A rod of very small diameter with a mass 2m and length 3L is placed along the xaxis with one end at the origin. An identical rod
is placed along the yaxis with one end at the origin, so that the two rods form an L-shape. What are the coordinates of the center of mass for these two rods
1 answer:
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The time interval for which the proton remains in the field is -
Δt = .
We have a proton entering a uniform magnetic field which is in a direction perpendicular to the proton's velocity.
We have to determine time interval during which the proton is in the field.
<h3>What is the magnitude of force on the charged particle moving in a uniform magnetic field?</h3>The magnitude of force on the charged particle moving in a uniform magnetic field is given by -
F = qvB sinθ
According to the question, we have -
Entering Velocity (v) = 20 i m/s
Magnetic field intensity (B) = 0.3 T
Leaving velocity (u) = - 20 j m/s
Now -
The entering and leaving velocity vectors have 90 degrees difference between them. Therefore, only a quarter of distance of the complete circular path of radius 'R' is traced by the proton. Therefore -
d = =
Since, the radius of circular path is not given, we will assume it R.
Therefore, time for which proton remained in the field is -
t =
Hence, the time interval for which the proton remains in the field is -
Δt =
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Answer:
The “terminal speed” of the ball bearing is 5.609 m/s
Explanation:
Radius of the steel ball R = 2.40 mm
Viscosity of honey η = 6.0 Pa/s
While calculating the terminal speed in liquids where density is high the stokes law is used for viscous force and buoyant force is taken into consideration for effective weight of the object. So the expression for terminal speed (Vt)
Substitute the given values to find "terminal speed"
The “terminal speed” of the ball bearing is 5.609 m/s