An electron moves in a circular path perpendicular to a uniform magnetic field with a magnitude of 1.95 mT. If the speed of the
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<span>c. one person exerts more force than the other so that the forces are unbalanced.
In fact, the door is initially at rest. In order to move the door, a net force different from zero should be applied, according to Newton's second law:
</span>
<span>where the term on the left is the resultant of the forces acting on the door, m is the door mass and a its acceleration.
In order to move the door, the acceleration must be different from zero. But this means that the resultant of the forces acting on it must be different from zero: this is possible only if the forces applied by the two persons are unbalanced, i.e. one person exerts more force than the other.</span>
<span>c. one person exerts more force than the other so that the forces are unbalanced.
In fact, the door is initially at rest. In order to move the door, a net force different from zero should be applied, according to Newton's second law:
</span>
<span>where the term on the left is the resultant of the forces acting on the door, m is the door mass and a its acceleration.
In order to move the door, the acceleration must be different from zero. But this means that the resultant of the forces acting on it must be different from zero: this is possible only if the forces applied by the two persons are unbalanced, i.e. one person exerts more force than the other.</span>
Answer:
the magnitude of the average contact force exerted on the leg is 3466.98 N
Explanation:
Given the data in the question;
Initial velocity of hand v₀ = 5.25 m/s
final velocity of hand v = 0 m/s
time interval t = 2.65 ms = 0.00265 s
mass of hand m = 1.75 kg
We calculate force on the hand F
using equation for impulse in momentum
F × t = m( v - v₀ )
we substitute
F × 0.00265 = 1.75( 0 - 5.25 )
F × 0.00265 = 1.75( - 5.25 )
F × 0.00265 = -9.1875
F = -9.1875 / 0.00265
F = -3466.98 N
Next we determine force on the leg F
Using Newton's third law of motion
for every action, there is an equal opposite reaction;
so, F = - F
we substitute
F = - ( -3466.98 N )
F = 3466.98 N
Therefore, the magnitude of the average contact force exerted on the leg is 3466.98 N