Without counting wind resistance, They will both reach the ground at the same time. If we apply the concept of kinematics, such as the equation vf^2=vi^2 + 2ad. This equation doesn't count how big or how heavy the mass is, it only focuses on how fast where they in the start and how far are both of them from the ground. So if they both have the same distance and same initial veloctity, then they will reach the ground at the same time.
For example, Try dropping a pen and a paper(Vertically) at the same height, you'll see they'll reach the ground at the same time.
If you count wind resistance, the heavier ball will hit the ground faster, because the air molecules will resist the lighter ball compared to the heavier ball.
Answer:
1. False
2. True
3. True
Explanation:
1- False —> The relation between electric potential and electric field is given such that
Therefore, for a uniform E field, electric potential is linearly proportional to the distance.
2- True —> The electric field lines always cross the equipotential lines perpendicularly.
3- True —> In order to be a potential difference, one source of electric field is enough. The electric potential will decrease radially according to the following formula:
There is no test charge in the formula, only the source charge. Even when there is no test charge, the potential difference between points in space can exist.
Acceleration is the rate at which an object picks up speed. deceleration is the rate at which an object loses speed.
Answer:
The molecules in the gas would be completely stationary
Explanation:
The temperature of a gas is proportional to the average kinetic energy of its molecules, according to the equation (valid for an ideal monoatomic gas):
where
is the average kinetic energy of the molecules
k is the Boltzmann constant
T is the absolute temperature of the gas
For a gas at 0 Kelvin, T = 0: this means that the average kinetic energy of the molecules is also zero. But this means that the motion of the molecules has completely stopped, so the molecules are completely still.