What is the voltage of A simple circuit with one battery
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We know that arc length (x(t)) is given with the following formula:
Where r is the radius of the barrel. We must keep in mind that as barrel rolls its radius decreases because less and less tape is left on it.
If we say that the thickness of the tape is D then with every full circle our radius shrinks by d. We can write this down mathematically:
When we plug this back into the first equation we get:
We must solve this quadratic equation.
The final solution is:
It is rather complicated solution. If we asume that the tape has no thickness we get simply:
Where r is the radius of the barrel. We must keep in mind that as barrel rolls its radius decreases because less and less tape is left on it.
If we say that the thickness of the tape is D then with every full circle our radius shrinks by d. We can write this down mathematically:
When we plug this back into the first equation we get:
We must solve this quadratic equation.
The final solution is:
It is rather complicated solution. If we asume that the tape has no thickness we get simply:
The given vectors quantities can be described by their properties of both
magnitude and direction.
- a. The drawing of the vector extending from point (0, 0) to (190, 0) on the coordinate plane is attached.
- b. The velocity vector extending from (0, 0) to (108.76, 50.714) on the coordinate plane is attached.
- c. The displacement vector extending from (0, 0) to (30·√2, 30·√2) is attached.
Reasons:
a. The magnitude of the vector = 190 N
The direction in which the vector acts = East
Therefore, in vector form, we have;
= 190 × cos(0)·i + 190 × sin(0)·j = 190·i
The vector can be represented by an horizontal line, 190 units long
Coordinate points on the vector = (0, 0) and (190, 0)
The drawing of the vector with the above points using MS Excel is attached.
b. Magnitude of the velocity vector = 120 km/hr. 25° North of east
Solution;
The vector form of the velocity is; = 120 × cos(25)·i + 120×sin(25)·j, which gives;
= 120 × cos(25)·i + 120×sin(25)·j ≈ 108.76·i + 50.714·j
≈ 108.76·i + 50.714·j
Therefore, points that define the vector are; (0, 0) and (108.76, 50.714)
The drawing of the vector is attached
c. The magnitude of the vector = 60 m
The direction of the vector is southwest = West 45° south
The vector form of the displacement is = 60 × cos(45°)·i + 60 × sin(45°)·j
Which gives;
= 30·√2·i + 30·√2·j
Points on the vector are therefore; (0, 0), and (30·√2, 30·√2)
The drawing of the vector is attached
Learn more about vectors here:
