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makvit [3.9K]
4 years ago
15

A copper Wire has a length of 160 m,and a diameter of 1.0 mm,if ,the wire is connected to a 1.5-Volt battery ,How much Current f

lows through the wire?????
Physics
1 answer:
makkiz [27]4 years ago
3 0
Some guidance notes which may help.To calculate the current flow, Ohm's law can be used. This can be written as current=voltage/resistance, or I=V/R. V is 1.5V.R for the copper wire quoted would be calculated as R = resistivity x length/cross sectional area. The area would be calculated from the formula area = pi x diameter squared/4So, R=resistivity x length divided by (pi x diameter squared/4)Until is the resistivity of copper is known, that's about as far as can be gone.Any further questions, please ask.
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A 1.0 C charged object and a 2.0 C charged object are separated by 100 m. Where should a -1.0x10-3 C charged object be placed on
horsena [70]

Answer:

x = 41.2 m

Explanation:

The electric force is a vector magnitude, so it must be added as vectors, remember that the force for charges of the same sign is repulsive and for charges of different sign it is negative.

In this case the fixed charges (q₁ and q₂) are positive and separated by a distance (d = 100m), the charge (q₃ = -1.0 10⁻³ C)) is negative so the forces are attractive, such as loads q₃ must be placed between the other two forces subtract

             F = F₁₃ - F₂₃

let's write the expression for each force, let's set a reference frame on the charge q1

           F₁₃ = k \frac{q_1 q_3}{x^2}

           F₂₃ = k \frac{q_2 q_3}{(d-x)^2}

they ask us that the net force be zero

           F = 0

           0 = F₁₃ - F₂₃

           F₁₃ = F₂₃

          k \frac{q_1 q_3}{x^2} =k \frac{q_2 q_3}{(d-x)^2}

          \frac{q_1}{x^2} = \frac{q_2}{(d-x)^2 }q1 / x2 = q2 / (d-x) 2

       

           (d-x)² = \frac{q_2}{q_1} x²

we substitute

           (100 - x)² = 2/1  x²

           100- x = √2  x

           100 = 2.41 x

           x = 41.2 m

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What is the flow of energy from the falling water to the steam?
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Answer:

The flow of energy from falling water to the steam is;

a) Mechanical → Mechanical → Electrical → Thermal → Thermal

Explanation:

1) Mechanical → Mechanical

The water in the pipe before it falls possesses potential energy which it converts into kinetic energy as it falls from height

2) Mechanical → Mechanical

The water falling from the pipe stream unto the turbine wheel transfers its kinetic (mechanical) energy due to its motion on to the turbine wheel to give the wheel rotational motion

3) Mechanical → Electrical

The kinetic (mechanical) energy from the rotating turbine wheel is converted into electrical energy in the electrical generator which transported through the electrical circuit

4) Electrical → Thermal

The electrical energy from the electric current is then converted into thermal energy as the current passes through the resistors in the heating filament

5) Thermal → Thermal

The heated filament transfers thermal energy to the the water in the beaker by conduction which raises the temperature of the water such that as the water acquires more thermal energy it turns into steam

Therefore, we have the flow of energy from the falling water to steam as follows;

1) Mechanical 2) Mechanical 3) Electrical 4) Thermal 5) Thermal

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