Answer:
correct answer is b
Explanation:
The frequency of a wave depends on the properties of medium density and the elasticity properties change the amplitude depends on the energy carried by the wave, that is, the amplitude is proportional to the height of the wave (oscillation).
Consequently the amplitude of independent of the frequency because it depends on different factors.
Therefore when changing the amplitude the wavelength remains constant
the correct answer is b
Answer: a circluar motion.
Explanation:
The total kinetic energy before the collision is not equal to the total kinetic energy after the collision. A portion of the kinetic energy is converted to other forms of energy such as sound energy and thermal energy. A collision in which total system kinetic energy is not conserved is known as an inelastic collision.
Answer: Option (b) is the correct answer.
Explanation:
As Kristina is over training, therefore, there is pain in her muscles because human body also requires rest in order to work or function properly.
That is why, gym trainers suggest to give one day off in a week for your workout sessions so that body muscles should recover.
Thus, we can conclude that if Kristina is over training, then recovery training principle should Kristina consider before continuing her program.
Answer:
Part a)
![\Delta U = 4\times 10^5eV](https://tex.z-dn.net/?f=%5CDelta%20U%20%3D%204%5Ctimes%2010%5E5eV)
Part b)
![\Delta U = -4\times 10^5eV](https://tex.z-dn.net/?f=%5CDelta%20U%20%3D%20-4%5Ctimes%2010%5E5eV)
Explanation:
Part a)
Change in potential energy of a charge is given as
![\Delta U = q\Delta V](https://tex.z-dn.net/?f=%5CDelta%20U%20%3D%20q%5CDelta%20V)
here we know that
for proton
also we have
![\Delta V = 400 kV](https://tex.z-dn.net/?f=%5CDelta%20V%20%3D%20400%20kV)
now we have
![\Delta U = e(400 kV)](https://tex.z-dn.net/?f=%5CDelta%20U%20%3D%20e%28400%20kV%29)
![\Delta U = 4\times 10^5eV](https://tex.z-dn.net/?f=%5CDelta%20U%20%3D%204%5Ctimes%2010%5E5eV)
Part b)
Change in potential energy of a charge is given as
![\Delta U = q\Delta V](https://tex.z-dn.net/?f=%5CDelta%20U%20%3D%20q%5CDelta%20V)
here we know that
for proton
also we have
![\Delta V = 400 kV](https://tex.z-dn.net/?f=%5CDelta%20V%20%3D%20400%20kV)
now we have
![\Delta U = -e(400 kV)](https://tex.z-dn.net/?f=%5CDelta%20U%20%3D%20-e%28400%20kV%29)
![\Delta U = -4\times 10^5eV](https://tex.z-dn.net/?f=%5CDelta%20U%20%3D%20-4%5Ctimes%2010%5E5eV)