Answer:
v1=18.46m/s
v2=29.8cm/s
Explanation:
We know that
![A=10cm\\T=2s](https://tex.z-dn.net/?f=A%3D10cm%5C%5CT%3D2s)
the equation of the motion is
![x=Acos(\omega t)\\](https://tex.z-dn.net/?f=x%3DAcos%28%5Comega%20t%29%5C%5C)
we can calculate w by using
![\omega=\frac{2\pi}{T}=\frac{2\pi}{2}=\pi](https://tex.z-dn.net/?f=%5Comega%3D%5Cfrac%7B2%5Cpi%7D%7BT%7D%3D%5Cfrac%7B2%5Cpi%7D%7B2%7D%3D%5Cpi)
Hence, we have that
![x=10cm*cos(\pi t)\\](https://tex.z-dn.net/?f=x%3D10cm%2Acos%28%5Cpi%20t%29%5C%5C)
the speed will be
![v=-\omega*Asin(\omega t)\\|v(0.8)|=|\pi*10cm*sin(\pi *0.8)|=18.46\frac{cm}{s}\\|v(1.4)|=|\pi*10cm*sin(\pi *1.4)|=29.8\frac{cm}{s}](https://tex.z-dn.net/?f=v%3D-%5Comega%2AAsin%28%5Comega%20t%29%5C%5C%7Cv%280.8%29%7C%3D%7C%5Cpi%2A10cm%2Asin%28%5Cpi%20%2A0.8%29%7C%3D18.46%5Cfrac%7Bcm%7D%7Bs%7D%5C%5C%7Cv%281.4%29%7C%3D%7C%5Cpi%2A10cm%2Asin%28%5Cpi%20%2A1.4%29%7C%3D29.8%5Cfrac%7Bcm%7D%7Bs%7D)
hope this helps!
<u>Answer:</u>
A perfect example of wave reflection is an <u>echo</u>.
<u>Explanation:</u>
A wave reflection takes place when waves cannot pass through a surface and in turn they bounce back. It is not necessary that wave reflections can only happen with sound waves, they can also take place in light waves. Also, the waves which are reflected have the same frequency as the original wave, but their direction is different. When a wave strikes an object in the same angle, they bounce back straight but when they hit an object with different angle, their direction changes.
Answer:
![V_{3.01}=-93.2m/s](https://tex.z-dn.net/?f=V_%7B3.01%7D%3D-93.2m%2Fs)
![V_{3.005}=-93.1m/s](https://tex.z-dn.net/?f=V_%7B3.005%7D%3D-93.1m%2Fs)
![V_{3.002}=-93.04m/s](https://tex.z-dn.net/?f=V_%7B3.002%7D%3D-93.04m%2Fs)
![V_{3.001}=-93.02m/s](https://tex.z-dn.net/?f=V_%7B3.001%7D%3D-93.02m%2Fs)
![V_{3}=-93m/s](https://tex.z-dn.net/?f=V_%7B3%7D%3D-93m%2Fs)
Explanation:
To calculate average velocity we need the position for both instants t0 and t1.
Now we will proceed to calculate all the positions we need:
![Y_{3}=-99m/s](https://tex.z-dn.net/?f=Y_%7B3%7D%3D-99m%2Fs)
![Y_{3.01}=-99.932m/s](https://tex.z-dn.net/?f=Y_%7B3.01%7D%3D-99.932m%2Fs)
![Y_{3.005}=-99.4655m/s](https://tex.z-dn.net/?f=Y_%7B3.005%7D%3D-99.4655m%2Fs)
![Y_{3.002}=-99.18608m/s](https://tex.z-dn.net/?f=Y_%7B3.002%7D%3D-99.18608m%2Fs)
![Y_{3.001}=-99.09302m/s](https://tex.z-dn.net/?f=Y_%7B3.001%7D%3D-99.09302m%2Fs)
Replacing these values into the formula for average velocity:
![V_{3-3.01}=\frac{Y_{3.01}-Y_{3}}{3.01-3}=-93.2m/s](https://tex.z-dn.net/?f=V_%7B3-3.01%7D%3D%5Cfrac%7BY_%7B3.01%7D-Y_%7B3%7D%7D%7B3.01-3%7D%3D%3Cstrong%3E-93.2m%2Fs%3C%2Fstrong%3E)
![V_{3-3.005}=\frac{Y_{3.005}-Y_{3}}{3.005-3}=-93.1m/s](https://tex.z-dn.net/?f=V_%7B3-3.005%7D%3D%5Cfrac%7BY_%7B3.005%7D-Y_%7B3%7D%7D%7B3.005-3%7D%3D%3Cstrong%3E-93.1m%2Fs%3C%2Fstrong%3E)
![V_{3-3.002}=\frac{Y_{3.002}-Y_{3}}{3.005-3}=-93.04m/s](https://tex.z-dn.net/?f=V_%7B3-3.002%7D%3D%5Cfrac%7BY_%7B3.002%7D-Y_%7B3%7D%7D%7B3.005-3%7D%3D%3Cstrong%3E-93.04m%2Fs%3C%2Fstrong%3E)
![V_{3-3.001}=\frac{Y_{3.001}-Y_{3}}{3.001-3}=-93.02m/s](https://tex.z-dn.net/?f=V_%7B3-3.001%7D%3D%5Cfrac%7BY_%7B3.001%7D-Y_%7B3%7D%7D%7B3.001-3%7D%3D%3Cstrong%3E-93.02m%2Fs%3C%2Fstrong%3E)
To know the actual velocity, we derive the position and we get:
![V=27-40t = -93m/s](https://tex.z-dn.net/?f=V%3D27-40t%20%3D%20-93m%2Fs)
Since you didn't tell us the choices, I can pick anything I like.
The one that always does it for me is " foaming brine " .
Answer:
The Force exerted on the student by crate will be 100 N.
Explanation:
As per the given Question student is trying to push the crate in right direction with the force of 100N.
And we know that, Newton's third law states that every action has equal and opposite reaction.
So from the Newton's 3rd law of motion it is very clear the student must be experiencing the same amount of force which he is applying.