Answer:
b, a, c
Explanation:
The middle one has the shortest wavelength, then it's the top one and the last one has the longest wavelength.
If object is not accelerating, the sum of all forces on the object will be equal to ZERO...
Answer:
The circular loop experiences a constant force which is always directed towards the center of the loop and tends to compress it.
Explanation:
Since the magnetic field, B points in my direction and the current, I is moving in a clockwise direction, the current is always perpendicular to the magnetic field and will thus experience a constant force, F = BILsinФ where Ф is the angle between B and L.
Since the magnetic field is in my direction, it is perpendicular to the plane of the circular loop and thus perpendicular to L where L = length of circular loop. Thus Ф = 90° and F = BILsin90° = BIL
According to Fleming's left-hand rule, the fore finger representing the magnetic field, the middle finger represent in the current and the thumb representing the direction of force on the circular loop.
At each point on the circular loop, the force is always directed towards the center of the loop and thus tends to compress it.
<u>So, the circular loop experiences a constant force which is always directed towards the center of the loop and tends to compress it.</u>
Answer:
54%
Explanation:
So, we have that the "magnitude of its displacement from equilibrium is greater than (0.66)A—''. Thus, the first step to take in answering this question is to write out the equation showing the displacement in simple harmonic motion which is = A cos w×t.
Therefore, we will have two instances t the displacement that is to say at a point 2π/w - a2 and the second point at a = a2.
Let us say that 2π/w = A, then, we have that a = A cos ^-1 (0.66)/2π. Also, we have that a2 = A/2 - A cos^- (0.66) / 2π.
The next thing to do is to calculate or determine the total length of of the required time. Thus, the total length is given as:
2a1 + ( A - 2a2) = 2A{ cos^-1 (0.66)}/ π.
Therefore, the total percentage of the period does the mass lie in these regions = 100 × {2a1 + ( A - 2a2) }/A = 2 { cos^-1 (0.66)}/ π × 100 = 54%.
Thus, the total percentage of the period does the mass lie in these regions = 54%.
