Answer: 580 x 10^-3 J
Explanation:
0.6mm is 0.6/1000 = 600*10^-6 m
The plate area is .17*.17 = 28.9*10^-3 m^2
Air:
The voltage that can be sustained by 0.60 mm of air dielectric is:
V = 3.0*10^6* 600*10^-6 = 1800 V
The capacitance is:
C = ε*A/d = 8.854*10^-12 * 28.9*10^-3/600*10^-6 = 426*10^-12 F = 426 pF
The energy stored in a capacitor is:
E = (1/2)*C*V^2 = (1/2)*426*10^-12*(1800)^2 = 691*10^-6 J
Teflon:
The voltage is:
V = 60*10^6* 600*10^-6 = 36*10^3 = 36 kV
According to the listed reference, the relative dielectric constant for teflon is 2.1, this figure multiplies the "ε" of free space.
The capacitance is:
C = ε*A/d = 2.1*8.854*10^-12 * 28.9*10^-3/600*10^-6 = 896*10^-12 F = 896 pF
It would have been easier to note that the capacitance is 2.1 times the air-dielectric case.
The maximum energy stored is:
E = (1/2)*C*V^2 = (1/2)* 896*10^-12* (36*10^3)^2 = 580*10^-3 J
I wanna say its A . I could be wrong but im almost 100 percent sure that its A wood
More compressed. moving up = apparent weight (i.e., your norma force) is greater. this means you’ll weighr more and push those springs down even more than you would at rest.
To model time-variant data, one must create a new entity in an m:n relationship with the original entity, is a False statement.
- Like the majority of software engineering initiatives, the ER process begins with gathering user requirements. What information must be retained, what questions must be answered, and what business rules must be implemented (For instance, if the manager column in the DEPARTMENT table is the only column, we have simply committed to having one manager for each department.)
- The end result of the E-R modeling procedure is an E-R diagram that can be roughly mechanically transformed into a set of tables. Tables will represent both entities and relationships; entity tables frequently have a single primary key, but the primary key for relationship tables nearly invariably involves numerous characteristics.
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