Answer:
reviewing the opinion of the two students we see that neither is right, since when the kinetic energy increases the potential energy decreases by the same value
Explanation:
For this exercise we must use the law of conservation of energy.
Starting point. Resting electron
Em₀ = U = eV
the potential difference and the electric field are related
V = - E d
Final point. When leaving the electric field
= K = ½ m v²
Em₀ = Em_{f}
e E d = ½ m v²
From this expression we see that when an electron moves from the initial point to the final point, the potential energy must decrease, for the total energy to be constant.
When reviewing the opinion of the two students we see that neither is right, since when the kinetic energy increases the potential energy decreases by the same value
The answer is : <span>Planetesimals, protoplanets, planets. This is the order of the celestial body from earliest to latest. </span><span>A </span>protoplanet<span> is a massive object that will eventually become a planet. </span><span>They are at first formed by the accumulation of </span>planetesimals<span> into </span>protoplanets<span>, then into planets.</span>
The net force is 12 N to the left.
Answer:
The highest percentage of change corresponds to the thinnest rod, the correct answer is a
Explanation:
For this exercise we are asked to change the length of the bar by the action of a force applied along its length, in this case we focus on the expression of longitudinal elasticity
F / A = Y ΔL/L
where F / A is the force per unit length, ΔL / L is the fraction of the change in length, and Y is Young's modulus.
In this case the bars are made of the same material by which Young's modulus is the same for all
ΔL / L = (F / A) / Y
the area of the bar is the area of a circle
A = π r² = π d² / 4
A = π / 4 d²
we substitute
ΔL / L = (F / Y) 4 /πd²
changing length
ΔL = (F / Y 4 /π) L / d²
The amount between paracentesis are all constant in this exercise, let's look for the longitudinal change
a) values given d and 3L
ΔL = cte 3L / d²
ΔL = cte L /d² 3
To find the percentage, we must divide the change in magnitude by its value and multiply by 100.
ΔL/L % = [(F /Y 4/π 1/d²) 3L ] / 3L 100
ΔL/L % = cte 100%
b) 3d and L value, we repeat the same process as in part a
ΔL = cte L / 9d²
ΔL = cte L / d² 1/9
ΔL / L% = cte 100/9
ΔL / L% = cte 11%
c) 2d and 2L value
ΔL = (cte L / d ½
)/ 2L
ΔL/L% = cte 100/4
ΔL/L% = cte 25%
d) value 4d and L
ΔL = cte L / d² 1/16
ΔL/L % = cte 100/16
ΔL/L % = cte 6.25%
The highest percentage of change corresponds to the thinnest rod, the correct answer is a
