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lozanna [386]
4 years ago
11

A certain electric furnace consumes 24 kw when it is connected to a 240-v line. what is the resistance of the furnace?

Physics
1 answer:
weeeeeb [17]4 years ago
3 0

One very handy electrical formula is

Power dissipated by a resistance = (Voltage)²/(resistance) .

24 kilowatts = (240 v)² / Resistance

Multiply each side by (Resistance):

(Resistance) x (24 kilowatts) = (240 v)²

Divide each side by (24 kilowatts):

Resistance = (240 v)² / (24,000 watts)

Resistance = (57,600 / 24,000) (volt² / volt · Amp)

Resistance = 2.4 (volt/Amp)

Resistance = 2.4 Ohms

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On the graph of voltage versus current, which line represents a 2.0 Ω resistor?​
Vikki [24]

Answer:

<h2>line B</h2>

Explanation:

According to ohm's law V = IR where;

V i sthe supply voltage (in volts)

I = supply current (in amperes)

R = resistance (in ohms)

In order to calculate the line that is equal to 2ohms, we need to calculate the slope of each line using the formula.

For line B, R = ΔV/ΔI

R = V₂-V₁/I₂-I₁

R = 14.0-4.0/7.0-2.0

R = 10.0/5.0

R = 2.0ohms

Since the slope of line B is equal to 2 ohms, this shows that the line B is the one that represents the 2ohms resistor.

3 0
4 years ago
A spring with a constant of 16 N/m has 98 J of energy stored in it when it is extended. How far is the spring extended?
Alchen [17]

The spring has been extended for 3.5 m

<u>Explanation:</u>

We have the formula,

PE =1/2 K X²

Rewrite the equation as

PE=1/2 K d²

multiply both the sides by 2/K to simplify the equation

2/k . PE= 1/2 K  d² . 2/K

√d²=√2PE/K

Cancelling the root value and now we have,

d=√2PE/k

d=√2×98 J / 16N/m

d=√12.25

d=3.5 m

The spring has been extended for 3.5 m

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6 0
4 years ago
What total energy (in J) is stored in the capacitors in the figure below (C1 = 0.900 µF, C2 = 16.0 µF) if 1.80 10-4 J is stored
Musya8 [376]

The total energy  stored in the capacitors is determined as  2.41 x 10⁻⁴ J.

<h3>What is the potential difference of the circuit?</h3>

The potential difference of the circuit is calculated as follows;

U = ¹/₂CV²

where;

  • C is capacitance of the capacitor
  • V is the potential difference

For a parallel circuit the voltage in the circuit is always the same.

The energy stored in 2.5 μf capacitor is known, hence the potential difference of the circuit is calculated as follows;

U = ¹/₂CV²

2U = CV²

V = √2U/C

V = √(2 x 1.8 x 10⁻⁴ / 2.5 x 10⁻⁶)

V = 12 V

The equivalent capacitance of C1 and C2 is calculated as follows;

1/C = 1/C₁ + 1/C₂

1/C = (1)/(0.9 x 10⁻⁶)  +  (1)/(16 x 10⁻⁶)

1/C = 1,173,611.11

C = 1/1,173,611.11

C = 8.52 x 10⁻⁷ C

The total capacitance of the circuit is calculated as follows;

Ct = 8.52 x 10⁻⁷ C   +   2.5 x 10⁻⁶ C

Ct = 3.35 x 10⁻⁶ C

The total energy of the circuit is calculated as follows;

U =  ¹/₂CtV²

U =  ¹/₂(3.35 x 10⁻⁶ )(12)²

U = 2.41 x 10⁻⁴ J

Learn more about energy stored in a capacitor here: brainly.com/question/14811408

#SPJ1

7 0
1 year ago
I fixed the pic can someone PLEASE HELP ME?
IrinaK [193]
The current through all the resistors will add up to 10A
4 0
3 years ago
Someone help with this also
Mumz [18]
I’m pretty sure it’s average speed= total distance and total time which is A.
3 0
3 years ago
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