Answer:
Explanation:
Given
mass of archer ![m=0.3\ kg](https://tex.z-dn.net/?f=m%3D0.3%5C%20kg)
Average force ![F_{avg}=201\ N](https://tex.z-dn.net/?f=F_%7Bavg%7D%3D201%5C%20N)
extension in arrow ![x=1.3\ m](https://tex.z-dn.net/?f=x%3D1.3%5C%20m)
Work done to stretch the bow with arrow
![W=F\cdot x](https://tex.z-dn.net/?f=W%3DF%5Ccdot%20x)
![W=201\times 1.3=261.3\ m](https://tex.z-dn.net/?f=W%3D201%5Ctimes%201.3%3D261.3%5C%20m)
This work done is converted into kinetic Energy of arrow
![W=\frac{1}{2}mv^2](https://tex.z-dn.net/?f=W%3D%5Cfrac%7B1%7D%7B2%7Dmv%5E2)
where v= velocity of arrow
![261.3=\frac{1}{2}\times 0.3\times v^2](https://tex.z-dn.net/?f=261.3%3D%5Cfrac%7B1%7D%7B2%7D%5Ctimes%200.3%5Ctimes%20v%5E2)
![v=\sqrt{1742}](https://tex.z-dn.net/?f=v%3D%5Csqrt%7B1742%7D)
![v=41.73\ m/s](https://tex.z-dn.net/?f=v%3D41.73%5C%20m%2Fs)
(b)if arrow is thrown vertically upward then this energy is converted to Potential energy
![W=mgh](https://tex.z-dn.net/?f=W%3Dmgh)
![261.3=0.3\times 9.8\times h](https://tex.z-dn.net/?f=261.3%3D0.3%5Ctimes%209.8%5Ctimes%20h)
![h=\frac{261.3}{0.3\times 9.8}](https://tex.z-dn.net/?f=h%3D%5Cfrac%7B261.3%7D%7B0.3%5Ctimes%209.8%7D)
![h=88.87\ m](https://tex.z-dn.net/?f=h%3D88.87%5C%20m)
The magnitude of the unknown height of the projectile is determined as 16.1 m.
<h3>
Magnitude of the height</h3>
The magnitude of the height of the projectile is calculated as follows;
H = u²sin²θ/2g
H = (36.6² x (sin 29)²)/(2 x 9.8)
H = 16.1 m
Thus, the magnitude of the unknown height of the projectile is determined as 16.1 m.
Learn more about height here: brainly.com/question/1739912
#SPJ1
C. Maintain correct Posture
Answer:
![\Delta K = 52J](https://tex.z-dn.net/?f=%5CDelta%20K%20%3D%2052J)
Explanation:
The change in kinetic energy will be simply the difference between the final and initial kinetic energies: ![\Delta K=K_f-K_i](https://tex.z-dn.net/?f=%5CDelta%20K%3DK_f-K_i)
We know that the formula for the kinetic energy for an object is:
![K=\frac{mv^2}{2}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7Bmv%5E2%7D%7B2%7D)
where <em>m </em>is the mass of the object and <em>v</em> its velocity.
For our case then we have:
![\Delta K = K_f-K_i=\frac{mv_f^2}{2}-\frac{mv_i^2}{2}=\frac{m(v_f^2-v_i^2)}{2}](https://tex.z-dn.net/?f=%5CDelta%20K%20%3D%20K_f-K_i%3D%5Cfrac%7Bmv_f%5E2%7D%7B2%7D-%5Cfrac%7Bmv_i%5E2%7D%7B2%7D%3D%5Cfrac%7Bm%28v_f%5E2-v_i%5E2%29%7D%7B2%7D)
Which for our values is:
![\Delta K = \frac{m(v_f^2-v_i^2)}{2} = \frac{(26Kg)((2m/s)^2-(0m/s)^2)}{2} = 52J](https://tex.z-dn.net/?f=%5CDelta%20K%20%3D%20%5Cfrac%7Bm%28v_f%5E2-v_i%5E2%29%7D%7B2%7D%20%3D%20%5Cfrac%7B%2826Kg%29%28%282m%2Fs%29%5E2-%280m%2Fs%29%5E2%29%7D%7B2%7D%20%3D%2052J)
Answer: It is both B and D
Select all that apply.
At night, thermal energy moves _____.
from space to the atmosphere
from the land to the atmosphere
from the atmosphere to the land
from the atmosphere to space