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The weight of the meterstick is:
and this weight is applied at the center of mass of the meterstick, so at x=0.50 m, therefore at a distance
from the pivot.
The torque generated by the weight of the meterstick around the pivot is:
To keep the system in equilibrium, the mass of 0.50 kg must generate an equal torque with opposite direction of rotation, so it must be located at a distance d2 somewhere between x=0 and x=0.40 m. The magnitude of the torque should be the same, 0.20 Nm, and so we have:
from which we find the value of d2:
So, the mass should be put at x=-0.04 m from the pivot, therefore at the x=36 cm mark.
and this weight is applied at the center of mass of the meterstick, so at x=0.50 m, therefore at a distance
from the pivot.
The torque generated by the weight of the meterstick around the pivot is:
To keep the system in equilibrium, the mass of 0.50 kg must generate an equal torque with opposite direction of rotation, so it must be located at a distance d2 somewhere between x=0 and x=0.40 m. The magnitude of the torque should be the same, 0.20 Nm, and so we have:
from which we find the value of d2:
So, the mass should be put at x=-0.04 m from the pivot, therefore at the x=36 cm mark.
Answer:
The magnitude of a uniform electric field that will stop these protons in a distance of 2 m is 1.01 x N/C
Explanation:
given information,
kinetic energy, KE = 3.25 x J
proton's mass, m = 1.673 x kg
charge, q = 1.602 x C
distance, d = 2 m
to find the electric field that will stop the proton, we can use the following equation:
E = F/q
= (KE/d) / q , KE = Fd --> F = KE/d
= KE/qd
= (3.25 x J) / (1.602 x
C)(2 m)
= 1.01 x N/C