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4 years ago

Identify each statement as referring to a series or parallel circuit.

Engineering

1 answer:

jolli1 [7]4 years ago

7 0

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

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In the LC-3 data path, the output of the address adder goes to both the MARMUX and the PCMUX, potentially causing two very diffe

dangina [55]

Answer:

no need for that

Explanation:

they are not the same at all

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3 years ago

Nancy ate a 500 Cal lunch. Neglecting efficiency issues (i.e., assuming 100% conversion of energy to work), to what height could

VladimirAG [237]

Answer:

$4265.04\ \text{m}$

$2.38\times 10^{10}\ \text{W}$

Explanation:

PE = Energy of food = 500 cal = $500\times4184=2.092\times10^6\ \text{J}$

m = Mass of object = 50 kg

g = Acceleration due to gravity = $9.81\ \text{m/s}^2$

Potential energy of food is given by

$PE=mgh\\\Rightarrow h=\dfrac{PE}{mg}\\\Rightarrow h=\dfrac{2.092\times 10^6}{50\times 9.81}\\\Rightarrow h=4265.04\ \text{m}$

Nancy could raise the weight to a maximum height of $4265.04\ \text{m}$ .

Mass of $H_2$ used per year = $25\times 10^{9}\ \text{kg/year}$

Energy of $H_2$ = $\dfrac{30\times10^9}{1000}=30\times 10^6\ \text{J/kg}$

Power

$P=25\times 10^{9}\ \text{kg/year}\times 30\times 10^6\ \text{J/kg}\\\Rightarrow P=7.5\times 10^{17}\ \text{J/year}\\\Rightarrow P=\dfrac{7.5\times 10^{17}}{365.25\times 24\times 60\times 60}\\\Rightarrow P=2.38\times 10^{10}\ \text{W}$

The power requirement is $2.38\times 10^{10}\ \text{W}$ .

6 0

3 years ago

The following are related to wastewater treatment:

Paraphin [41]

Answer:

A. - Primary wastewater treatment: the physical treatment process which is used to in order to get rid of suspended solids that can settle from wastewater.

- Secondary wastewater treatment: treatment processes remove waste organic ( which are once living/biological) material from wastewater, usually by the use of a biological treatment process.

-Tertiary wastewater treatment: Tertiary treatment isn't needed at all wastewater treatment plants, and the place it is needed, it may be different from one plant to another, depending on the type of water contamination that must be removed.

B. Primary treatment: Approximately 35% of the incoming biochemical oxygen demand (BOD).

Secondary treatment : Approximately 85% of biochemical oxygen demand (BOD).

C. - Activated Sludge, Rotating biological contractors.

Explanation:

A. Primary wastewater treatment refers to sedimentation, the physical treatment process which is used to in order to get rid of suspended solids that can settle from wastewater. The main goal of primary treatment is to remove organic and inorganic solids that settles by sedimentation, and by act of skimming removed materials end up floating.

Secondary wastewater treatment processes remove waste organic ( which are once living/biological) material from wastewater, usually by the use of a biological treatment process. The goal of secondary treatment is the further treatment of the effluent from primary treatment in order to get rid of the residual organics and suspended solids.

Advanced wastewater is quite in between. Tertiary treatment isn't needed at all wastewater treatment plants, and the place it is needed, it may be different from one plant to another, depending upon the type of water contamination that must be removed. Advanced treatment processes can be combined with primary or secondary treatment or used instead of secondary treatment.

B. Primary treatment: Approximately 35% of the incoming biochemical oxygen demand (BOD).

Secondary treatment : Approximately 85% of biochemical oxygen demand (BOD).

-Activated Sludge: The activated sludge process is a kind of wastewater treatment process for treating sewage/industrial wastewaters by the use of aeration and a biological floc which consists of bacteria and protozoa.

- Rotating biological contactor: Rotating biological contactors (RBCs) are fixed-film reactors that have similarities with biofilters in the sense that organisms are attached to support media.

Activated sludge is a suspended process, which is when the biomass is mixed with the sewage while in rotating biological contractor, the biomass grows on the media and then, the sewage passes over the surface.

6 0

4 years ago

Read 2 more answers

With increases in magnification, which of the following occur? a. The field of view decreases. b. The ambient illumination decre

Irina-Kira [14]

By increasing magnification you decrease the field of view.

The answer is A.

Hope this helps.

r3t40

7 0

3 years ago

Consider a voltage v = Vdc + vac where Vdc = a constant and the average value of vac = 0. Apply the integral definition of RMS t

Anna11 [10]

Answer:

Proof is as follows

Proof:

Given that , $V = V_{ac} + V_{dc}$

<u>for any function f with period T, RMS is given by</u>

<u /> $RMS = \sqrt{\frac{1}{T}\int\limits^T_0 {[f(t)]^{2} } \, dt }$ <u />

In our case, function is $V = V_{ac} + V_{dc}$

$RMS = \sqrt{\frac{1}{T}\int\limits^T_0 {[V_{ac} + V_{dc}]^{2} } \, dt }$

Now open the square term as follows

$RMS = \sqrt{\frac{1}{T}\int\limits^T_0 {[V_{ac}^{2} + V_{dc}^{2} + 2V_{dc}V_{ac}] } \, dt }$

Rearranging terms

$RMS = \sqrt{\frac{1}{T}\int\limits^T_0 {V_{dc}^{2} } \, dt + \frac{1}{T}\int\limits^T_0 {V_{ac}^{2} } \, dt + \frac{1}{T}\int\limits^T_0 {2V_{dc}V_{ac} } \, dt }$

You can see that

second term is square of RMS value of Vac

Third terms is average of VdcVac and given is that average of $V_{ac}V_{dc} = 0$

$RMS = \sqrt{\frac{1}{T}TV_{dc}^{2} + [RMS~~ of~~ V_{ac}]^2 }$

$RMS = \sqrt{V_{dc}^{2} + [RMS~~ of~~ V_{ac}]^2 }$

So it has been proved that given expression for root mean square (RMS) is valid

7 0

3 years ago