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Anna007 [38]
3 years ago
8

"1.0 kg mass is attached to the end of a spring. The mass has an amplitude of 0.10 m and vibrates 2.0 times per second. Find its

speed when it passes the equilibrium position."
Physics
2 answers:
AleksandrR [38]3 years ago
6 0

Answer:

haha lol

Explanation:

sorry but i dont know this

prisoha [69]3 years ago
4 0

Answer:

1.3m/s

Explanation:

Data given,

Mass,m=1.0kg,

Amplitude,A=0.10m,

Frequency,f=2.0Hz.

From the equation of a simple harmonic motion, the displacement of the object at a given time is define as

x=Acos\alpha \\

we can express the velocity by the derivative of the displacement,

Hence

V=-Awsin\alpha \\

at equilibrium, the velocity becomes

V=wA\\w=2\pi f

Hence if we substitute values we arrive at

V=2\pi fA\\V=2\pi *2*0.1\\V=1.3m/s

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Answer:

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A child whirls a 3.00 kg ball on a string .50 m from the axis of rotation in a horizontal circle. The ball makes 1 revolution in
melamori03 [73]

Answers:

a) 0.5 m/s^{2}

b) 1.5 N

Explanation:

a) The centripetal acceleration a_{c} of an object moving in a uniform circular motion is given by the following equation:  

a_{c}=\omega^{2} r  

Where:

\omega=1 \frac{rev}{s} is the angular velocity of the ball

r=0.5 m is the radius of the circular motion, which is equal to the length of the string

Then:

a_{c}=(1 \frac{rev}{s})^{2} 0.5 m  

a_{c}=0.5 m/s^{2} This is the centripetal acceleration of the ball

b) On the other hand, in this circular motion there is a force (centripetal force F) that is directed towards the center and is equal to the tension (T) in the string:

F=T=m. a_{c}

Where m=3 kg is the mass of the ball

Hence:

T=(3 kg)(0.5 m/s^{2})

T=1.5 N This is the tension in the string

7 0
3 years ago
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Which of these results in kinetic energy of an object? position motion mass volume
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Answer:

motion

Explanation:

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2 years ago
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What is Planck’s law?
Lyrx [107]

Answer:

The wavelength of the emitted radiation is inversely proportional to its frequency, or λ = c/ν. The value of Planck's constant is defined as 6.62607015 × 10−34 joule∙second.

Explanation:

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Hope this helps!

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