Answer:
Work, W = F * d, and
Work = change in kinetic energy, so W=deltaKE.
Hence,
deltaKE=F * d
(1/2)*m*v^2 =F * d
d=[(1/2)*m*v^2]/F
d=[(1/2)*0.6*20^2]/5
d=24 m.
Explanation:
Work = change in kinetic energy, so W=deltaKE.
Answer:
Part B)
v = 4.98 m/s
Explanation:
Part a)
As the ball is rolling on the inclined the the friction force will be static friction and the contact point of the ball with the plane is at instantaneous rest
The point of contact is not slipping on the ground so we can say that the friction force work done would be zero.
So here in this case of pure rolling we can use the energy conservation
Part b)
By energy conservation principle we know that
initial kinetic energy + initial potential energy = final kinetic energy + final potential energy
so we will have
![\frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = \frac{1}{2}mv'^2 + \frac{1}{2}I\omega'^2 + mgh](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7Dmv%5E2%20%2B%20%5Cfrac%7B1%7D%7B2%7DI%5Comega%5E2%20%3D%20%5Cfrac%7B1%7D%7B2%7Dmv%27%5E2%20%2B%20%5Cfrac%7B1%7D%7B2%7DI%5Comega%27%5E2%20%2B%20mgh)
here in pure rolling we know that
![v = R\omega](https://tex.z-dn.net/?f=v%20%3D%20R%5Comega)
now from above equation we have
![\frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mR^2)(\frac{v}{R})^2 = \frac{1}{2}mv'^2 + \frac{1}{2}(\frac{2}{5}mR^2)(\frac{v'}{R})^2 + mgh](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7Dmv%5E2%20%2B%20%5Cfrac%7B1%7D%7B2%7D%28%5Cfrac%7B2%7D%7B5%7DmR%5E2%29%28%5Cfrac%7Bv%7D%7BR%7D%29%5E2%20%3D%20%5Cfrac%7B1%7D%7B2%7Dmv%27%5E2%20%2B%20%5Cfrac%7B1%7D%7B2%7D%28%5Cfrac%7B2%7D%7B5%7DmR%5E2%29%28%5Cfrac%7Bv%27%7D%7BR%7D%29%5E2%20%2B%20mgh)
now we have
![\frac{1}{2}mv^2(1 + \frac{2}{5}) = \frac{1}{2}mv'^2(1 + \frac{2}{5}) + mgh](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7Dmv%5E2%281%20%2B%20%5Cfrac%7B2%7D%7B5%7D%29%20%3D%20%5Cfrac%7B1%7D%7B2%7Dmv%27%5E2%281%20%2B%20%5Cfrac%7B2%7D%7B5%7D%29%20%2B%20mgh)
now plug in all values in it
![\frac{1}{2}m(6^2)(\frac{7}{5}) = \frac{1}{2}mv'^2(\frac{7}{5}) + m(9.81)(0.80)](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7Dm%286%5E2%29%28%5Cfrac%7B7%7D%7B5%7D%29%20%3D%20%5Cfrac%7B1%7D%7B2%7Dmv%27%5E2%28%5Cfrac%7B7%7D%7B5%7D%29%20%2B%20m%289.81%29%280.80%29)
![25.2 = 0.7v^2 + 7.848](https://tex.z-dn.net/?f=25.2%20%3D%200.7v%5E2%20%2B%207.848)
![v = 4.98 m/s](https://tex.z-dn.net/?f=v%20%3D%204.98%20m%2Fs)
The distance r between the pions when the criteria are satisfied is 1.45 × 10⁻¹⁶ m
If we consider the potential energy (U) between the pions, then (U) can be expressed as:
![\mathbf{U = \dfrac{kq^2}{r} ---- (1)}](https://tex.z-dn.net/?f=%5Cmathbf%7BU%20%3D%20%5Cdfrac%7Bkq%5E2%7D%7Br%7D%20----%20%281%29%7D)
Given that at some instance, the potential energy becomes negligible compared to the final K.E.
