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Oxana [17]
4 years ago
7

Design circuits that demonstrate all of the principles listed below. Set up the circuits and take measurements to show that the

principle in question is indeed correct.
Principles of Series Circuits:
The voltages across each circuit element add to that of the battery.
The current through each circuit element is the same.
Higher resistances have higher voltage across them.
Resistors added in series to any circuit increase the resistance to current in the circuit.

Principles of Parallel Circuits:
Voltages across each circuit element are the same.
The current through each parallel circuit element adds to that going through the battery.
The higher resistance in a parallel circuit has less current.
Resistors added in parallel to any circuit reduce the resistance to current in the circuit.

General Principles:
A branch in a circuit that contains a short has no voltage across it.
A branch in a circuit that is open has all the voltage across it.
A battery is a constant voltage source.
A battery is not a constant current source.
Ammeters have very little internal resistance.
Voltmeters have very large internal resistance.

Engineering
1 answer:
Nata [24]4 years ago
3 0

<u>Explanation</u>:

For series

\Delta V=V_{1}+V_{2}+\ldots+V_{n}=I R_{1}+I R_{2}+\ldots+I R_{n}(\text {voltages add to the batter } y)

\(I=I_{1}=I_{2}=I_{n}\) (current is the same)

V=I R(\text {voltage is directly proportional to } R)

R_{e q}=R_{1}+R_{2}+\ldots+R_{n} \quad \text { (resistance increase) }

For parallel

\Delta V=\Delta V_{1}=\Delta V_{2}=\Delta V_{n} \quad(\text { same voltage })

I=I_{1}+I_{2}+\ldots+I_{n}(\text {current adds})

\(I=\frac{\Delta V}{R_{e q}} \quad(R \text { inversal } y \text { proportional to } I)\)

\frac{1}{R_{e q}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\ldots+\frac{1}{R_{n}}

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A piston-cylinder device contains helium gas. During a reversible, isothermal process, the entropy of the helium will _________
DiKsa [7]

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During a reversible, adiabatic process, the entropy of the helium will NEVER increase.

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An airplane is cruising at a velocity of 800 km/h in air whose density is 0.526 . The airplane has a wing planform area of 90 .
Aliun [14]

Complete question:

An airplane is cruising at a velocity of 800 km/h in air whose density is 0.526 kg/m³ . The airplane has a wing planform area of 90 m² . The lift and drag coefficients on cruising conditions are estimated to be 2.0 and 0.06, respectively. The power that needs to be supplied to provide enough trust to overcome wing drag is

Answer:

The power that needs to be supplied to provide enough trust to overcome wing drag is 15,600 kW.

(C) 15,600 kW

Explanation:

Given;

velocity of the airplane, v =  800 km/h = 222.22 m/s

density of air, ρ = 0.526 kg/m³

wing planform area, A = 90 m²

lift coefficients, CL = 2.0

drag coefficients, CD = 0.06

F_D = C_D*A* \rho*\frac{v^2}{2} = 0.06*90* 0.526*\frac{222.22^2}{2} =70,131.93 \ N

Power supplied = FD* V

Power = 70131.93*222.22 = 15,584,717.63 \ W = 15,584.72 \ kW

This is approximately 15,600 kW.

Therefore, The power that needs to be supplied to provide enough trust to overcome wing drag is 15,600 kW.

The correct option is C

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What are the main causes of injuries when using forklifts?
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4 years ago
A shaft made of aluminum is 40.0 mm in diameter at room temperature (21°C). Its coefficient of thermal expansion = 24.8 x 10-6 m
Tomtit [17]

Answer:

Temperature to which the shaft must be cooled, \theta_2 = -180.61 ^0C

Explanation:

Diameter of the shaft at room temperature, d₁ = 40 mm

Room temperature, θ₁ = 21°C

Coefficient of thermal expansion, \alpha = 24.8 * 10^{-6} / ^0 C

The shaft is reduced in size by 0.20 mm:

Δd = - 0.20 mm

The temperature to which the shaft must be cooled, θ₂ = ?

The coefficient of thermal expansion is given by the equation:

\alpha = \frac{\triangle d}{d_1 * \triangle \theta}\\\\24.8 * 10^{-6} = \frac{-0.20}{40 * \triangle \theta}\\\\\triangle \theta = \frac{-0.20 }{24.8 * 10^{-6} * 40} \\\\\triangle \theta = - 201.61 ^0 C\\\triangle \theta = \theta_2 - \theta_1\\\\- 201.61 = \theta_2 - 21\\\\\theta_2 = -201.61 + 21\\\\\theta_2 = -180.61 ^0C

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