The displacement ∆S of the particle during the interval from t = 2sec to 4sec is; 210 sec
<h3>How to find the displacement?</h3>
We are given the velocity equation as;
s' = 40 - 3t²
Thus, the speed equation will be gotten by integration of the velocity equation to get;
s = ∫40 - 3t²
s = 40t - ¹/₂t³
Thus, the displacement between times of t = 2 sec and t = 4 sec is;
∆S = [40(4) - ¹/₂(4)³] - [40(2) - ¹/₂(2)³]
∆S = 210 m
Read more about Displacement at; brainly.com/question/4931057
#SPJ1
Answer:
cpct gvxjjxjhdfjokjdzfjiyddzzsjhxf
Answer:
Maximum number of vehicle = 308
Explanation:
See the attached file for the calculation.
Answer: D. All of the choice A, B and C are correct.
Answer:
The required diameter of the fuse wire should be 0.0383 cm to limit the current to 0.53 A with current density of 459 A/cm²
.
Explanation:
We are given current density of 459 A/cm² and we want to limit the current to 0.53 A in a fuse wire. We are asked to find the corresponding diameter of the fuse wire.
Recall that current density is given by
j = I/A
where I is the current flowing through the wire and A is the area of the wire
A = πr²
but r = d/2 so
A = π(d/2)²
A = πd²/4
so the equation of current density becomes
j = I/πd²/4
j = 4I/πd²
Re-arrange the equation for d
d² = 4I/jπ
d = √4I/jπ
d = √(4*0.53)/(459π)
d = 0.0383 cm
Therefore, the required diameter of the fuse wire should be 0.0383 cm to limit the current to 0.53 A with current density of 459 A/cm²
.
