The strength of the fireman in vertical direction will be given by F = m * g. Then, the work done will be given by definition by W = F * d. Substituting the expression of the Force in that of the work, we have that the work will be W = m * g * d. Substituting the given values and assuming that g = 10m / s ^ 2, we have a total work of W = (73) * (10) * (9) = 6570 J
Dropping the ball as this will decrease flight time than the one thrown
The correct answer to this question would be trees.
Answer:
![R=0.023m](https://tex.z-dn.net/?f=R%3D0.023m)
Explanation:
From the question we are told that:
Mass ![m=1.16*10^{-26}](https://tex.z-dn.net/?f=m%3D1.16%2A10%5E%7B-26%7D)
Potential difference ![V=523V](https://tex.z-dn.net/?f=V%3D523V)
Magnitude ![m=0.370 T](https://tex.z-dn.net/?f=m%3D0.370%20T)
Generally the equation for Velocity is mathematically given by
![\frac{1}{2}mv^2=ev](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7Dmv%5E2%3Dev)
![v=\frac{2ev}{m}](https://tex.z-dn.net/?f=v%3D%5Cfrac%7B2ev%7D%7Bm%7D)
![v=\frac{2*1.6*10^{-19}*542}{1.16*10^{-26}}](https://tex.z-dn.net/?f=v%3D%5Cfrac%7B2%2A1.6%2A10%5E%7B-19%7D%2A542%7D%7B1.16%2A10%5E%7B-26%7D%7D)
![v=12.22*10^4m/s](https://tex.z-dn.net/?f=v%3D12.22%2A10%5E4m%2Fs)
Generally the equation for Force is mathematically given by
![F=qvBsin \theta](https://tex.z-dn.net/?f=F%3DqvBsin%20%5Ctheta)
Where
![qVB=m\frac{v^2}{R}](https://tex.z-dn.net/?f=qVB%3Dm%5Cfrac%7Bv%5E2%7D%7BR%7D)
![F=m\frac{v^2}{R}sin\theta](https://tex.z-dn.net/?f=F%3Dm%5Cfrac%7Bv%5E2%7D%7BR%7Dsin%5Ctheta)
Therefore
![R=\frac{mv}{qB sin \theta}](https://tex.z-dn.net/?f=R%3D%5Cfrac%7Bmv%7D%7BqB%20sin%20%5Ctheta%7D)
![R=\frac{1.6*10^{-26}*12.2*10^{4}}{1.60*10^{-19}*0.394 sin 90}](https://tex.z-dn.net/?f=R%3D%5Cfrac%7B1.6%2A10%5E%7B-26%7D%2A12.2%2A10%5E%7B4%7D%7D%7B1.60%2A10%5E%7B-19%7D%2A0.394%20sin%2090%7D)
![R=0.023m](https://tex.z-dn.net/?f=R%3D0.023m)
Work energy theorem states that net work done by forces on an object equals a change in kinetic energy
therefore
a) you push on a box and it moves