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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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Two horizontal metal plates, each 10.0 cm square, are aligned 1.00 cm apart with one above the other. They are given equal-magni
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a) motion is PARABOLIC, b) positive particle is accelerated towards the negative plate,  c)  x = 6.19 10⁹ m

Explanation:

This exercise looks at the motion of a positively charged particle in an electric field.

a) Since the field is vertical the acceleration in this direction is

            F = m a

the electric force is

           F = q E

we substitute

          q E = m a

           a = qE / m

the mass of the particle is m = 2.00 10-16 kg

           a = 1.6 10⁻¹⁹ 2.02 10³ / 2.00 10⁻¹⁶ kg

           a = 1,616 m / s²

           

on the x-axis there are no relationships because there are no forces.

Since the particle has velocities in both axes, its motion is PARABOLIC,

b) the positive particle is accelerated towards the negative plate,

The field is descending, for which the event is down

c) where  hit the particle on the x-axis

they indicate that the particle leaves the center of the negative plate, for which we will fix our reference system at this point.

Let's find the components of the initial velocity.

           sin θ = v_{oy} / v

           cos θ = v₀ₓ / v

           v_{oy} = v₀ sin θ

           v₀ₓ = v₀ cos θ

           v_{oy) = 1.02 10⁵ sin 37 = 0.6139 10⁵ m / s

           v₀ₓ = 1.02 10⁵ cos 37 = 0.8146 10⁵ m / s

Let's find the time it takes to hit the negative plate

            y = y₀I + v_{oy} t + ½ a and t2

in this case the positions are y = y₀ = 0 and the accelerations

a = - 1,616m/s2,

we substitute

            0 = 0 + v_{oy} t - ½ a_y t²

            v_{oy}= ½ a_y t

            t = 2v_{oy} / a_y

let's calculate

           t = 2 0.6139 10⁵ / 1.616

           t = 7.597 10⁴s

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           x = v₀ₓ t

           x = 0.8145 10⁵ 7.597 10⁴

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the particle falls off the plate

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