There is no such thing as an AC Generator, if its a generator, its DC. I think you are talking about an AC Alternator, in which case the answer would be -wrong answer, deleted-
a 1.25 kg block is attached to a spring with spring constant 17.0 n/m . while the block is sitting at rest, a student hits it with a hammer and almost instantaneously gives it a speed of 46.0 cm/s .The amplitude of the subsequent oscillations 48.13 cm/s
a 1.25 kilogram block is fastened to a spring with a 17.0 newtons per meter spring constant. Given that K is equal to 14 Newtons per meter and mass equals 10.5 kg. The block is then struck with a hammer by a student while it is at rest, giving it a speedo of 46.0 cm for a brief period of time. The required energy provided by the hammer, which is half mv squared, is transformed into potential energy as a result of the succeeding oscillations. This is because we know that energy is still available for consultation. So access the amplitude here from here. He will therefore be equal to and by. Consequently, the Newton's spring constant is 14 and the value is 10.5. The velocity multiplied by 0.49
Speed at X equals 0.35 into amplitude, or vice versa. At this point, the spirit will equal half of K X 1 squared plus half. Due to the fact that this is the overall energy, square is equivalent to half of a K square or an angry square. amplitude is 13 and half case 14 x one is 0.35. calculate that is equal to initial velocities of 49 squares and masses of 10.5. This will be divided in half and start at about 10.5 into the 49-square-minus-14. 13.42 into the entire square in 20.35. dividing by 10.5 and taking the square as a result. 231 6.9 Six centimeters per square second. 10.5 into 49 sq. 14. 2 into a 13.42 square entire. then subtract 10.5 from the result to get the square. So that is 48.13cm/s.
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This is incomplete question Complete Question is:
a 1.25 kg block is attached to a spring with spring constant 17.0 n/m . while the block is sitting at rest, a student hits it with a hammer and almost instantaneously gives it a speed of 46.0 cm/s . what are The amplitude of the subsequent oscillations?
Answer:
281.25 J
Explanation:
We are told that the two objects with masses m and 3m.
Also that energy stored in the spring is 375 joules.
Now, initially the centre of mass of the system took place at rest, it means v1 = v and v2 = v/3
Thus, from principle of conservation of energy, we have;
½mv² + ½(3m)(v/3)² = 375J
(m + 3m/9)½v² = 375
(4/3)m × ½v² = 375
Multiply both sides by ¾ to get;
½mv² = 375 × ¾
½mv² = 281.25 J
Therefore, energy of lighter body is 281.25 J
Answer:
F = 2,894 N
Explanation:
For this exercise let's use Newton's second law
F = m a
The acceleration is centripetal
a = v² / r
Angular and linear variables are related.
v = w r
Let's replace
F = m w² r
The radius r and the length of the rope is related
cos is = r / L
r = L cos tea
Let's replace
F = m w² L cos θ
Let's reduce the magnitudes to the SI system
m = 101.7 g (1 kg / 1000g) = 0.1017 kg
θ = 5 rev (2π rad / rev) = 31,416 rad
w = θ / t
w = 31.416 / 5.1
w = 6.16 rad / s
F = 0.1017 6.16² 0.75 cos θ
F = 2,894 cos θ
The maximum value of F is for θ equal to zero
F = 2,894 N
