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Citrus2011 [14]
3 years ago
12

4. Some early mornings , dew drops can be found on grass or a car parked outside,

Physics
1 answer:
horsena [70]3 years ago
3 0

In some early mornings , dew drops can be found on grass or a car parked outside, but not on other materials such as the sidewalk because  the night -time temperature on grass and the car went below the dew point, but  the temperature of the concrete did not drop enough to reach the dew point  level

Dew can be formed on any object when the temperature of the object drop. When this happen, the object will be cool which will eventually cool the surrounding air around the object.

Dew drops is as a result of condensation in the air. When the cool air causes the air vapor to convert to liquid. The dew will form when the temperature of the object balances with the dew point in the surrounding environment.

In some early mornings , dew drops can be found on grass or a car parked outside, but not on other materials such as the sidewalk because  the night -time temperature on grass and the car went below the dew point, but  the temperature of the concrete did not drop enough to reach the dew point  level

Therefore the correct option is therefore A

Learn more here : brainly.com/question/13834972

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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
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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}

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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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