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Answer:
Explanation:
For this problem to be easier to calculate, we can represent the triangle as a right triangle whose right angle is located at the origin of a coordinate system. (See picture attached).
With this disposition of the triangle, we can start finding our integral. The hydrostatic force can be set as an integral with the following shape:
γhxdy
we know that γ=62.5 lb/
from the drawing, we can determine the height (or depth under the water) of each differential area is given by:
h=8-y
x can be found by getting the equation of the line, which we'll get by finding the slope of the line and using one of the points to complete the equation:
when substituting the x and y-values given on the graph, we get that the slope is:
once we got this slope, we can substitute it in the point-slope form of the equation:
which yields:
which simplifies to:
we can now solve this equation for x, so we get that:
with this last equation, we can substitute everything into our integral, so it will now look like this:
Now that it's all written in terms of y we can now simplify it, so we get:
we can now proceed and evaluate it.
When using the power rule on each of the terms, we get the integral to be:
By using the fundamental theorem of calculus we get:
When solving we get:
Answer:
Explanation:
Volume of the insulating shell is,
Charge density of the shell is,
Here,
B)
The electric field is
For 0 <r<R the electric field is zero, because the electric field inside the conductor is zero.
C)
For R <r <2R According to gauss law
substitute
D)
The net charge enclosed for each r in this range is positive and the electric field is outward
E)
For r>2R
Charge enclosed is zero, so electric field is zero