"1.0 kg mass is attached to the end of a spring. The mass has an amplitude of 0.10 m and vibrates 2.0 times per second. Find its
2 answers:
Answer:
1.3m/s
Explanation:
Data given,
Mass,m=1.0kg,
Amplitude,A=0.10m,
Frequency,f=2.0Hz.
From the equation of a simple harmonic motion, the displacement of the object at a given time is define as
we can express the velocity by the derivative of the displacement,
Hence
at equilibrium, the velocity becomes
Hence if we substitute values we arrive at
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Answer:
Work out = 28.27 kJ/kg
Explanation:
For R-134a, from the saturated tables at 800 kPa, we get
= 171.82 kJ/kg
Therefore, at saturation pressure 140 kPa, saturation temperature is
= -18.77°C = 254.23 K
At saturation pressure 800 kPa, the saturation temperature is
= 31.31°C = 304.31 K
Now heat rejected will be same as enthalpy during vaporization since heat is rejected from saturated vapour state to saturated liquid state.
Thus, =
= 171.82 kJ/kg
We know COP of heat pump
COP =
=
= 6.076
Therefore, Work out put, W =
= 171.82 / 6.076
= 28.27 kJ/kg
Answer:
The momentum is 1.94 kg m/s.
Explanation:
To solve this problem we equate the potential energy of the spring with the kinetic energy of the ball.
The potential energy of the compressed spring is given by
,
where is the length of compression and
is the spring constant.
And the kinetic energy of the ball is
When the spring is released all of the potential energy of the spring goes into the kinetic energy of the ball; therefore,
solving for we get:
And since momentum of the ball is ,
Putting in numbers we get: