They'll still be magnets, but they'll never be able to touch each other where they were cut.
I hope this helps you! :-)
Answer:
The car C has KE = 100, PE = 0
Explanation:
The principle of conservation of energy states that although energy can be transformed from one form to another, the total energy of the given system remains unchanged.
The energy that a body possesses due to its motion or position is known as mechanical energy. There are two kinds of mechanical energy: kinetic energy, KE and potential energy, PE.
Kinetic energy is the energy that a body possesses due to its motion.
Potential energy is the energy a body possesses due to its position.
From the principle of conservation of energy, kinetic energy can be transformed into potential energy and vice versa, but in all cases the energy is conserved or constant.
In the diagram above, the cars at various positions of rest or motion are transforming the various forms of mechanical energy, but the total energy is conserved at every point. At the point A, energy is all potential, at B, it is partly potential partly kinetic energy, However, at the point C, all the potential energy has been converted to kinetic energy. At D, some of the kinetic energy has been converted to potential energy as the car climbs up the hill.
Therefore, the car C has KE = 100, PE = 0
Answer:
The near point of an eye with power of +2 dopters, u' = - 50 cm
Given:
Power of a contact lens, P = +2.0 diopters
Solution:
To calculate the near point, we need to find the focal length of the lens which is given by:
Power, P = ![\frac{1}{f}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bf%7D)
where
f = focal length
Thus
f = ![\frac{1}{P}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7BP%7D)
f =
= + 0.5 m
The near point of the eye is the point distant such that the image formed at this point can be seen clearly by the eye.
Now, by using lens maker formula:
![\frac{1}{f} = \frac{1}{u} + \frac{1}{u'}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bf%7D%20%3D%20%5Cfrac%7B1%7D%7Bu%7D%20%2B%20%5Cfrac%7B1%7D%7Bu%27%7D)
where
u = object distance = 25 cm = 0.25 m = near point of a normal eye
u' = image distance
Now,
![\frac{1}{u'} = \frac{1}{f} - \frac{1}{u}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bu%27%7D%20%3D%20%5Cfrac%7B1%7D%7Bf%7D%20-%20%5Cfrac%7B1%7D%7Bu%7D)
![\frac{1}{u'} = \frac{1}{0.5} - \frac{1}{0.25}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bu%27%7D%20%3D%20%5Cfrac%7B1%7D%7B0.5%7D%20-%20%5Cfrac%7B1%7D%7B0.25%7D)
![\frac{1}{u'} = \frac{1}{f} - \frac{1}{u}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bu%27%7D%20%3D%20%5Cfrac%7B1%7D%7Bf%7D%20-%20%5Cfrac%7B1%7D%7Bu%7D)
Solving the above eqn, we get:
u' = - 0.5 m = - 50 cm