Use the circuit diagram to decide if the lightbulb will light. Justify your answer.

Physics

2 answers:

podryga [215]3 years ago

5 0

Answer:

The lightbulb will NOT light.

Explanation:

You put me in a difficult position. I can't help it, but the "sample answer" is by far the best way to explain this, briefly and correctly. There's no other choice but to copy it.

This is a short circuit. The branch without the bulb has almost no resistance, so all the current will flow through that branch instead of flowing through the bulb.

If the lower switch were opened, THEN we would have a series circuit. Current would no longer have any other choice but to flow through the bulb, and the bulb would light.

kogti [31]3 years ago

3 0

Answer:

Sample Response: The lightbulb will not light because this is a short circuit. The branch without the bulb has almost no resistance, so all the current will flow through that branch instead of flowing through the bulb.

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Explanation:

This problem can be solved by doing the total energy balance, i.e:

initial (KE + PE) = final (KE + PE). { KE = Kinetic Energy and PE = Potential Energy}

Since the rod comes to a halt at the topmost position, the KE final is 0. Therefore, all the KE initial is changed to PE, i.e, ΔKE = ΔPE.

Now, at the initial position (the rod hanging vertically down), the bottom-most end is given a velocity of v0. The initial angular velocity(ω) of the rod is given by ω = v/r , where v is the velocity of a particle on the rod and r is the distance of this particle from the axis.

Now, taking v = v0 and r = length of the rod(L), we get ω = v0/ 0.8 rad/s

The rotational KE of the rod is given by KE = 0.5Iω², where I is the moment of inertia of the rod about the axis of rotation and this is given by I = 1/3mL², where L is the length of the rod. Therefore, KE = 1/2ω²1/3mL² = 1/6ω²mL². Also, ω = v0/L, hence KE = 1/6m(v0)²

This KE is equal to the change in PE of the rod. Since the rod is uniform, the center of mass of the rod is at its center and is therefore at a distane of L/2 from the axis of rotation in the downward direction and at the final position, it is at a distance of L/2 in the upward direction. Hence ΔPE = mgL/2 + mgL/2 = mgL. (g = 9.8 m/s²)

Now, 1/6m(v0)² = mgL ⇒ v0 = $\sqrt{6gL}$