Answer:
Speed = 575 m/s
Mechanical energy is conserved in electrostatic, magnetic and gravitational forces.
Explanation:
Given :
Potential difference, U = ![$-3.45 \times 10^{-3} \ V$](https://tex.z-dn.net/?f=%24-3.45%20%5Ctimes%2010%5E%7B-3%7D%20%5C%20V%24)
Mass of the alpha particle, ![$m_{\alpha} = 6.68 \times 10^{-27} \ kg$](https://tex.z-dn.net/?f=%24m_%7B%5Calpha%7D%20%3D%206.68%20%5Ctimes%2010%5E%7B-27%7D%20%5C%20kg%24)
Charge of the alpha particle is, ![$q_{\alpha} = 3.20 \times 10^{-19} \ C$](https://tex.z-dn.net/?f=%24q_%7B%5Calpha%7D%20%3D%203.20%20%5Ctimes%2010%5E%7B-19%7D%20%5C%20C%24)
So the potential difference for the alpha particle when it is accelerated through the potential difference is
![$U=\Delta Vq_{\alpha}$](https://tex.z-dn.net/?f=%24U%3D%5CDelta%20Vq_%7B%5Calpha%7D%24)
And the kinetic energy gained by the alpha particle is
![$K.E. =\frac{1}{2}m_{\alpha}v_{\alpha}^2 $](https://tex.z-dn.net/?f=%24K.E.%20%3D%5Cfrac%7B1%7D%7B2%7Dm_%7B%5Calpha%7Dv_%7B%5Calpha%7D%5E2%20%24)
From the law of conservation of energy, we get
![$K.E. = U$](https://tex.z-dn.net/?f=%24K.E.%20%3D%20U%24)
![$\frac{1}{2}m_{\alpha}v_{\alpha}^2 = \Delta V q_{\alpha}$](https://tex.z-dn.net/?f=%24%5Cfrac%7B1%7D%7B2%7Dm_%7B%5Calpha%7Dv_%7B%5Calpha%7D%5E2%20%3D%20%5CDelta%20V%20q_%7B%5Calpha%7D%24)
![$v_{\alpha} = \sqrt{\frac{2 \Delta V q_{\alpha}}{m_{\alpha}}}$](https://tex.z-dn.net/?f=%24v_%7B%5Calpha%7D%20%3D%20%5Csqrt%7B%5Cfrac%7B2%20%5CDelta%20V%20q_%7B%5Calpha%7D%7D%7Bm_%7B%5Calpha%7D%7D%7D%24)
![$v_{\alpha} = \sqrt{\frac{2(3.45 \times 10^{-3 })(3.2 \times 10^{-19})}{6.68 \times 10^{-27}}}$](https://tex.z-dn.net/?f=%24v_%7B%5Calpha%7D%20%3D%20%5Csqrt%7B%5Cfrac%7B2%283.45%20%5Ctimes%2010%5E%7B-3%20%7D%29%283.2%20%5Ctimes%2010%5E%7B-19%7D%29%7D%7B6.68%20%5Ctimes%2010%5E%7B-27%7D%7D%7D%24)
![$v_{\alpha} \approx 575 \ m/s$](https://tex.z-dn.net/?f=%24v_%7B%5Calpha%7D%20%5Capprox%20575%20%5C%20m%2Fs%24)
The mechanical energy is conserved in the presence of the following conservative forces :
-- electrostatic forces
-- magnetic forces
-- gravitational forces
Total thermal energy is the answer to your question.
By how much would its speed reading increase with each second of fall? ... Ex 3.24 For a freely falling object dropped from rest, what is its acceleration at the end of the 5th second ... Pb 3.3 A ball is thrown straight up with an initial speed of 30 m/s. How high does it go, and how long is it in the air (neglecting air resistance)?.
Explanation:
Given that,
Object-to-image distance d= 71 cm
Image distance = 26 cm
We need to calculate the object distance
![u -v= d](https://tex.z-dn.net/?f=u%20-v%3D%20d)
![u=71+26=97\ cm](https://tex.z-dn.net/?f=u%3D71%2B26%3D97%5C%20cm)
We need to calculate the focal length
Using formula of lens
![\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7Bf%7D%3D%5Cdfrac%7B1%7D%7Bv%7D-%5Cdfrac%7B1%7D%7Bu%7D)
put the value into the formula
![\dfrac{1}{f}=\dfrac{1}{-26}+\dfrac{1}{97}](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7Bf%7D%3D%5Cdfrac%7B1%7D%7B-26%7D%2B%5Cdfrac%7B1%7D%7B97%7D)
![\dfrac{1}{f}=-\dfrac{71}{2522}](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7Bf%7D%3D-%5Cdfrac%7B71%7D%7B2522%7D)
![f=-35.52\ cm](https://tex.z-dn.net/?f=f%3D-35.52%5C%20cm)
The focal length of the lens is 35.52.
(B). Given that,
Object distance = 95 cm
Focal length = 29 cm
We need to calculate the distance of the image
Using formula of lens
![\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7Bf%7D%3D%5Cdfrac%7B1%7D%7Bv%7D-%5Cdfrac%7B1%7D%7Bu%7D)
Put the value in to the formula
![\dfrac{1}{-29}=\dfrac{1}{v}-\dfrac{1}{95}](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B-29%7D%3D%5Cdfrac%7B1%7D%7Bv%7D-%5Cdfrac%7B1%7D%7B95%7D)
![\dfrac{1}{v}=\dfrac{1}{-29}-\dfrac{1}{95}](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7Bv%7D%3D%5Cdfrac%7B1%7D%7B-29%7D-%5Cdfrac%7B1%7D%7B95%7D)
![\dfrac{1}{v}=-\dfrac{124}{2755}](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7Bv%7D%3D-%5Cdfrac%7B124%7D%7B2755%7D)
![v=-22.21\ cm](https://tex.z-dn.net/?f=v%3D-22.21%5C%20cm)
We need to calculate the magnification
Using formula of magnification
![m=\dfrac{v}{u}](https://tex.z-dn.net/?f=m%3D%5Cdfrac%7Bv%7D%7Bu%7D)
![m=\dfrac{22.21}{95}](https://tex.z-dn.net/?f=m%3D%5Cdfrac%7B22.21%7D%7B95%7D)
![m=0.233](https://tex.z-dn.net/?f=m%3D0.233)
The magnification is 0.233.
The image is virtual.
Hence, This is the required solution.