Answer:
The change in momentum is
Explanation:
From the question we are told that
The mass of the probe is
The location of the prob at time t = 22.9 s is
The momentum at time t = 22.9 s is
The net force on the probe is
Generally the change in momentum is mathematically represented as
The initial time is 22.6 s
The final time is 22.9 s
Substituting values
(a) 4.06 cm
In a simple harmonic motion, the displacement is written as
(1)
where
A is the amplitude
is the angular frequency
is the phase
t is the time
The displacement of the piston in the problem is given by
(2)
By putting t=0 in the formula, we find the position of the piston at t=0:
(b) -14.69 cm/s
In a simple harmonic motion, the velocity is equal to the derivative of the displacement. Therefore:
(3)
Differentiating eq.(2), we find
And substituting t=0, we find the velocity at time t=0:
(c) -101.13 cm/s^2
In a simple harmonic motion, the acceleration is equal to the derivative of the velocity. Therefore:
Differentiating eq.(3), we find
And substituting t=0, we find the acceleration at time t=0:
(d) 5.00 cm, 1.26 s
By comparing eq.(1) and (2), we notice immediately that the amplitude is
A = 5.00 cm
For the period, we have to start from the relationship between angular frequency and period T:
Using and solving for T, we find
Answer:
Approximately .
Explanation:
This question suggests that the rotation of this object slows down "uniformly". Therefore, the angular acceleration of this object should be constant and smaller than zero.
This question does not provide any information about the time required for the rotation of this object to come to a stop. In linear motions with a constant acceleration, there's an SUVAT equation that does not involve time:
,
where
The angular analogue of that equation will be:
, where
For this object:
The question is asking for an angular acceleration with the unit . However, the angular displacement from the question is described with the number of revolutions. Convert that to radians:
.
Rearrange the equation and solve for
:
.