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Whitepunk [10]
3 years ago
7

A particle executes simple harmonic motion with an amplitude of 1.69 cm. At what positive displacement from the midpoint of its

motion does its speed equal one half of its maximum speed
Physics
1 answer:
m_a_m_a [10]3 years ago
5 0

Answer: 0.0146m

Explanation: The formula that defines the velocity of a simple harmonic motion is given as

v = ω√A² - x²

Where v = linear velocity, A = amplitude = 1.69cm = 0.0169m, x = displacement.

The maximum speed of a simple harmonic motion is derived when x = A, hence v = ωA

One half of maximum speed = speed of motion

3ωA/2 = ω√A² - x²

ω cancels out on both sides of the equation, hence we have that

A/2 = √A² - x²

(0.0169)/2 = √(0.0169² - x²)

0.00845 = √(0.0169² - x²)

By squaring both sides, we have that

0.00845² = 0.0169² - x²

x² = 0.0169² - 0.00845²

x² = 0.0002142

x = √0.0002142

x = 0.0146m

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3 years ago
A constant force of 8.0 N is exerted for 4.0 s on a 16-kg object initially at rest. The change in speed of this object will be:
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The change in speed of this object is 3m/s

According to Newton's second law;

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Substitute the given parameters into the formula

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A car travels in a straight line for 5 h at a constant speed of 72 km/h. What is it’s acceleration
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Answer:

a \approx \: 0.001 \: m {s}^{ - 2}

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