An unknown mass is on a flat frictionless surface and a 2 kg object is attached to it with a string going over pulleyt mounted t
1 answer:
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a) 120 s
b) v = 0.052R [m/s]
Explanation:
a)
The period of a revolution in a simple harmonic motion is the time taken for the object in motion to complete one cycle (in this case, the time taken to complete one revolution).
The graph of the problem is missing, find it in attachment.
To find the period of revolution of the book, we have to find the time between two consecutive points of the graph that have exactly the same shape, which correspond to two points in which the book is located at the same position.
The first point we take is t = 0, when the position of the book is x = 0.
Then, the next point with same shape is at t = 120 s, where the book returns at x = 0 m.
Therefore, the period is
T = 120 s - 0 s = 120 s
b)
The tangential speed of the book is given by the ratio between the distance covered during one revolution, which is the perimeter of the wheel, and the time taken, which is the period.
The perimeter of the wheel is:
where R is the radius of the wheel.
The period of revolution is:
Therefore, the tangential speed of the book is:
Answer:
(A). The speed of the ions is
(B). The radius of curvature of a singly charged lithium ion is
Explanation:
Given that,
Electric field = 60000 N/C
Magnetic field = 0.0500 T
(A). We need to calculate the velocity
For no deflection
(B). We need to calculate the radius
Using magnetic force balance by centripetal force
Put the value into the formula
Hence, (A). The speed of the ions is
(B). The radius of curvature of a singly charged lithium ion is