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Semmy [17]
2 years ago
10

"\bold{What \: is \: the \: speed? }" alt="\bold{What \: is \: the \: speed? }" align="absmiddle" class="latex-formula">​

Physics
2 answers:
scoray [572]2 years ago
8 0

Refer to the above attachment

STALIN [3.7K]2 years ago
5 0

A mouse runs a distance of 2 meters in 15 seconds.

It's Speed will be 0.13 m/sec.

Average speed =

\frac{distance}{time}

Therefore,

Average Speed =

\frac{2}{15} = 0.1333333 m/sec

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Read 2 more answers
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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