What is the net displacement of the particle between 0 seconds and 80seconds?
1 answer:
Answer:
Option (A)
Explanation:
Displacement of a particle on a velocity time graph is represented by the area between the line representing velocity and x-axis (time).
Displacement of a particle from t = 0 o t = 40 seconds = Area of ΔAOB
Area of triangle AOB =
=
= 80 m
Similarly, displacement of the particle from t = 40 to t = 80 seconds = Area of ΔBCD
Area of ΔBCD =
= 80 m
Total displacement of the particle from t = 0 to t = 80 seconds,
= 80 + 80
= 160 m
Option (A) will be the answer.
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a) The length of the arm of the centrifuge is 10.9 m
b) The angular acceleration is
Explanation:
a)
In a uniform circular motion, the centripetal acceleration is given by
where:
is the angular speed of the circular motion
r is the radius of the circle
For the centrifuge in this problem, we have:
is the angular speed
The centripetal acceleration is 3.2 times the acceleration due to gravity (), so:
Therefore, we can re-arrange the previous equation to find r, the radius of the circle (which corresponds to the length of the arm of the centrifuge):
b)
In the second part of the exercise, the centrifuge speeds up from an initial angular speed of 0 to a final angular speed of 1.7 rad/s. The total acceleration experienced at the final moment is
So, 4.4 times the acceleration due to gravity.
The total acceleration is the resultant of the centripetal acceleration () and the tangential acceleration (
):
We know that:
a = 4.4g
So, we can find the tangential acceleration:
The angular acceleration is related to the tangential acceleration by
where r = 10.9 m is the length of the centrifuge. Substituting,
Learn more about centripetal and angular acceleration here:
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