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xz_007 [3.2K]
3 years ago
13

A student look at ocean waves coming into the beach. An oceanwavd with more energy will

Physics
2 answers:
nlexa [21]3 years ago
8 0
It is A, they will have a greater height. I just had this question so it's right. 
pshichka [43]3 years ago
5 0
<span>C. Travel toward the beach faster</span>
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How do you find the missing terms in a number sentence?
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Determine the rule that the numbers in the sequence obey, and then utilize this rule to determine the missing term to complete a number sequence. The rule that the numbers in the example above adhere to is "Add 8 and then remove 2." Therefore, 32 is the term that is missing from the given number sequence.

<h3>What is a number sentence?</h3>
  • A number sentence is an equation or inequality that is articulated using numbers and mathematical symbols in the context of mathematics education.
  • In the US, Canada, UK, Australia, New Zealand, South Africa, and other countries, the phrase is used in the instruction of elementary-level mathematics.
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Learn more about number sentences here:

brainly.com/question/13349953

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5 0
1 year ago
In an oscillating lc circuit, when 75.0% of the total energy is stored in the inductor's magnetic field, (a) what multiple of th
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4 0
3 years ago
An ideal spring hangs from the ceiling. A 2.15 kg mass is hung from the spring, stretching the spring a distance d = 0.0895 m fr
Igoryamba

Answer:

The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.

Explanation:

Given that,

Mass = 2.15 kg

Distance = 0.0895 m

Amplitude = 0.0235 m

We need to calculate the spring constant

Using newton's second law

F= mg

Where, f = restoring force

kx=mg

k=\dfrac{mg}{x}

Put the value into the formula

k=\dfrac{2.15\times9.8}{0.0895}

k=235.41\ N/m

We need to calculate the kinetic energy of the mass

Using formula of kinetic energy

K.E=\dfrac{1}{2}mv^2

Here, v = A\omega

K.E=\dfrac{1}{2}m\times(A\omega)^2

Here, \omega=\sqrt{\dfrac{k}{m}}^2

K.E=\dfrac{1}{2}m\times A^2\sqrt{\dfrac{k}{m}}^2

K.E=\dfrac{1}{2}kA^2

Put the value into the formula

K.E=\dfrac{1}{2}\times235.41\times(0.0235)^2

K.E=0.06500\ J

Hence, The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.

8 0
3 years ago
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