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anzhelika [568]
3 years ago
7

Explain how the weight of an object is an application of newton's second law

Physics
1 answer:
Arisa [49]3 years ago
4 0
Weight is a measure of the force of gravity acting on an object. According to Newton’s laws of motion, force is directly proportional to both mass and acceleration, and the equation for force is F=m*a, where m is mass and a is acceleration. We can use this equation to solve for weight.
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Vocabulary word for the total amount of charge in a closed system remains constant
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An oscillator consists of a block attached to a spring (k = 500 N/m). At some time t, the position (measured from the system's e
Alex_Xolod [135]

Answer:

a) \omega = 10.407\,\frac{rad}{s}, b) m = 4.617\,kg, c) A = 1.355\,m

Explanation:

a) The system have a simple armonic motion, whose position function is:

x(t) = A\cdot \cos (\omega\cdot t + \phi)

The velocity function is determined by deriving the position function in terms of time:

v(t) = -\omega \cdot A \cdot \sin(\omega\cdot t + \phi)

The acceleration function is found by deriving again:

a(t) = -\omega^{2} \cdot A \cdot \cos (\omega\cdot t + \phi)

Let assume that t = 0\,s. The following nonlinear system is built:

A\cdot \cos \phi = 0.660\,m

-\omega \cdot A \cdot \sin \phi = -12.3\,\frac{m}{s}

-\omega^{2}\cdot A \cdot \sin \phi = -128\,\frac{m}{s^{2}}

System can be reduced by divinding the second and third expressions by the first expression:

\omega \cdot \tan \phi = 18.636\,\frac{1}{s}

\omega^{2}\cdot \tan \phi = 193.94\,\frac{1}{s^{2}}

Now, the last expression is divided by the first one:

\omega = 10.407\,\frac{rad}{s}

b) The mass of the block is:

m = \frac{k}{\omega^{2}}

m = \frac{500\,\frac{N}{m} }{(10.407\,\frac{rad}{s})^{2} }

m = 4.617\,kg

c) The phase angle is:

\phi = \tan^{-1} \left(\frac{18.636\,\frac{1}{s} }{\omega}  \right)

\phi \approx 0.338\pi

The amplitude is:

A = \frac{0.660\,m}{\cos 0.338\pi}

A = 1.355\,m

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Which of the following determines the value of a piece of information?
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The answer is Its origin
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