How much force must a locomotive exert on a 12840-kg boxcar to make it accelerate forward at 0.490 m/s2?
1 answer:
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Answer:
The new distance is d = 0.447 d₀
Explanation:
The electric out is given by Coulomb's Law
F = k q₁ q₂ / r²
This electric force is in balance with tension.
We reduce the charge of sphere B to 1/5 of its initial value (=q₂ = q₂ / 5) than new distance (d = n d₀)
dat
q₁ =
q₂ =
r = d₀
In order for the deviation to maintain the electric force it should not change, so we apply the Coulomb equation for the two points
F = k q₁ q₂ / d₀²
F = k q₁ (q₂ / 5) / (n d₀)²
.k q₁ q₂ / d₀² = q₁ q₂ / (5 n² d₀²)
5 n² = 1
n = √ 1/5
n = 0.447
The new distance is
d = 0.447 d₀
Answer:
The second system must be set in motion seconds later
Explanation:
The oscillation time, T, for a mass, m, attached to spring with Hooke's constant, k, is:
One oscillation takes T secondes, and that is equivalent to a 2π phase. Then, a difference phase of π/2=2π/4, is equivalent to a time t=T/4.
If the phase difference π/2 of the second system relative to the first oscillator. The second system must be set in motion seconds later