To create the shapes, stars are arranged on a piece of cardboard in the desired configuration. If the stars are placed in a smiley face pattern on the cardboard, for example, they will explode into a smiley face in the sky. In fact, you may see several smiley faces in the sky at one time.
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Answer:
meter per second
Explanation:
It could be any other unit such as yard or feet, put it will be whatever measure per second or whatever time.
Examples
feet per second
miles per hour
Part a)
per day electricity power consumed when 100 W bulb is used for 8 hours
for one year consumption
now the cost will be given
now when other energy efficient light is used
for one year consumption
now the cost will be given
Answer:
1.
2.
3.The results from part 1 and 2 agree when r = R.
Explanation:
The volume charge density is given as
We will investigate this question in two parts. First r < R, then r > R. We will show that at r = R, the solutions to both parts are equal to each other.
1. Since the cylinder is very long, Gauss’ Law can be applied.
The enclosed charge can be found by integrating the volume charge density over the inner cylinder enclosed by the imaginary Gaussian surface with radius ‘r’. The integration of E-field in the left-hand side of the Gauss’ Law is not needed, since E is constant at the chosen imaginary Gaussian surface, and the area integral is
where ‘h’ is the length of the imaginary Gaussian surface.
2. For r> R, the total charge of the enclosed cylinder is equal to the total charge of the cylinder. So,
3. At the boundary where r = R:
As can be seen from above, two E-field values are equal as predicted.
Answer:
b. The current stays the same.
Explanation:
In the case given current is supplied by the battery to a bulb . Here, we should know that bulb also apply resistance to the flow of current .
Now, when an identical bulb is connected in parallel to the original bulb .
Therefore, both the resistance( bulb) are in parallel.
We know, when two resistance are in parallel , current through them is same and voltage is divided between them.
Therefore, in this case current stays same in the original bulb.
Hence, this is the required solution.