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Anit [1.1K]
2 years ago
11

An object is spun around in circular motion such that its period is 0.2 seconds.

Physics
1 answer:
kherson [118]2 years ago
3 0

Answer:

a. 5 Hertz

b. 35 rotations

Explanation:

a. Frequency is given by f=\frac{1}{T} where T is the period, in seconds. Given that T=0.2, the frequency of this object's rotation is:

f=\frac{1}{T}=\frac{1}{0.2}=\boxed{5\text{ Hz}}

b. Since one rotation, or period, is completed in 0.2 seconds, divide the amount of time (7 seconds) by the time it takes to complete one revolution (0.2 seconds):

\frac{7}{0.2}=\boxed{35 \text{ rotations}}

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A box slides down a ramp inclined at 24◦ to the horizontal with an acceleration of 1.7 m/s 2 . The acceleration of gravity is 9.
dsp73

Answer:

<h3>0.445</h3>

Explanation:

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\mu = \frac{F_f}{R}

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\mu = \frac{Wsin \theta}{W cos\theta} \\\mu = \frac{sin \theta}{cos\theta} \\\mu = tan \theta

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A pressure vessel that has a volume of 10m3 is used to store high-pressure air for operating a supersonic wind tunnel. If the ai
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Answer:

The mass of air stored in the vessel is 235.34 kilograms.

Explanation:

Let supossed that air inside pressure vessel is an ideal gas, The density of the air (\rho), measured in kilograms per cubic meter, is defined by following equation:

\rho = \frac{P\cdot M}{R_{u}\cdot T} (1)

Where:

P - Pressure, measured in kilopascals.

M - Molar mass, measured in kilomoles per kilogram.

R_{u} - Ideal gas constant, measured in kilopascal-cubic meters per kilomole-Kelvin.

T - Temperature, measured in Kelvin.

If we know that P = 2026.5\,kPa, M = 28.965\,\frac{kg}{kmol}, R_{u} = 8.314\,\frac{kPa\cdot m^{2}}{kmol\cdot K} and T = 300\,K, then the density of air is:

\rho = \frac{(2026.5\,kPa)\cdot \left(28.965\,\frac{kg}{kmol} \right)}{\left(8.314\,\frac{kPa\cdot m^{2}}{kmol\cdot K} \right)\cdot (300\,K)}

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The mass of air stored in the vessel is derived from definition of density. That is:

m = \rho \cdot V (2)

Where m is the mass, measured in kilograms.

If we know that \rho = 23.534\,\frac{kg}{m^{3}} and V = 10\,m^{3}, then the mass of air stored in the vessel is:

m = \left(23.534\,\frac{kg}{m^{3}} \right)\cdot (10\,m^{3})

m = 235.34\,kg

The mass of air stored in the vessel is 235.34 kilograms.

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