The y-position of a damped oscillator as a function of time is shown in the figure.
1 answer:
The length of time that an oscillator is allowed to oscillate, as well as its damping coefficient.
- t= 1.33s
- b = 0.0426 s^-1
In most situations, the equation may be expressed numerically as
When we look at the data and see that there are cycles between the timestamps t= 0s and t= 4s, we may conclude that it finishes three cycles once every four seconds. As a result, the length of time that is going to be necessary to finish one cycle of damping will be
t =4/3sec
t= 1.33s
In most situations, the equation for amplitude may be expressed analytically as
A=A_0e^{-bt}
Therefore
3=5e^{-12b}
Therefore
-0.511 = -12b lne
In conclusion, damping refers to an influence that either operates from inside an oscillatory system or acts on it and has the consequence of reducing or halting the system from oscillating. This impact might occur from either side of the system. In physical systems, damping is produced by processes that cause the energy that is stored in an oscillation to be lost. These processes are called dissipative. The collective name for these processes is "damping agents." The damping coefficient may thus be written as
Learn more about the damping coefficient by reading up on it.
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Answer:
Explanation:
The Compton Shift in wavelength when the photons are scattered is given by the following equation:
(1)
Where:
is a constant whose value is given by
, being
the Planck constant,
the mass of the electron and
the speed of light in vacuum.
the angle between incident phhoton and the scatered photon.
We are told the maximum Compton shift in wavelength occurs when a photon isscattered through :
(2)
(3)
Now, let's find the angle that will produce a fourth of this maximum value found in (3):
(4)
(5)
If we want ,
must be equal to 1:
(6)
Finding :
Finally:
This is the scattering angle that will produce
Answer:
24.084 m/s
Explanation:
From the law of conservation of linear momentum
Total momentum before collision equals to the total momentum after collision
Since momentum=mv where m is mass and v is velocity
where
is the mass of the truck,
is velocity of the truck,
is the common velocity of moving and standing truck after collision and
is the mass of the standing truck
Making the subject we obtain
Substituting as 25000 Kg,
as 22.3 m/s,
as 2000 Kg we obtain
Therefore, assuming no friction and considering that after collision they still move eastwards hence common velocity and initial truck velocities are positive
The truck was moving at 24.084 m/s