Physics, 09.09.2020 23:01 christopherschool04

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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 from: Quest

cell differentiation is how generic embryonic cells become specialized cells. this occurs through a process called gene expression. gene expression is the specific combination of genes that are turned on or off (expressed or repressed), and this is what dictates how a cell functions.

explanation:

Answer from: Quest

we have been given a quadratic function $f(x)=(x-5)^{2} +2$ and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is $[2,\infty)$ .

if we restrict the domain of this function to either $(-\infty,5]$ or $[5,\infty)$ , it will become one to one function.

let us know find its inverse.

$y=(x-5)^{2}+2$

upon interchanging x and y, we get:

$x=(y-5)^{2}+2$

let us now solve this function for y.

$(y-5)^{2}=x-2\\ y-5=\pm \sqrt{x-2}\\ y=5\pm \sqrt{x-2}\\$

hence, the inverse function would be $f^{-1}(x)=5+\sqrt{x-2}$ if we restrict the domain of original function to $[5,\infty)$ and the inverse function would be $f^{-1}(x)=5-\sqrt{x-2}$ if we restrict the domain to $(-\infty,5]$ .

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