Answer:
Part a)
![x = 15.76 m](https://tex.z-dn.net/?f=x%20%3D%2015.76%20m)
Part b)
![y = 7.94 m](https://tex.z-dn.net/?f=y%20%3D%207.94%20m)
Part c)
![x = 26.16 m](https://tex.z-dn.net/?f=x%20%3D%2026.16%20m)
Part d)
![y = 7.49 m](https://tex.z-dn.net/?f=y%20%3D%207.49%20m)
Part e)
![x = 83.23 m](https://tex.z-dn.net/?f=x%20%3D%2083.23%20m)
Part f)
![y = -75.6 m](https://tex.z-dn.net/?f=y%20%3D%20-75.6%20m)
Explanation:
As we know that catapult is projected with speed 19.9 m/s
so here we have
![v_x = 19.9 cos39.9](https://tex.z-dn.net/?f=v_x%20%3D%2019.9%20cos39.9)
![v_x = 15.3 m/s](https://tex.z-dn.net/?f=v_x%20%3D%2015.3%20m%2Fs)
similarly we have
![v_y = 19.9 sin39.9](https://tex.z-dn.net/?f=v_y%20%3D%2019.9%20sin39.9)
![v_y = 12.76 m/s](https://tex.z-dn.net/?f=v_y%20%3D%2012.76%20m%2Fs)
Part a)
Horizontal displacement in 1.03 s
![x = v_x t](https://tex.z-dn.net/?f=x%20%3D%20v_x%20t)
![x = (15.3)(1.03)](https://tex.z-dn.net/?f=x%20%3D%20%2815.3%29%281.03%29)
![x = 15.76 m](https://tex.z-dn.net/?f=x%20%3D%2015.76%20m)
Part b)
Vertical direction we have
![y = v_y t - \frac{1]{2}gt^2](https://tex.z-dn.net/?f=y%20%3D%20v_y%20t%20-%20%5Cfrac%7B1%5D%7B2%7Dgt%5E2)
![y = (12.76)(1.03) - 4.9(1.03)^2](https://tex.z-dn.net/?f=y%20%3D%20%2812.76%29%281.03%29%20-%204.9%281.03%29%5E2)
![y = 7.94 m](https://tex.z-dn.net/?f=y%20%3D%207.94%20m)
Part c)
Horizontal displacement in 1.71 s
![x = v_x t](https://tex.z-dn.net/?f=x%20%3D%20v_x%20t)
![x = (15.3)(1.71)](https://tex.z-dn.net/?f=x%20%3D%20%2815.3%29%281.71%29)
![x = 26.16 m](https://tex.z-dn.net/?f=x%20%3D%2026.16%20m)
Part d)
Vertical direction we have
![y = v_y t - \frac{1]{2}gt^2](https://tex.z-dn.net/?f=y%20%3D%20v_y%20t%20-%20%5Cfrac%7B1%5D%7B2%7Dgt%5E2)
![y = (12.76)(1.71) - 4.9(1.71)^2](https://tex.z-dn.net/?f=y%20%3D%20%2812.76%29%281.71%29%20-%204.9%281.71%29%5E2)
![y = 7.49 m](https://tex.z-dn.net/?f=y%20%3D%207.49%20m)
Part e)
Horizontal displacement in 5.44 s
![x = v_x t](https://tex.z-dn.net/?f=x%20%3D%20v_x%20t)
![x = (15.3)(5.44)](https://tex.z-dn.net/?f=x%20%3D%20%2815.3%29%285.44%29)
![x = 83.23 m](https://tex.z-dn.net/?f=x%20%3D%2083.23%20m)
Part f)
Vertical direction we have
![y = v_y t - \frac{1]{2}gt^2](https://tex.z-dn.net/?f=y%20%3D%20v_y%20t%20-%20%5Cfrac%7B1%5D%7B2%7Dgt%5E2)
![y = (12.76)(5.44) - 4.9(5.44)^2](https://tex.z-dn.net/?f=y%20%3D%20%2812.76%29%285.44%29%20-%204.9%285.44%29%5E2)
![y = -75.6 m](https://tex.z-dn.net/?f=y%20%3D%20-75.6%20m)
Lifting hands and the down by one student at a time best describe the presentation of the transverse wave by students. Option D is correct.
<h3>
What is a Transverse wave?</h3>
- The wave in which the oscillation of particles is is perpendicular to the direction of energy transfer.
- The students can make a transverse wave by raising their hands up and then down, one student at a time.
- The raised hand represents the oscillation of particles while the sequence of the raising hand represents the direction of energy transfer.
Therefore, lifting hands and the down by one student at a time best describe the presentation of the transverse wave by students.
Learn more about Transverse waves:
brainly.com/question/3813804
Answer:
Velocity of truck will be 20.287 m /sec
Explanation:
We have given mass of the truck m = 4000 kg
Radius of the turn r = 70 m
Coefficient of friction ![\mu =0.6](https://tex.z-dn.net/?f=%5Cmu%20%3D0.6)
Centripetal force is given ![F=\frac{mv^2}{r}](https://tex.z-dn.net/?f=F%3D%5Cfrac%7Bmv%5E2%7D%7Br%7D)
And frictional force is equal to ![F_{frictional}=\mu mg](https://tex.z-dn.net/?f=F_%7Bfrictional%7D%3D%5Cmu%20mg)
For body to be move these two forces must be equal
So ![\frac{mv^2}{r}=\mu mg](https://tex.z-dn.net/?f=%5Cfrac%7Bmv%5E2%7D%7Br%7D%3D%5Cmu%20mg)
![v=\sqrt{\mu rg}=\sqrt{0.6\times 70\times 9.8}=20.287m/sec](https://tex.z-dn.net/?f=v%3D%5Csqrt%7B%5Cmu%20rg%7D%3D%5Csqrt%7B0.6%5Ctimes%2070%5Ctimes%209.8%7D%3D20.287m%2Fsec)
Convection, because it is the process of heat transfer from one location to the next by the movement of fluids. The moving fluid carries energy within it.
Violet light is at the end of the visible light section of the electromagnetic spectrum. Ultraviolet rays are directly next to violet rays on the EM Spectrum.