looked up BBC Radio 4
on-line, and I was surprised. They broadcast on several different frequencies,
and one of them is 198 kHz !
Answer:
hi what is your question?? say in English please
<span>Germanium
To determine which melts first, convert their melting temperatures so they're both expressed on same scale. It doesn't matter what scale you use, Kelvin, Celsius, of Fahrenheit. Just as long as it's the same scale for everything. Since we already have one substance expressed in Kelvin and since it's easy to convert from Celsius to Kelvin, I'll use Kelvin. So convert the melting point from Celsius to Kelvin for Gold by adding 273.15
1064 + 273.15 = 1337.15 K
So Germanium melts at 1210K and Gold melts at 1337.15K. Germanium has the lower melting point, so it melts first.</span>
Answer:
Velocity = 3.25[m/s]
Explanation:
This problem can be solved if we use the Bernoulli equation: In the attached image we can see the conditions of the water inside the container.
In point 1, (surface of the water) we have the atmospheric pressure and at point 2 the water is coming out also at atmospheric pressure, therefore this members in the Bernoulli equation could be cancelled.
The velocity in the point 1 is zero because we have this conditional statement "The water surface drops very slowly and its speed is approximately zero"
h2 is located at point 2 and it will be zero.
![(P_{1} +\frac{v_{1}^{2} }{2g} +h_{1} )=(P_{2} +\frac{v_{2}^{2} }{2g} +h_{2} )\\P_{1} =P_{2} \\v_{1}=0\\h_{2} =0\\v_{2}=\sqrt{0.54*9.81*2}\\v_{2}=3.25[m/s]](https://tex.z-dn.net/?f=%28P_%7B1%7D%20%2B%5Cfrac%7Bv_%7B1%7D%5E%7B2%7D%20%7D%7B2g%7D%20%2Bh_%7B1%7D%20%29%3D%28P_%7B2%7D%20%2B%5Cfrac%7Bv_%7B2%7D%5E%7B2%7D%20%7D%7B2g%7D%20%2Bh_%7B2%7D%20%29%5C%5CP_%7B1%7D%20%3DP_%7B2%7D%20%5C%5Cv_%7B1%7D%3D0%5C%5Ch_%7B2%7D%20%3D0%5C%5Cv_%7B2%7D%3D%5Csqrt%7B0.54%2A9.81%2A2%7D%5C%5Cv_%7B2%7D%3D3.25%5Bm%2Fs%5D)
