The net charge on the said dielectric is 0
1C = 6.25 * 10¹⁸ electrons; hence:
-12.5x10¹⁸ free electrons = (-12.5x10¹⁸ free electrons * (1C/ 6.25 * 10¹⁸ electrons))
-12.5x10¹⁸ free electrons = -2 C
A positively charged electric has a charge of 2 Coulombs, hence:
Q₁ = 2C
12.5x10¹⁸ free electrons are added to it, hence:
Q₂ = -2C
Hence the net charge on the said dielectric is 0
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Answer:
Technician A is wrong
Technician B is right
Explanation:
voltage drop of 0.8 volts on the starter ground circuit is not within specifications. Voltage drop should be within the range of 0.2 V to 0.6 V but not more than that.
A spun bearing can seize itself around the crankshaft journal causing it not to move. As the car ignition system is turned on, the stater may draw high current in order to counter this seizure.
Si depende en que casos se y de que trate su trabajo pero tampoco lo pueden obligar pero como ya dije depende
Answer: The overhead percentage is 7.7%.
Explanation:
We call overhead, to all those bytes that are delivered to the physical layer, that don't carry real data.
We are told that we have 700 bytes of application data, so all the other bytes are simply overhead, i.e. , 58 bytes composed by the transport layer header, the network layer header, the 14 byte header at the data link layer and the 4 byte trailer at the data link layer.
So, in order to assess the overhead percentage, we divide the overhead bytes between the total quantity of bytes sent to the physical layer, as follows:
OH % = (58 / 758) * 100 = 7.7 %
