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san4es73 [151]
3 years ago
10

if you changed the direction of electron flow by switching the connections to the battery what would happen

Physics
1 answer:
Rudiy273 years ago
7 0
If you changed the flow you would technically change the elections to a positive charge
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You want to produce three 2.00-mm-diametercylindrical wires, each with a resistance of 1.00 Ω at room temperature. One wire is g
Vlada [557]

Answer:

(a) L =  128.75 m

(b) L =  182.56 m

(c) L =  114.28 m

(d) Mass of Gold = 7.68 kg = 7680 gram

(e) Cost of Gold Wire = $ 307040

Explanation:

The resistance of the wire is given as:

R = ρL/A

where,

R = Resistance

ρ = resistivity

L = Length

A = cross-sectional area

(a)

For Gold Wire:

ρ = 2.44 x 10⁻⁸ Ω.m

A = πd²/4 = π(2 x 10⁻³ m)²/4 = 3.14 x 10⁻⁶ m²

R = 1 Ω

Therefore,

1 Ω = (2.44 x 10⁻⁸ Ω.m)L/(3.14 x 10⁻⁶ m²)

L = (1 Ω)(3.14 x 10⁻⁶ m²)/(2.44 x 10⁻⁸ Ω.m)

<u>L =  128.75 m</u>

<u></u>

(b)

For Copper Wire:

ρ = 1.72 x 10⁻⁸ Ω.m

A = πd²/4 = π(2 x 10⁻³ m)²/4 = 3.14 x 10⁻⁶ m²

R = 1 Ω

Therefore,

1 Ω = (1.72 x 10⁻⁸ Ω.m)L/(3.14 x 10⁻⁶ m²)

L = (1 Ω)(3.14 x 10⁻⁶ m²)/(1.72 x 10⁻⁸ Ω.m)

<u>L =  182.56 m</u>

<u></u>

(c)

For Aluminum Wire:

ρ = 2.75 x 10⁻⁸ Ω.m

A = πd²/4 = π(2 x 10⁻³ m)²/4 = 3.14 x 10⁻⁶ m²

R = 1 Ω

Therefore,

1 Ω = (2.75 x 10⁻⁸ Ω.m)L/(3.14 x 10⁻⁶ m²)

L = (1 Ω)(3.14 x 10⁻⁶ m²)/(2.75 x 10⁻⁸ Ω.m)

<u>L =  114.28 m</u>

<u></u>

(d)

Density = Mass/Volume

Mass = (Density)(Volume)

Volume of Gold = AL = (3.14 x 10⁻⁶ m²)(128.75 m) = 4.04 x 10⁻⁴ m³

Therefore,

Mass of Gold = (1.9 x 10⁴ kg/m³)(4.04 x 10⁻⁴ m³)

<u>Mass of Gold = 7.68 kg = 7680 gram</u>

<u></u>

(e)

Cost of Gold Wire = (Unit Price of Gold)(Mass of Gold)

Cost of Gold Wire = ($ 40/gram)(7680 grams)

<u>Cost of Gold Wire = $ 307040</u>

7 0
3 years ago
A room has a wall with an R value of 18 F sq.ft. hr/BTU. The room is 15 feet long and 11 feet wide with walls that are 9 ft high
seropon [69]

Answer:

The heat transferred through the wall that day is  13728 BTUs

Explanation:

Here, we have the area of the wall given as

Area of wall = 2 × Length × Height + 2 × Width × Height

Length = 15 feet

Width = 11 Feet and

Height = 9 feet

Therefore, the area = 2×15×9 + 2×11×9 = 468 ft²

Temperature difference is given by

Average outside temperature - Wall temperature = 40 - 18 = 22 °F

Therefore the heat transferred through the wall that day (24 hours) at 18 sq.ft. hr/BTU is given by;

468 × 22 × 24/18 = 13728 = 13728 BTUs.

5 0
3 years ago
Which best describes the result when the concentration of a solution is increased? The amount of solute is increased, and the co
oee [108]

The amount of solute is increased, and the conductivity of the solution is increased

Explanation:

The ability of a solution to conduct electricity is conductivity.The more concentrated a solution is the higher the conductivity.This is because the concentration of ions will increase.Increasing solute increases the concentration of a solution thus rising the conductivity.

Learn More

Conductivity of a solution :brainly.com/question/9486572

Keywords : concentration, solution, solute, conductivity

#LearnwithBrainly

8 0
3 years ago
Measurements made on an operating centrifugal pump indicate that for a flow rate of 240 gpm, 6 HP are being consumed. The operat
Ludmilka [50]

Answer:

62.1 ft

Explanation:

We know that 1 gpm= 0.0022 ft^{3}/s hence 240 gpm will be

240*0.0022=0.528 ft^{3}/s

From the formula of obtaining efficiency

\eta=\frac {\gamma Q h_a}{500W_{in}} and making h_a the subject we have

h_a=\frac {\eta*550*W_{in}}{\gamma Q} where Q is flow rate, \gamma is specific weight of water, h_a is actual head rise and W_{in} is the power input to the compressor

Substituting 0.62 for \eta, 6hp for W_{in}, 62.4 lb/ft^{3} for \gamma and 0.528 ft^{3}/s for Q then

h_a=\frac {0.62*550*6}{62.4*0.528}=\frac {2046}{32.9472}=62.09936\approx 62.1 ft

Therefore, actual head rise of the water being pumped is 62.1 ft

6 0
3 years ago
A student observes a chemical reaction.
ankoles [38]

Answer:4Al + 2O2 --> 3Al2O3

Explanation:THIS IS YOUR ANSWER

4 0
3 years ago
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