A potential difference of 3.00 nV is set up across a 2.00 cm length of copper wire that has a radius of 2.00 mm. How much charge
drifts through a cross section in 3.00 ms? The resistivity of copper is 8 rho 1.69 10 m − = ⋅ Ω⋅ .
2 answers:
Answer:
3.35×10^-7C
Explanation:
First we must put down the formula for charge drifting at a time t. But current is related to voltage by ohm's law as shown in the image attached. From the relation of ohm's law and the formula for resistivity, the expression that can be used to obtain the charge flowing at a time t, can be derived. Substitution of values gives the answer.
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Answer:
The outline of the energy transfer are;
a) Kinetic energy → Clockwork spring → Potential energy
b) Potential energy in clockwork car → Clockwork spring coil unwound → Clockwork car run
c) Chemical potential energy → Batteries in the car → Electric motors → Kinetic energy
Please find attached the drawings of the energy transfer created with MS Visio
Explanation:
The energy transfer diagrams are diagrams that can be used to indicate the part of a system where energy is stored and the form and location to which the energy is transferred
a) The energy transfer diagram for the winding up a clockwork car is given as follows;
Mechanical kinetic energy is used to wind up (turn) the clockwork car such that the kinetic energy is transformed into potential energy and stored in the wound up clockwork as follows;
Kinetic energy → Clockwork spring → Potential energy
b) Letting a wound up clockwork car run results in the conversion of mechanical potential energy into kinetic (energy due tom motion) energy as follows;
Potential energy in clockwork car → Clockwork spring coil unwound → Clockwork car run
c) The energy stored in the battery of a battery powered car is chemical potential energy. When the battery powered car runs, the chemical potential energy produces an electromotive force which is converted into kinetic energy as electric current flows from the batteries
Therefore, we have;
Chemical potential energy → Batteries in the car → Electric motors → Kinetic energy
Answer:
Electric field, E = 0.064 V/m
Explanation:
It is given that,
Resistivity of silver wire,
Current density of the wire,
We need to find the magnitude of the electric field inside the wire. The relationship between electric field and the current density is given by :
E = 0.0636 V/m
or
E = 0.064 V/m
So, the magnitude of electric field inside the wire is 0.064 V/m. Hence, this is the required solution.