Answer:
30N is the correct answer
African American, Native American, and asian
Answer:
W = 181.26 J
Explanation:
Given that,
The force acting on the cart, F = 20 N
It is at an angle of 25 degrees above the horizontal to move a crate a distance of 10m across the floor.
We need to find work done by the student. The work done by the student is given by :
![W=Fd\cos\theta\\\\W=20\times 10\times \cos25\\W=181.26\ J](https://tex.z-dn.net/?f=W%3DFd%5Ccos%5Ctheta%5C%5C%5C%5CW%3D20%5Ctimes%2010%5Ctimes%20%5Ccos25%5C%5CW%3D181.26%5C%20J)
So, the required work done is 181.26 J.
Answer:
0.614 nm
Explanation:
Energy of the nth state of one dimensional infinite wall is,
![E=\frac{n^{2} h^{2} }{8mL^{2} }](https://tex.z-dn.net/?f=E%3D%5Cfrac%7Bn%5E%7B2%7D%20h%5E%7B2%7D%20%7D%7B8mL%5E%7B2%7D%20%7D)
Given the energy to excite an electron from ground state to first excited state is,
![\Delta E=3eV\\\Delta E=3(1.6\times10^{-19})J](https://tex.z-dn.net/?f=%5CDelta%20E%3D3eV%5C%5C%5CDelta%20E%3D3%281.6%5Ctimes10%5E%7B-19%7D%29J)
And the Plank's constant is, ![h=6.626\times10^{-34}Js](https://tex.z-dn.net/?f=h%3D6.626%5Ctimes10%5E%7B-34%7DJs)
Mass of electron,![m=9.1\times10^{-31}kg](https://tex.z-dn.net/?f=m%3D9.1%5Ctimes10%5E%7B-31%7Dkg)
Now the energy will of a 1 dimensional infinite wall which excite an electron from ground state to first excited state will be,
![\Delta E=\frac{2^{2} h^{2} }{8mL^{2} } -\frac{1^{2} h^{2} }{8mL^{2} }](https://tex.z-dn.net/?f=%5CDelta%20E%3D%5Cfrac%7B2%5E%7B2%7D%20h%5E%7B2%7D%20%7D%7B8mL%5E%7B2%7D%20%7D%20-%5Cfrac%7B1%5E%7B2%7D%20h%5E%7B2%7D%20%7D%7B8mL%5E%7B2%7D%20%7D)
Put all the variables in above equation and rearrange it for L.
![L^{2} =\frac{3((6.626\times10^{-34}Js)^{2} )}{8\times9.1\times10^{-31}kg\times3(1.6\times10^{-19})J} \\L^{2} =0.376922012\times10^{-18} m^{2}\\ L=0.6139\times10^{-9}m\\ L=0.614nm](https://tex.z-dn.net/?f=L%5E%7B2%7D%20%3D%5Cfrac%7B3%28%286.626%5Ctimes10%5E%7B-34%7DJs%29%5E%7B2%7D%20%29%7D%7B8%5Ctimes9.1%5Ctimes10%5E%7B-31%7Dkg%5Ctimes3%281.6%5Ctimes10%5E%7B-19%7D%29J%7D%20%5C%5CL%5E%7B2%7D%20%3D0.376922012%5Ctimes10%5E%7B-18%7D%20m%5E%7B2%7D%5C%5C%20L%3D0.6139%5Ctimes10%5E%7B-9%7Dm%5C%5C%20L%3D0.614nm)
Therefore the width of the box is 0.614 nm.