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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:
Net force on the block is 32 N.
Acceleration of the object is 6.4 m/s².
Explanation:
Let the acceleration of the object be m/s².
Given:
Mass of the block is,
Force of pull is,
Frictional force on the block is,
The free body diagram of the object is shown below.
From the figure, the net force in the forward direction is given as:
Now, from Newton's second law of motion, net force is equal to the product of mass and acceleration. So,
Therefore, the acceleration of the object in the forward direction is 6.4 m/s².
