The time period resulting in oscillations will be 1.986 seconds.
<h3>What is the period of oscillation?</h3>
The period is the amount of time it takes for a particle to perform one full oscillation. T is the symbol for it. Taking the reciprocal of the frequency yields the frequency of the oscillation.
The time period of the oscillation is;
Hence the time period resulting oscillations will be 1.986 seconds.
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Answer:
t1 = t2 + 3.02 V = 41.5
V t1 - 1/2 g t1^2 = V t2 - 1/2 g t2^2
Both stones reach the same height after the specified times
V (t1 - t2) = g/2 (t1^2 - t2^2) = g/2 (t1 - t2) (t1 + t2)
2 V / g = t1 + t2 = 2t1 + 3.02
t1 = V / g - 1.51 = 41.5 / 9.8 -1.51 = 2.72 s
t2 = t1 + 3.02 = 5.74 sec
Check:
41.5 * 2.72 - 4.9 * 2.72^2 = 76.6 m
41.5 * 5.74 - 4.9 * 5.74^2 = 76.8 m
Speed of second stone = 41.5 - 9.8 * 2.72 = 14.8 m/s
Answer:
Explanation:
you mean deceleration right ? because the acceleration is 250m/s
Answer:
1.) 274.5v
2.) 206.8v
Explanation:
1.) Given that In one part of the lab activities, students connected a 2.50 µF capacitor to a 746 V power source, whilst connected a second 6.80 µF capacitor to a 562 V source.
The potential difference and charge across EACH capacitor will be
V = Voe
Where Vo = initial voltage
e = natural logarithm = 2.718
For the first capacitor 2.50 µF,
V = Vo × 2.718
746 = Vo × 2.718
Vo = 746/2.718
Vo = 274.5v
To calculate the charge, use the below formula.
Q = CV
Q = 2.5 × 10^-6 × 274.5
Q = 6.86 × 10^-4 C
For the second capacitor 6.80 µF
V = Voe
562 = Vo × 2.718
Vo = 562/2.718
Vo = 206.77v
The charge on it will be
Q = CV
Q = 6.8 × 10^-6 × 206.77
Q = 1.41 × 10^-3 C
B.) Using the formula V = Voe again
165 = Vo × 2.718
Vo = 165 /2.718
Vo = 60.71v
Q = C × 60.71
Q = C
Answer:
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