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# Tutor profile: Walter W.

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Walter W.
Experienced Physics Teacher
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## Questions

### Subject:Physics (Newtonian Mechanics)

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Question:

An object has three forces acting on it. As a result the object moves to the right in translational equilibrium. What does this mean?

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Walter W.

Translational equilibrium means that there is no net force acting on the object. The vector sum of the three forces acting on the object equals zero. Therefore the object is not undergoing an acceleration. This brings us to the conclusion that the object is moving at a constant speed.

### Subject:Physics (Electricity and Magnetism)

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Question:

An apparatus consists of two oppositely charged parallel conducting plates, each with area A = 0.25 $$m^{2}$$, seperated by a distance of 0.010 m. Each plate has a hole at its center through which electrons can pass. High velocity electrons produced by an electron source enter the top plate with speed $$v_{0}$$ = 5.40 x $$10^{6}$$ m/s, take 1.49 ns to travel between the plates, and leave the bottom plate with speed $$v_{f}$$ = 8.02 x $$10^{6}$$ m/s. (a) Which of the plates, top or bottom, is negatively charged? (b) Calculate the magnitude of the electric field between the plates. (c) Calculate the magnitude of the charge on each plate.

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Walter W.

(a) Since the electron accelerates downward towards the bottom plate (from law of charges) the bottom plate must have a positive charge, making the upper plate with a negative charge. Therefore the electric field (E) points upward between the plates. (b) E = $$\frac{F}{q}$$ = $$\frac{ma}{q}$$ = $$\frac{m(v_{f} - v_{i})/t}{q}$$ E = $$\frac{(9.11 x 10^{-31} kg)(8.02 x 10^6 m/s - 5.4 x 10^6 m/s)}{(1.6 x 10^{-19} C)(1.49 x 10^{-9} s)}$$ = 10000N/C (c) E = $$\frac{Q}{\epsilon_{0}A}$$ Q = E$$\epsilon_{0}$$A Q = (10000N/C)(8.85 x $$10^{-12}$$$$C^{2}$$/N$$m^{2}$$)(0.25$$m^{2}$$) Q = 2.2 x $$10^{-8}$$C

### Subject:Physics

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Question:

How much tension must a rope withstand if it is to accelerate a 1200 kg mass vertically upward at 0.80 m/$$s^{2}$$?

Inactive
Walter W.

We need to identify the forces acting on the 1200 kg mass by using a free-body diagram. A point represents the mass and the forces acting on the mass are represented by arrows. In this problem we have one force pointing up representing the tension in the rope $$F_{T}$$, and one force pointing down representing the weight $$F_{G}$$ = mg. Next we apply Newton's 2nd Law, $$\sum_{} F=ma$$ (remember: direction of the arrows from the free body diagram determines if the force is positive (up) or negative (down)) $$F_{T}$$ - $$F_{G}$$ = ma Solve for $$F_{T}$$ $$F_{T}$$ = ma + $$F_{G}$$ $$F_{T}$$ = ma + mg Now plug in the appropriate quantities. $$F_{T}$$ = (1200 kg)(0.80 m/$$s^{2}$$) + (1200 kg)(9.8 m/$$s^{2}$$) $$F_{T}$$ = 12720 kg m/$$s^{2}$$ or 12720 N

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