As such the conservation of the total energy in the system can be given as:
Again, if we consider the ratio of the potential energy to the kinetic energy to be about 0.01, then:
![\mathbf{\dfrac{U}{K}= 0.01} \\ \\ \mathbf{U = 0.01 K----(2)}](https://tex.z-dn.net/?f=%5Cmathbf%7B%5Cdfrac%7BU%7D%7BK%7D%3D%200.01%7D%20%5C%5C%20%5C%5C%20%5Cmathbf%7BU%20%3D%200.01%20K----%282%29%7D)
∴
Equating both equations (1) and (2) together, we have:
![\mathbf{\dfrac{kq^2}{r} = 0.01 K}](https://tex.z-dn.net/?f=%5Cmathbf%7B%5Cdfrac%7Bkq%5E2%7D%7Br%7D%20%3D%200.01%20K%7D)
![\mathbf{\dfrac{kq^2}{r} = 0.01 \Bigg [ m_{o \pi}c^2 \Big [\dfrac{1}{\sqrt{1 - \dfrac{v_{\pi}^2}{c^2} }} \Big ] \Bigg] }](https://tex.z-dn.net/?f=%5Cmathbf%7B%5Cdfrac%7Bkq%5E2%7D%7Br%7D%20%3D%200.01%20%5CBigg%20%5B%20m_%7Bo%20%5Cpi%7Dc%5E2%20%5CBig%20%5B%5Cdfrac%7B1%7D%7B%5Csqrt%7B1%20-%20%5Cdfrac%7Bv_%7B%5Cpi%7D%5E2%7D%7Bc%5E2%7D%20%7D%7D%20%5CBig%20%5D%20%5CBigg%5D%20%7D)
![\mathbf{r =\dfrac{kq^2}{ 0.01 \Bigg [ m_{o \pi}c^2 \Big [\dfrac{1}{\sqrt{1 - \dfrac{v_{\pi}^2}{c^2} }} \Big ] \Bigg] }}](https://tex.z-dn.net/?f=%5Cmathbf%7Br%20%3D%5Cdfrac%7Bkq%5E2%7D%7B%200.01%20%5CBigg%20%5B%20m_%7Bo%20%5Cpi%7Dc%5E2%20%5CBig%20%5B%5Cdfrac%7B1%7D%7B%5Csqrt%7B1%20-%20%5Cdfrac%7Bv_%7B%5Cpi%7D%5E2%7D%7Bc%5E2%7D%20%7D%7D%20%5CBig%20%5D%20%5CBigg%5D%20%7D%7D)
where:
- r = distance
- k = Columb's constant
- q = charge on a proton
- m_o = rest mass of each pion in the previous question
- c = velocity of light
= calculated velocity of proton in the previous question
Replacing their values in the above equation, the distance (r) between the pions is calculated as:
![\mathbf{r =\dfrac{(9\times 10^9 \ N.m^2/C^2) (1.6022 \times 10^{-19} \ C)^2}{ 0.01 \Bigg [ (2.5 \times 10^{-28\ kg } )\times (3\times 10^8 \ m/s)^2 \Big [\dfrac{1}{\sqrt{1 - \dfrac{(2.97 \times 10^8 \ m/s)^2}{(3 \times 10^8 \ m/s)^2} }} \Big ] \Bigg] }}](https://tex.z-dn.net/?f=%5Cmathbf%7Br%20%3D%5Cdfrac%7B%289%5Ctimes%2010%5E9%20%5C%20N.m%5E2%2FC%5E2%29%20%281.6022%20%5Ctimes%2010%5E%7B-19%7D%20%5C%20C%29%5E2%7D%7B%200.01%20%5CBigg%20%5B%20%282.5%20%5Ctimes%2010%5E%7B-28%5C%20kg%20%7D%20%29%5Ctimes%20%283%5Ctimes%2010%5E8%20%5C%20m%2Fs%29%5E2%20%5CBig%20%5B%5Cdfrac%7B1%7D%7B%5Csqrt%7B1%20-%20%5Cdfrac%7B%282.97%20%5Ctimes%2010%5E8%20%5C%20m%2Fs%29%5E2%7D%7B%283%20%5Ctimes%2010%5E8%20%5C%20m%2Fs%29%5E2%7D%20%7D%7D%20%5CBig%20%5D%20%5CBigg%5D%20%7D%7D)
distance (r) = 1.45 × 10⁻¹⁶ m
Therefore, we can conclude that the distance r between the pions when the criteria are satisfied is 1.45 × 10⁻¹⁶ m
Learn more about electric potential energy here:
brainly.com/question/21808222?referrer=searchResults
Answer:
Concave lenses are used in eyeglasses that correct myopia or nearsightedness.
Answer:
599 N/m
Explanation:
Let k (N/m) be the spring constant. According to the law of energy conservation, the total mechanical energy (kinetic and elastic) any any point along the spring must be the same:
![kx^2/2 + mv^2/2 = constant](https://tex.z-dn.net/?f=kx%5E2%2F2%20%2B%20mv%5E2%2F2%20%3D%20constant)
![k0.12^2/2 + 0.73*7.2^2/2 = k0.23^2/2 + 0.73*4.5^2/2](https://tex.z-dn.net/?f=k0.12%5E2%2F2%20%2B%200.73%2A7.2%5E2%2F2%20%3D%20k0.23%5E2%2F2%20%2B%200.73%2A4.5%5E2%2F2)
![0.0144k + 37.8432 = 0.0529k + 14.7825](https://tex.z-dn.net/?f=0.0144k%20%2B%2037.8432%20%3D%200.0529k%20%2B%2014.7825)
![0.0385k = 23.0607](https://tex.z-dn.net/?f=0.0385k%20%3D%2023.0607)
![k = 599 N/m](https://tex.z-dn.net/?f=k%20%3D%20599%20N%2Fm)