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igor_vitrenko [27]
3 years ago
11

A wire with resistance R is connected to the terminals of a 6.0 V battery. What is the potential difference between the ends of

the wire and the current I through it if the wire has the following resistances?
a) 1.0Ω
b) 2.0Ω
c) 3.0Ω
Physics
1 answer:
Bas_tet [7]3 years ago
7 0

Answer:

Potential difference = 6.0 V

I for 1.0Ω = 6 A

I for 2.0Ω = 3 A

I for 3.0Ω = 2 A

Explanation:

Potential difference (ΔV) = Current (I) x Resistance (R)

The potential difference is constant and equals 6.0 V, hence;

I = ΔV/R

When R = 1.0, I =6/1 = 6 amperes

When R = 2.0, I = 6/2 = 3 amperes

When R = 3.0, I = 6/3 = 2 amperes

<em>The potential difference is 6.0 V and the current is 6, 3, and 2 amperes for a resistance of 1.0, 2.0 and 3.0Ω respectively.</em>

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1) 5.5 N

When the ball is at the bottom of the circle, the equation of the forces is the following:

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m \frac{v^2}{R} is the centripetal force, which points upward, with

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R = 1.1 m being the radius of the circular trajectory

Substituting numbers and re-arranging the formula, we find T:

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The minimum velocity for the ball to keep the circular motion occurs when the centripetal force is equal to the weight of the ball, and the tension in the string is zero; therefore:

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