What are the choices ?
Without some directed choices, I'm, free to make up any
reasonable statement that could be said about Kevin in this
situation. A few of them might be . . .
-- Kevin will have no trouble getting back in time for dinner.
-- Kevin will have no time to enjoy the scenery along the way.
-- Some simple Physics shows us that Kevin is out of his mind.
He can't really do that.
-- Speed = (distance covered) / (time to cover the distance) .
If time to cover the distance is zero, then speed is huge (infinite).
-- Kinetic energy = (1/2) (mass) (speed)² .
If speed is huge (infinite), then kinetic energy is huge squared (even more).
There is not enough energy in the galaxy to push Kevin to that kind of speed.
-- Mass = (Kevin's rest-mass) / √(1 - v²/c²)
-- As soon as Kevin reaches light-speed, his mass becomes infinite.
-- It takes an infinite amount of energy to push him any faster.
-- If he succeeds somehow, his mass becomes imaginary.
-- At that point, he might as well turn around and go home ...
if he ever reached Planet-Y, nobody could see him anyway.
<span> B. A person moving a ball through a stream of water</span>
Answer:
energy is equal to 1000 J
Explanation:
When the jumper is in the tent, he has a given height, this height gives him a gravitational potential energy, which forms his initial mechanical energy of 1000 J. After jumping, this energy is converted into elastic energy of the rope plus a remainder of potential energy gravitational, it does not reach the ground, but as the friction is negligible the total mechanical energy is conserved, therefore its energy is equal to 1000 J
This is a case of energy transformation, but the total value of mechanical energy does not change
Answer:
A. The target nucleus split into two nuclei, each with fewer nucleons than the original.
Explanation:
Explanation:
Given that,
Object-to-image distance d= 71 cm
Image distance = 26 cm
We need to calculate the object distance
We need to calculate the focal length
Using formula of lens
put the value into the formula
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
Put the value in to the formula
We need to calculate the magnification
Using formula of magnification
The magnification is 0.233.
The image is virtual.
Hence, This is the required solution.
