Suppose you spray your sister with water from a garden hose. The water is supplied to the hose at a rate of 0.625×10−3 m3/s and
1 answer:
Answer:
<em>0.153 m/s</em>
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Explanation:
The flowrate Q = 0.625 x 10-3 m^3-/s
The diameter of the nozzle d = 5.19 x 10^-3 m
the velocity V = ?
The cross-sectional area of the flow A =
==> (3.142 x 5.19 x 10^-3)/4 = 4.077 x 10^-3 m^2
From the continuity equation,
Q = AV
V = Q/A = (0.625 x 10-3)/(4.077 x 10^-3) = <em>0.153 m/s</em>
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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:
Where:
- Position of the mass with respect to the equilibrium position, measured in centimeters.
- Amplitude of the mass-spring system, measured in centimeters.
- Angular frequency, measured in radians per second.
- Time, measured in seconds.
- Phase, measured in radians.
The acceleration experimented by the mass is obtained by deriving the position equation twice:
Where the maximum acceleration of the system is represented by .
The natural frequency of the mass-spring system is:
Where:
- Spring constant, measured in newtons per meter.
- Mass, measured in kilograms.
If and
, the natural frequency is:
Lastly, the maximum acceleration of the system is:
The maximum acceleration of the system is 359.970 centimeters per square second.