To respond to the question, we need the data provided with the question.
Answer:
Radius of the solenoid is 0.93 meters.
Explanation:
It is given that,
The magnetic field strength within the solenoid is given by the equation,
, t is time in seconds
The induced electric field outside the solenoid is 1.1 V/m at a distance of 2.0 m from the axis of the solenoid, x = 2 m
The electric field due to changing magnetic field is given by :
x is the distance from the axis of the solenoid
, r is the radius of the solenoid
r = 0.93 meters
So, the radius of the solenoid is 0.93 meters. Hence, this is the required solution.
Impulse = Ft=mΔv => Δv = Ft/m = 4.28/0.18 = 23.78 m/s
But,
Δv = v1-v2, where v1 = initial velocity = 16 m/s, v2 = final velocity
Therefore,
v1 - v2 = 23.78 => v2 = v1 - 23.78 => v2 = 16 - 23.78 = -7.78 m/s
The velocity of ball after the force is 7.78 m/s in the direction of the force.
The maximum static force that can be applied is equal to the normal force*the frictional force. the normal force on the box is equal to mg since the floor is flat using 9.81m/s^2 for gravity 12kg*9.81m/s^2 = 118N multiplying the normal force by the frictional force you get a 118*.42= 49.6N so overcome the force of static friction on the box a minimum of 49.6N would need to be applied.
Answer:
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