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jenyasd209 [6]
3 years ago
14

If the period of a spring is 5 seconds, what is the frequency?

Physics
2 answers:
masha68 [24]3 years ago
8 0
Frequency(f) = 1/Period(T)
f = 1/5 = 0.2 Hz
Anastasy [175]3 years ago
3 0

Answer:0.2Hz

Explanation:

Frequency is defined as the number of cycles completed by a wave in one seconds. Frequency is therefore measured in cycles/second or Hertz(Hz).

Period (T) is the time taken to complete an oscillation. The relationship between both is that frequency (F) is the reciprocal of the period. Mathematically, F = 1/T

From the question, Period (T) is 5seconds

Frequency = 1/5

F = 0.2Hz

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What is the energy equivalent of an object with a mass of 1. 05 g? 3. 15 Ă— 105 J 3. 15 Ă— 108 J 9. 45 Ă— 1013 J 9. 45 Ă— 1016 J
Molodets [167]

Considering the equivalence between mass and energy given by the expression of Einstein's theory of relativity, the correct answer is the last option: the energy equivalent of an object with a mass of 1.05 kg is 9.45×10¹⁶ J.

The equivalence between mass and energy is given by the expression of Einstein's theory of relativity, where the energy of a body at rest (E) is equal to its mass (m) multiplied by the speed of light (c) squared:

E=m×c²

This indicates that an increase or decrease in energy in a system correspondingly increases or decreases its mass, and an increase or decrease in mass corresponds to an increase or decrease in energy.  

In other words, a change in the amount of energy E, of an object is directly proportional to a change in its mass m.

In this case, you know:

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Replacing:

E= 1.05 kg× (3×10⁸ m/s)²

Solving:

<u><em>E= 9.45×10¹⁶ J</em></u>

Finally, the correct answer is the last option: the energy equivalent of an object with a mass of 1.05 kg is 9.45×10¹⁶ J.

Learn more:

  • brainly.com/question/9477556
5 0
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Answer:

D. Calculate the area under the graph.

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Going even further, one can make the rectangular bars arbitrarily narrow and cover the area under the curve with more and more of these. In fact, in the limit, this is something called a Riemann sum and leads to the definition of the Riemann integral. Using calculus, the area under a curve (hence the distance in this case) can be calculated precisely, under certain existence criteria.

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