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bija089 [108]
3 years ago
9

Electric resistance ____________ with a(n) ____________in the length of a wire and as a result current flow ___________.

Physics
1 answer:
maria [59]3 years ago
3 0
The answer is D. Electric resistance increases with an increase in the length of a wire and as a result current flow decreases. There is a direct relationship between the length of the wire and the resistance. The longer the wire, the more resistance there will be. Additionally, from Ohm's Law, current is inversely proportional to resistance. This means as the current increases, resistance decreases or vice versa.
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explain what would be the same and what would be different about the phases and view of Earth to an inhabitant of the near side
guapka [62]

Answer:

Both the Earth and the moon are always half illuminated by the sun. But from either world, at any given time, you can see varying portions of that lighted half – or various phases of the Earth or moon. ... Instead, from a given point on the moon's near side, you'd always see Earth hanging in your sky.

Explanation:

8 0
3 years ago
What quantity is represented by a unit called newton (N) ?
bixtya [17]

Newton is the SI unit of force . Newton is the name of a British scientist and the name of unit is to honour him. The unit is actually Kg.m/s 2 The unit can be derived by the formula. Take the example of weight. It's formula is W = mg . We know that the unit of mass is kg and gravity is m/s 2 so the unit of weight becomes kg.m/s 2 This unit is known as a Newton. It is always given a capital letter because it is someone's name. Other units that are always capitalised (upper case) are Ampere (Amp), Watt, Volt, Coulomb, Kelvin, Celsius, Fahrenheit, Curie, Roentgen because they are also people's names.

5 0
3 years ago
The work function for silver is 4.73 eV. (a) Convert the value of the work function from electron volts to joules.
pogonyaev

Answer:

W=7.56\times 10^{-19}\ J

Explanation:

Given that,

The work function for silver is 4.73 eV.

We need to find the value of the work function from electron volts to joules.

We know that,

1\ eV=1.6\times 10^{-19}\ J

For 4.73 eV,

4.73\ eV=1.6\times 10^{-19}\times 4.73\\\\=7.56\times 10^{-19}\ J

So, the work function for silver is 7.56\times 10^{-19}\ J.

6 0
3 years ago
Which bmi score listed below is considered over weight?
valentinak56 [21]
I don't see any answer choices, but if I remember correctly, the overweight BMI score is 20-22 and up. Probably 22 and up.
5 0
4 years ago
Ch 31 HW Problem 31.63 10 of 15 Constants In an L-R-C series circuit, the source has a voltage amplitude of 116 V , R = 77.0 Ω ,
Degger [83]

Answer:

a. I = 0.76 A

b. Z = 150.74

c. RL₁ = 34.41  ,  RL₂ = 602.58

d. RL₂ = 602.58

Explanation:

V₁ = 116 V , R₁ = 77.0 Ω , Vc = 364 V ,  Rc = 473 Ω

a.

Using law of Ohm

V = I * R

I = Vc / Rc =  364 V / 473 Ω

I = 0.76 A

b.

The impedance of the circuit in this case the resistance, capacitance and inductor

V = I * Z

Z = V / I

Z = 116 v / 0.76 A

Z = 150.74

c.

The reactance of the inductor can be find using

Z² = R² + (RL² - Rc²)

Solve to RL'

RL = Rc (+ / -) √ ( Z² - R²)

RL = 473 (+ / -)  √ 150.74² 77.0²

RL = 473 (+ / -)  (129.58)

RL₁ = 34.41  ,  RL₂ = 602.58

d.

The higher value have the less angular frequency  

RL₂ = 602.58

ω = 1 / √L*C

ω = 1 / √ 602.58 * 473

f = 285.02 Hz

6 0
4 years ago
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