As per Newton's law rate of change in momentum is net force
so we can write it as


now we know that




from above equation

so he will experience 900 N force in above case
The answer is Trend Line.
Answer:
4 m/s² down
Explanation:
We'll begin by calculating the net force acting on the object.
The net force acting on the object from the left and right side is zero because the same force is applied on both sides.
Next, we shall determine the net force acting on the object from the up and down side. This can be obtained as follow:
Force up (Fᵤ) = 15 N
Force down (Fₔ) = 25 N
Net force (Fₙ) =?
Fₙ = Fₔ – Fᵤ
Fₙ = 25 – 15
Fₙ = 10 N down
Finally, we shall determine the acceleration of the object. This can be obtained as follow:
Mass (ml= 2.5 Kg
Net force (Fₙ) = 10 N down
Acceleration (a) =?
Fₙ = ma
10 = 2.5 × a
Divide both side by 2.5
a = 10 / 2.5
a = 4 m/s² down
Therefore, the acceleration of the object is 4 m/s² down
Answer:
The maximum velocity is 0.377 m/s
Explanation:
Please, the solution is in the Word file attached
The radius of a nucleus of hydrogen is approximately

, while we can use the Borh radius as the distance of an electron from the nucleus in a hydrogen atom:

The radius of a dime is approximately

: if we assume that the radius of the nucleus is exactly this value, then we can find how far is the electron by using the proportion

from which we find

So, if the nucleus had the size of a dime, we would find the electron approximately 500 meters away.