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An electric heater carries a current of 13.5 A when operating at a voltage of 120V . What is the resistance of the heater?

Aloiza [94]

An electric heater that carries a current of 13.5 A when operating at a voltage of 120V , will have the resistance of the heater as 8.89 ohm.

Ohm's law can be described as the law in physics that focus on how voltage or potential difference between two points is directly proportional to the current or electricity that is moving in the direction of the resistance.

An this is also directly proportional to the resistance of the circuit, hence it can be calculated using the Ohm's law is V=IR.

from V=IR

R=V/I

= 120/13.55

=8.89 ohm.

Therefore, electric heater that carries a current of 13.5 A when operating at a voltage of 120V , will have the resistance of the heater as 8.89 ohm.

Learn more about ohms law on:

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4 0

1 year ago

A charge of 7.00 mC is placed at opposite corners corner of a square 0.100 m on a side and a charge of -7.00 mC is placed at oth

andrew-mc [135]

Answer:

4.03\times10^{7}N[/tex], 135°

Explanation:

charge, q = 7 mC = 0.007 C

charge, - q = - 7 mC = - 0.007 C

d = 0.1 m

Let the force on charge placed at C due to charge placed at D is FD.

$F_{D}=\frac{kq^{2}}{DC^{2}}$

$F_{D}=\frac{9 \times10^{9}\times 0.007 \times 0.007}{0.1^{2}}=4.41 \times 10^{7}N$

The direction of FD is along C to D.

Let the force on charge placed at C due to charge placed at B is FB.

$F_{B}=\frac{kq^{2}}{BC^{2}}$

$F_{B}=\frac{9 \times10^{9}\times 0.007 \times 0.007}{0.1^{2}}=4.41 \times 10^{7}N$

The direction of FB is along C to B.

Let the force on charge placed at C due to charge placed at A is FA.

$F_{A}=\frac{kq^{2}}{AC^{2}}$

$F_{D}=\frac{9 \times10^{9}\times 0.007 \times 0.007}{0.1 \times\sqrt{2} \times 0.1 \times\sqrt{2}}=2.205 \times 10^{7}N$

The direction of FA is along A to C.

The net force along +X axis

$F_{x}=F_{A}Cos45-F_{D}$

$F_{x}=2.205\times10^{7}Cos45-4.41\times10^{7}=-2.85\times10^{7}N$

The net force along +Y axis

$F_{y}=F_{B}-F_{A}Sin45$

$F_{y}=4.41\times10^{7}-2.205\times10^{7}Sin45=2.85\times10^{7}N$

The resultant force is given by

$F=\sqrt{F_{x}^{2}+F_{y}^{2}}=\sqrt{(-2.85\times10^{7})^{2}+(2.85\times10^{7})^{2}}$

$F = 4.03\times10^{7}N$

The angle from x axis is Ф

tan Ф = - 1

Ф = -45°

Angle from + X axis is 180° - 45° = 135°

5 0

3 years ago

Which changes to observed time and length occur when objects approach the speed of light?

mixas84 [53]

Answer:

Time moves slower and length decreases.

Explanation:

3 0

4 years ago

How do storms on Jupiter differ from storm systems on Earth?

Vladimir [108]

Answer:

As, Jupiter are one of the largest planet in the solar system and the large amount of the mass of the jupiter are consisted with the gases. The storm tracks in the symmetrical path at the proper latitude in the system. But the storm tracks on the earth in the system where planet are highly variable.

8 0

3 years ago

Online masses every 8 a.m. and 2:00 p.m. will continue

nikdorinn [45]

online is hard toooooooooo

6 0

3 years ago