An electron is released 7.2 cm from a very long nonconducting rod with a uniform 6.3 μc/m. what is the magnitude of the electron
's initial acceleration?
1 answer:
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Answer:
B. It begins to rotate counterclockwise
Explanation:
It can happen that the merry-go-round remains at rest in the case when the person is standing on the axle of the merry go round whose axis is fixed to some rigid support, but here the person is standing at the center of the merry-go-round not at the axle hence the according to the conservation of angular momentum. Angular momentum is given as:
for the conservation of momentum
where:
moment of inertia of the merry-go-round
angular velocity of the merry-go-round
moment of inertia of the bicycle wheel
angular velocity of the wheel
As the person tries to stop the wheel which is rotating then the person feels an opposing force which tries to moves the whole system in its direction.
Answer:
r₁ = 20.5 cm
Explanation:
In this exercise we can use the conservation of energy
the gravitational power energy is always attractive, the electrical power energy is repulsive if the charges are of the same sign
starting point.
Em₀ = U_g + U_e + K =
the two in the kinetic energy is because they are two particles
final point. When it is detained
Em_f = U_g + U_e =
the energy is conserved
Em₀ = em_f
the charges and masses of the two particles are equal
sustitute the values
-6.67-11 (4.5 10-3) ² / 0.25 - 9, 109 (30 10-9) ² / 0.25 + 4.5 10-3 4² = - 6.67 10- 11 (4.5 10-3) ² / r1 -9 109 (30 10-9) ² / r1
-5.4 10⁻¹⁵ + 3.24 10⁻⁵ - 7.2 10⁻⁵ = -1.35 10⁻¹⁵ / r₁ + 8.1 10⁻⁶ / r₁
We can see that the terms that correspond to the gravitational potential energy are much smaller than the terms of the electric power, which is why we depress them.
3.24 10⁻⁵ - 7.2 10⁻⁵ = 8.1 10⁻⁶ / r₁
-3.96 10⁻⁵ = 8.1 10⁻⁶ / r₁
r₁ = 8.1 10⁻⁶ /3.96 10⁻⁵
r₁ = 2.045 10⁻¹ m
r₁ = 20.5 cm