Potential energy = mgh
Potential energy = 10 x 9.8 x 1.3
Potential energy = 127.4 J
Answer:
The speed of the 270g cart after the collision is 0.68m/s
Explanation:
Mass of air track cart (m1) = 320g
Initial velocity (u1) = 1.25m/s
Mass of stationary cart (m2) = 270g
Velocity after collision (V) = m1u1/(m1+m2) = 320×1.25/(320+270) = 400/590 = 0.68m/s
To solve this problem it is necessary to apply an energy balance equation in each of the states to assess what their respective relationship is.
By definition the energy balance is simply given by the change between the two states:

Our states are given by



In this way the energy balance for the states would be given by,



Therefore the states of energy would be
Lowest : 0.9eV
Middle :7.5eV
Highest: 8.4eV
230 Newton
Electric charge consists of two types i.e. positively electric charge and negatively electric charge.There was a famous scientist who investigated about this charges. His name is Coulomb and succeeded in formulating the force of attraction or repulsion between two charges i.e. :
F = electric force (N)
k = electric constant (N m² / C²)
q = electric charge (C)
r = distance between charges (m)
The value of k in a vacuum = 9 x 10⁹ (N m² / C²)
F = k(q1 q2)/ r^2
Distance between protons = d = 10⁻¹⁵ m
charge of proton = q = 1.6 × 10⁻¹⁹ C
Here q1=q2
electric force = F =230N
Coulomb's Law. Two protons in an atomic nucleus are typically separated by a distance of 2×10−15m. The electric repulsive force between the protons is huge, but the attractive nuclear force is even stronger and keeps the nucleus from bursting apart.
2 Nuclei and the Need for an Attractive Nuclear Force. The Coulomb force also acts within atomic nucleii, whose characteristic dimension is 10 m, which is called a fermi. There are two protons in a He nucleus, which repel each other because of the Coulomb force.
Find more about electric force of repulsion between nuclear protons
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It depends the amount if mass the object haves and the amount of force it is applied to the object.