A 40-kg uniform semicircular sign 1.6 m in diameter is supported by two wires as shown. What
1 answer:
Answer:
T1 = 131.4 [N]
T2 = 261 [N]
Explanation:
To solve this problem we must make a sketch of how will be the semicircle, for this reason we conducted an internet search, to find the scheme of the problem. This scheme is attached in the first image.
Then we make a free body diagram, with this free body diagram, we raise the forces that act on the body. Since it is a problem involving static equilibrium, the sum of forces in any direction and moments must be equal to zero.
By performing a sum of forces on the Y axis equal to zero we can find an equation that relates the forces of tension T1 & T2.
The second equation can be determined by summing moments equal to zero, around the point of application of the T1 force. In this way we find the T2 force.
The value of T2, is replaced in the first equation and we can find the value for T1.
Therefore
T1 = 131.4 [N]
T2 = 261 [N]
The free body diagram and the developed equations can be seen in the second attached image.
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Using the ground as reference point, the mechanical energy of the brick when it is at 5 m from the ground is just potential energy (because the brick is initially at rest, so it doesn't have kinetic energy):
when the brick is at h'=3 m from the ground, its mechanical energy is now sum of kinetic energy and potential energy:
where v is the velocity of the brick. Since E is conserved, it must be equal to the initial energy (98.1 J), so we can solve this equation to find v:
(a) 12.0 V
In this problem, the capacitor is connected to the 12.0 V, until it is fully charged. Considering the capacity of the capacitor, , the charged stored on the capacitor at the end of the process is
When the battery is disconnected, the charge on the capacitor remains unchanged. But the capacitance, C, also remains unchanged, since it only depends on the properties of the capacitor (area and distance between the plates), which do not change. Therefore, given the relationship
and since neither Q nor C change, the voltage V remains the same, 12.0 V.
(b) (i) 24.0 V
In this case, the plate separation is doubled. Let's remind the formula for the capacitance of a parallel-plate capacitor:
where:
is the permittivity of free space
is the relative permittivity of the material inside the capacitor
A is the area of the plates
d is the separation between the plates
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and since Q (the charge) does not change (the capacitor is now isolated, so the charge cannot flow anywhere), the new voltage is
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which means