Let the key is free falling, therefore from equation of motion
.
Take initial velocity, u=0, so
.
As velocity moves with constant velocity of 3.5 m/s, therefore we can use formula
From above substituting t,
.
Now substituting all the given values and g = 9.8 m/s^2, we get
.
Thus, the distance the boat was from the point of impact when the key was released is 10.60 m.
The magnitude of the magnetic force on the proton is 1.25× 10—¹³ N.
Speed of the proton = 5.02 × 10 ⁶ m /a
Angel of between the velocity and the magnetic force = 60 °
The magnitude of magnetic field B = 0.180 T
The magnitude of the magnetic force on the proton is,
Therefore, the magnitude of the magnetic force on the proton is 1.25× 10—¹³ N.
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