Answer:
define the problem, do background research and specify requirements
Answer:
1. Parallel circuit
2. Parallel circuit
3. Series circuit
4. Series circuit
5. Parallel circuit
6. Parallel circuit
Explanation:
1. In a parallel circuit, there are multiple paths for current to flow. The path each current takes depends on the resistance of the resistors on that path.
2. In a parallel circuit, current splits up into various paths to get the total current through the circuit, the current flow through each resistor is added.
3. In a series circuit the voltage across each resistor is not the same. to get back the total voltage, the voltages across each resistor needs to be added.
4. Series circuits have voltage drops across each resistor. this makes the voltage across each resistor depend on the resistance of the resistor.
5. In parallel circuits voltage is the same across each resistor because they are all connected directly to the same source.
6. In parallel circuits, the power is the same in each resistor of equal resistance since the voltage across each resistor is the same
Answer:
35psig
Explanation:
Pressure is a force that acts on a surface area. The psig measure the pressure pounds per square inch gauge. It measure the difference in pressure between supply tank and outside air. This pressure is relative to atmospheric pressure. When charging an R-410A system, the vapor pressure should be at least 35 psig before switching to liquid charging.
Answer: a) 0.24E+20 m-3. b) p-type extrinsic.
Explanation:
The current density in a semiconductor is composed by two types of charge carriers: electrons and holes.
This parameter, is proportional to the Electric field within the semiconductor, being the proportionality constant, the electrical conductivity of the material, that takes into account the charge carrier concentrations, and the mobility for each type.
The expression for electrical conductivity is as follows:
σ = q . ne . µe + q . np . µp
Replacing by the given values, and the value of q (charge of an electron), we can get the only unknown that remains, ne , as follows:
ne =( σ – (q . np . µp)) / q µe = (13 (Ω.m)-1 – (1.6E-19) coul(4.0E+20) m-3.0.18) m2/V-s /( (1.6E-19).0.38) coul.m2/V-s
ne = 0.24E+20.
As ne is smaller than np, this means that the semiconductor behaves like a p-type extrinsic one.
Answer:
A. energy transformations
Explanation:
