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3 years ago

What was the first prototype of the artificial tree like?

Engineering

1 answer:

olga_2 [115]3 years ago

7 0

Answer:

An artificial Christmas tree is an artificial pine or fir tree manufactured for the specific purpose of use as a Christmas tree. The earliest artificial Christmas trees were wooden, tree-shaped pyramids or feather trees, both developed by Germans. Most modern trees are made of polyvinyl chloride (PVC) but many other types of trees have been and are available, including aluminum Christmas trees and fiber-optic illuminated Christmas trees.

Explanation:

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2 years ago

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

Nata [24]

<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

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

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

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

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

3 0

3 years ago

2) The switch in the circuit below has been closed a long time. At t=0, it is opened.

saul85 [17]

Answer:

il(t) = e^(-100t)

Explanation:

The current from the source when the switch is closed is the current through an equivalent load of 15 + 50║50 = 15+25 = 40 ohms. That is, it is 80/40 = 2 amperes. That current is split evenly between the two parallel 50-ohm resistors, so the initial inductor current is 2/2 = 1 ampere.

The time constant is L/R = 0.20/20 = 0.01 seconds. Then the decaying current is described by ...

il(t) = e^(-t/.01)

il(t) = e^(-100t) . . . amperes

8 0

2 years ago

You hang a heavy ball with a mass of 42 kg from a silver rod 2.7 m long by 1.9 mm by 2.6 mm. You measure the stretch of the rod,

nadezda [96]

Answer:

Explanation:

cross sectional area A = 1.9 x 2.6 x 10⁻⁶ m²

= 4.94 x 10⁻⁶ m²

stress = 42 x 9.8 / 4.94 x 10⁻⁶

= 83.32 x 10⁶ N/m²

strain = .002902 / 2.7

= 1.075 x 10⁻³

Young's modulus = stress / strain

= 83.32 x 10⁶ / 1.075 x 10⁻³

= 77.5 x 10⁹ N/m²

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2 years ago

A thin-walled cylinder of average radius 50 mm, and wall thickness 1.0 mm is of length 500 mm, and is built into a wall at one e

kotykmax [81]

Answer:

Explanation:

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3 years ago