Kinetic energy and potential energy pair is the quantity in which one will increase then other will decrease
As we know that sum of kinetic energy and potential energy will always remain conserved
So here we will have
so here as we move away from mean position the kinetic energy will decrease while at the same time potential energy will increase.
So the pair of potential energy and kinetic energy will satisfy the above condition
It would be the first option.
Explanation-
The number of protons is equal to the atomic number the number of neutrons is the mass minus the atomic number.
Answer:
His first law states that every object will remain at rest or in uniform motion in a straight line unless compelled to change its state by the action of an external force. In other words its inertia.
Explanation:
By using the second law of Newton, the frictional force is 200N.
We need to know about the second law of Newton (force) to solve this problem. The total force applied an object is proportional to the mass of object and acceleration. It can be defined as
∑F = m . a
where F is force, m is mass and a is acceleration.
From the question above, we know that
F1 = 200N
v = constant therefore (a = 0 m/s²)
By using second law of Newton, we get
∑F = m . a
F1 - Ffriction = m . 0
200 - Ffriction = 0
Ffriction = 200 N
Hence, the frictional force is 200N.
Find more on force at: brainly.com/question/25239010
#SPJ4
The weight of a column of air with cross-sectional area 4. 5 m^2 extending from earth's surface to the top of the atmosphere is, 4.56*10^5N.
To find the answer, we have to know about the pressure.
<h3>How to find the weight of a column of air?</h3>
- As we know that the expression of pressure as,
where; F is the force, here it is equal to the weight of the air column, and A is the area of cross section.
- It is given that, the air column is extending from earth's surface to the top of the atmosphere, thus, the pressure will be atmospheric pressure,
- From this, the value of weight will be,
Thus, we can conclude that, the weight of a column of air with cross-sectional area 4. 5 m^2 extending from earth's surface to the top of the atmosphere is, 4.56*10^5N.
Learn more about the pressure here:
brainly.com/question/12830237
#SPJ4
