Rank these electromagnetic waves on the basisof their speed (in vacuum). Rank from fastest to slowest. To rank items as equivale
1 answer:
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Answer:
The answer is below
Explanation:
We are to check if the statement is true of false. If it is false, we correct the statement.
Solution:
Acceleration is the time rate of change of velocity. It is the ratio of the change in velocity to the change in time. The acceleration can be gotten from a velocity time graph by finding the slope of the graph.
The x coordinate represent the time and the y coordinate velocity.
5) Graph A passes through the point (0, 0) and (4, 24). Therefore the acceleration (slope) is:
Acceleration =
This is correct.
6) Graph B is a straight line of 12 m/s. It passes through (0, 12) and (4, 12). Hence:
Acceleration =
This is false.
Therefore the acceleration of graph B is 0 m/s².
1) Magnitude of the force:
The magnetic field generated by a current-carrying wire is
where
is the vacuum permeability
I is the current in the wire
r is the distance at which the field is calculated
Using I=135 A, the current flowing in each wire, we can calculate the magnetic field generated by each wire at distance
,
which is the distance at which the other wire is located:
Then we can calculate the magnitude of the force exerted on each wire by this magnetic field, which is given by:
2) direction of the force:
The two currents run in opposite direction: this means that the force between them is repulsive. This can be determined by using the right hand rule. Let's apply it to one of the two wires, assuming they are in the horizontal plane, and assuming that the current in the wire on the right is directed northwards:
- the magnetic field produced by the wire on the left at the location of the wire on the right is directed upward (the thumb of the right hand is directed as the current, due south, and the other fingers give the direction of the magnetic field, upward)
Now let's apply the right-hand rule to the wire on the right:
- index finger: current --> northward
- middle finger: magnetic field --> upward
- thumb: force --> due east --> so the force is repulsive
A similar procedure can be used on the wire on the left, finding that the force exerted on it is directed westwards, so the force between the two wires is repulsive.
The magnetic field generated by a current-carrying wire is
where
I is the current in the wire
r is the distance at which the field is calculated
Using I=135 A, the current flowing in each wire, we can calculate the magnetic field generated by each wire at distance
which is the distance at which the other wire is located:
Then we can calculate the magnitude of the force exerted on each wire by this magnetic field, which is given by:
2) direction of the force:
The two currents run in opposite direction: this means that the force between them is repulsive. This can be determined by using the right hand rule. Let's apply it to one of the two wires, assuming they are in the horizontal plane, and assuming that the current in the wire on the right is directed northwards:
- the magnetic field produced by the wire on the left at the location of the wire on the right is directed upward (the thumb of the right hand is directed as the current, due south, and the other fingers give the direction of the magnetic field, upward)
Now let's apply the right-hand rule to the wire on the right:
- index finger: current --> northward
- middle finger: magnetic field --> upward
- thumb: force --> due east --> so the force is repulsive
A similar procedure can be used on the wire on the left, finding that the force exerted on it is directed westwards, so the force between the two wires is repulsive.