Light is emitted by electrons when they drop from one energy level to a lower level. Which transition results in the emission of
1 answer:
You might be interested in
Answer:
Option c) are perpendicular to the electric field
Explanation:
Equipotential surfaces are perpendicular to the electric field. the electric field lines are projected outwards from the equipotential surface, i.e., the lines of the electric field are at 90 to the equipotential surface.
Equipotential surface are those surfaces that have the same potential at any point on the surface. Thus the potential difference at any point on the surface is zero due to same potential.
Any charge particle on this surface will move in a perpendicular direction to the Coulombian force. No work is done by the force on a particle moving on an equipotential surface.
Answer:
The magnitude of actual velocity is <u>496.67 mph</u> and its direction is <u>51.54° with the x axis in the third quadrant</u>.
Explanation:
Given:
Speed of jumbo jet in southwesterly direction = 550 mph
Velocity of jet stream from west to east direction
First let us draw a vectorial representation of the above velocity vectors.
Consider the south direction as negative y axis and west direction as negative x axis.
From the diagram,
The velocity of the jet can be represented as:
Similarly, the velocity of the stream is,
Now, the vector sum of the above two vectors gives the actual velocity of the aircraft. So, the resultant velocity is given as:
Now, magnitude is given as the square root of sum of the squares of the 'i' and 'j' components. So,
As the horizontal and vertical components of actual velocity negative, the resultant vector makes an angle with the x axis in the third quadrant.
The direction is given as:
Therefore, the magnitude of actual velocity is 496.67 mph and its direction is 51.54° with the x axis in the third quadrant.
