A car battery with a 12 V emf and an internal resistance of 0.037 Ω is being charged with a current of 45 A. Note that in this p
1 answer:
Answer:
Part a)
Part b)
Part c)
Part d)
Explanation:
Part a)
As we know that the during the charging process of the battery the terminal voltage of the cell is given as
Part b)
Thermal energy dissipated in the battery is due to its internal resistance
so it is given as
here we have
Part c)
rate of energy conversion in the in the battery is given as
Part d)
percentage of the power conversion is given as
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Answer:
35.14°C
Explanation:
The equation for linear thermal expansion is , which means that a bar of length
with a thermal expansion coefficient
under a temperature variation
will experiment a length variation
.
We have then = 0.481 foot,
= 1671 feet and
= 0.000013 per centigrade degree (this is just the linear thermal expansion of steel that you must find in a table), which means from the equation for linear thermal expansion that we have a
= 22.14°. As said before, these degrees are centigrades (Celsius or Kelvin, it does not matter since it is only a variation), and the foot units cancel on the equation, showing no further conversion was needed.
Since our temperature on a cool spring day was 13.0°C, our new temperature must be = 35.14°C
Complete Question
The complete question is shown on the uploaded image
Answer:
The nuclei experience a force that will move it to the right of the conductor rod while the electrons experience a force that will move it to the left side of the conductor rod.
Explanation:
The force that act on the charges(both the positive and the negative charge ) is Mathematically expressed as
F = qE => E = F/q
where F is the force, q is the charge and E is the electric field
we can see that the force is directly proportional to the electric field,so an increase in the electric field would increase the force.
we can also see that the electric field is given as force per unit charge and generally the direction of this field is taken to be the direction of the force it would exert on a positive test charge
Now from the question we are being told that the external electric field is the direction of the positive x axis
Hence this field would drive the positive charge i.e the nuclei to the right.
In order to further explain let consider this
Generally the electric field is always radially outward originating from a positive point charge and radially in toward a negative point charge.
what this means for this question, is that the positive point charge is on the left side of the electric field while the negative point charge is at the right side of the field.
According to Coulomb's law which states that unlike term attract while like terms repel, it means that the electron would move to the left of the conductor rod while the nuclei would move to the right side of the conductor rod.
