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Anuta_ua [19.1K]
3 years ago
13

An object at rest has no net force acting on it. (5 points) True False

Physics
1 answer:
ankoles [38]3 years ago
3 0
The answer to An object at rest has no net force acting on it is True
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Which statement describes a controlled experiment?
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In a controlled experiment, an independent variable (the cause) is systematically manipulated and the dependent variable (the effect) is measured; any extraneous variables are controlled. The researcher can operationalize (i.e. define) the variables being studied so they can be objectivity measured.

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Effects of the earths orbit around the sun
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<h3><u>Effects of the earths orbit around the sun:</u></h3>

The earth moves around the sun in an elliptical orbit, Johannes Kepler, a "German mathematician, and astronomer" described this elliptical orbit first. The orbit is close to being a circle but not a circle. Earth orbiting the sun mainly effects on seasons on earth.

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4 years ago
A 0.40 kg mass hangs on a spring with a spring constant of 12 N/m. The system oscillated with a constant amplitude of 12 cm. Wha
Vaselesa [24]

Answer:

The maximum acceleration of the system is 359.970 centimeters per square second.

Explanation:

The motion of the mass-spring system is represented by the following formula:

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

Where:

x(t) - Position of the mass with respect to the equilibrium position, measured in centimeters.

A - Amplitude of the mass-spring system, measured in centimeters.

\omega - Angular frequency, measured in radians per second.

t - Time, measured in seconds.

\phi - Phase, measured in radians.

The acceleration experimented by the mass is obtained by deriving the position equation twice:

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

Where the maximum acceleration of the system is represented by \omega^{2}\cdot A.

The natural frequency of the mass-spring system is:

\omega = \sqrt{\frac{k}{m} }

Where:

k - Spring constant, measured in newtons per meter.

m - Mass, measured in kilograms.

If k = 12\,\frac{N}{m} and m = 0.40\,kg, the natural frequency is:

\omega = \sqrt{\frac{12\,\frac{N}{m} }{0.40\,kg} }

\omega \approx 5.477\,\frac{rad}{s}

Lastly, the maximum acceleration of the system is:

a_{max} = \left(5.477\,\frac{rad}{s})^{2}\cdot (12\,cm)

a_{max} = 359.970\,\frac{cm}{s^{2}}

The maximum acceleration of the system is 359.970 centimeters per square second.

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