A racing car going a 20 m/s stops in a distance of 20 m.What is its acceleration?
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The right hand rule to find the direction of the magnetic field for a falling bar is:
- The charge is positive the magnetic field is outgoing, horizontally and towards us.
- The charge of the bar is negative, the magnetic field is incoming, that is horizontal away from us.
The magnetic force is given by the vector product of the velocity and the magnetic field.
F = q v x B
Where the bolds indicate vectors, F is the force, q the charge on the particle, v the velocity and B the magnetic field.
In the vector product, the vectors are perpendicular, which is why the right-hand rule has been established, see attached:
- The thumb points in the direction of speed.
- Fingers extended in the direction of the magnetic field.
- The palm is in the direction of the force if the charge is positive and in the opposite direction if the charge is negative.
They indicate that the bar is dropped, therefore its speed is vertical and downwards, it moves to the left therefore this is the direction of the force, we use the right hand rule, the magnetic field must be horizontal, we have two possibilities:
- If the charge is positive the magnetic field is outgoing, horizontally and towards us.
- If the charge of the bar is negative, the magnetic field is incoming, that is, horizontal away from us
In conclusion using the right hand rule we can find the direction of the magnetic field for a falling bar is:
- The charge of the bar is negative, the magnetic field is incoming, that is horizontal away from us.
- The charge is positive the magnetic field is outgoing, horizontally and towards us.
Learn more about the right hand rule here: brainly.com/question/12847190
To solve this problem we will apply the concepts related to Ohm's law and Electric Power. By Ohm's law we know that resistance is equivalent to,
Here,
V = Voltage
I = Current
While the power is equivalent to the product between the current and the voltage, thus solving for the current we have,
Applying Ohm's law
Therefore the equivalent resistance of the light string is