We need the power law for the change in potential energy (due to the Coulomb force) in bringing a charge q from infinity to distance r from charge Q. We are only interested in the ratio U₁/U₂, so I'm not going to bother with constants (like the permittivity of space).
<span>The potential energy of charge q is proportional to </span>
<span>∫[s=r to ∞] qQs⁻²ds = -qQs⁻¹|[s=r to ∞] = qQr⁻¹, </span>
<span>so if r₂ = 3r₁ and q₂ = q₁/4, then </span>
<span>U₁/U₂ = q₁Qr₂/(r₁q₂Q) = (q₁/q₂)(r₂/r₁) </span>
<span>= 4•3 = 12.</span>
The static friction exerted on the block by the incline is
.
The given parameters;
- <em>mass of the block, = M</em>
- <em>coefficient of static friction in section 1, = </em>
<em /> - <em>angle of inclination of the plane, = θ</em>
<em />
The normal force on the block is calculated as follows;
Fₙ = Mgcosθ
The static friction exerted on the block by the incline is calculated as follows;

Thus, the static friction exerted on the block by the incline is 
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the electric force decreases because the distance has an indirect relationship to the force
Explanation:
The electric force between two objects is given by

where
k is the Coulomb's constant
q1 and q2 are the charges of the two objects
r is the distance between the two objects
As we can see from the formula, the magnitude of the force is inversely proportional to the square of the distance: so, when the distance between the object increases, the magnitude of the force decreases.
Answer: There will be 75258 nuclei left at 6 pm.
Explanation:
a) half-life of the radioactive substance:
Half life is the amount of time taken by a radioactive material to decay to half of its original value.


b) Expression for rate law for first order kinetics is given by:

where,
k = rate constant
t = time for decomposition = 6 hours ( from 12 noon to 6 pm)
A = activity at time t = ?
= initial activity = 600, 000


Thus there will be 75258 nuclei left at 6 pm.