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Delvig [45]
3 years ago
12

A clothes dryer uses about 9 amps of current from a 240 volt line. How much power does it use?

Physics
2 answers:
Basile [38]3 years ago
4 0
 240 x 9 = power
power= 2160 watts


Otrada [13]3 years ago
3 0

Answer:

Power used by dryer is 2160 watts

Explanation:

It is given that,

Current used by clothes dryer, I = 9 amps

Voltage used by dryer, V = 240 volts

We have to find the power used by it. Mathematically, the relationship between voltage, current and power is given by :

P=V\times I

Putting the values of I and V in above equation as :

P=240\ V\times 9\ A

P = 2160 watts

Hence, the power used by dryer is 2160 watts.

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A radio telescope has a circular collecting dish of diameter 5.0 m. It is used to observe two distant
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Skskdidododododoorrororo
3 0
2 years ago
A hot cube of iron was heated up using 1500 J of thermal energy and was placed in a beaker of water. Before it was heated, the i
Mariana [72]

Answer:

451.13 J/kg.°C

Explanation:

Applying,

Q = cm(t₂-t₁)............... Equation 1

Where Q = Heat, c = specific heat capacity of iron, m = mass of iron, t₂= Final temperature, t₁ = initial temperature.

Make c the subject of the equation

c = Q/m(t₂-t₁).............. Equation 2

From the question,

Given: Q = 1500 J, m = 133 g = 0.113 kg, t₁ = 20 °C, t₂ = 45 °C

Substitute these values into equation 2

c = 1500/[0.133(45-20)]

c = 1500/(0.133×25)

c = 1500/3.325

c = 451.13 J/kg.°C

4 0
2 years ago
Please help on this one?
Nonamiya [84]

30 or c because 60-30=30

3 0
3 years ago
A 72.9-kg base runner begins his slide into second base when moving at a speed of 4.02 m/s. The coefficient of friction between
elena-14-01-66 [18.8K]

Answer:

-589.05 J

Explanation:

Using work-kinetic energy theorem, the work done by friction = kinetic energy change of the base runner

So, W = ΔK

W = 1/2m(v₁² - v₀²) where m = mass of base runner = 72.9 kg, v₀ = initial speed of base runner = 4.02 m/s and v₁ = final speed of base runner = 0 m/s(since he stops as he reaches home base)

So, substituting the values of the variables into the equation, we have

W = 1/2m(v₁² - v₀²)

W = 1/2 × 72.9 kg((0 m/s)² - (4.02 m/s)²)

W = 1/2 × 72.9 kg(0 m²/s² - 16.1604 m²/s²)

W = 1/2 × 72.9 kg(-16.1604 m²/s²)

W = 1/2 × (-1178.09316 kgm²/s²)

W = -589.04658 kgm²/s²

W = -589.047 J

W ≅ -589.05 J

4 0
2 years ago
Storage of unlimited quantities of energy in batteries is possible. TrueFalse
dangina [55]

False

It is impossible to get infinite energy let alone to put it inside one battery

3 0
3 years ago
Read 2 more answers
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