Answer:
W = 1562.5 J
Explanation:
Path 1:
W₁ = F₁*d₁ = 385 N * 2.5 m = 962.5 J
Path 2:
W₂ = F₂*d₂ = 130 N * 10 m = 1300 J
Path 3:
W₃ = F₃*d₃ = (-350 N) * 2 m = - 700 J (opposite to the motion)
We get
W = W₁ + W₂ + W₃ = 962.5 J + 1300 J + (- 700 J) = 1562.5 J
Answer:
0.230 s
Explanation:
The period is the length of time from one peak to the next. If it takes the oscillator 0.115 s to go from the lowest point to the highest point, then it takes another 0.115 s to return to the lowest point. So the period is 0.115 + 0.115 = 0.230 seconds.
Explanation:
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Answer:
ΔK.E = 2.5 × 10⁻³ J
Explanation:
Given data in the question, we have:
Charge of the particle, q = 5.0 μC = 5 × 10 ⁻⁶ C
Initial speed of the particle, v = 55 m/s
The potential difference, ΔV = 500 V
Now, the gain in kinetic energy is given as
ΔK.E = q × ΔV
on substituting the values in the above formula, we get
ΔK.E = 5 × 10 ⁻⁶ C × 500 V
or
ΔK.E = 2.5 × 10⁻³ J
