If you know the answer please answer the following question down in the picture below.
1 answer:
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Answer:
Jolt can expect to replace 160 units.
Explanation:
given data:
standard deviation 1.7
average lfe of connectors = 18.2 months
number of connectors = 5000
z(19) = 0.4706
P(x< 19) = P(z< 0.4706)
P(x< 19) = 0.6810 from z table
therefore 68.1% fails within 19 months.
hence 0.3190 {100-68.1} will fail during that period.
number of unit of Jolt that need to be replace, on average is 0.319*500 = 160 units.
Answer:
Explanation:
A parallel-plate capacitors consist of two parallel plates charged with opposite charge.
Since the distance between the plates (1 cm) is very small compared to the side of the plates (19 cm), we can consider these two plates as two infinite sheets of charge.
The electric field between two infinite sheets with opposite charge is:
where
is the surface charge density, where
Q is the charge on the plate
A is the area of the plate
is the vacuum permittivity
In this problem:
- The side of one plate is
L = 19 cm = 0.19 m
So the area is
Here we want to find the maximum charge that can be stored on the plates such that the value of the electric field does not overcome:
Substituting this value into the previous formula and re-arranging it for Q, we find the charge: