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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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An electric oven has a resistance of 201 ohms and a voltage of 220 V. How much current does it draw?
asambeis [7]

1.1 A. An electric oven with a resistance of 201Ω and a voltage of 220V drwa a current of 1.1 A.

The easiest way to solve this problem is using the Ohm's Law I = V/R.

An electric oven has R = 201Ω, and a drop of voltage V = 220v, solve using I = V/R:

I = 220V / 201Ω

I = 1.09 A ≅ 1.1 A

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In this lab, you observed how different factors such as velocity, gradient, and ____ , or amount of water in a stream, affect th
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volume and erosion

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Read 2 more answers
A space probe produces a radio signal pulse. If the pulse reaches Earth 12.3 seconds after it is emitted by the probe, what is t
adoni [48]

Answer:

Distance = 3.69 × 10^9 m

The distance from the probe to Earth is 3.69 × 10^9 m

Explanation:

Distance from the probe to the Earth can be derived using the simple motion formula;

Distance = speed × time .....1

Since a radio signal uses an electromagnetic wave to transfer signal, it has the same speed as the speed of light.

Speed of radio signal = speed of light = 3.0 × 10^8 m/s

time taken to reach the earth = 12.3 seconds

Substituting the values of speed and time into equation 1;

Distance = 3.0 × 10^8 m/s × 12.3 s

Distance = 36.9 × 10^8 m

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Note: all electromagnetic radiation have the same speed which is equal to 3.0 × 10^8 m/s

5 0
2 years ago
Consider steady-state conditions for one-dimensional conduction in a plane wall having a thermal conductivity k = 50 W/m · K and
tatuchka [14]

Answer:

solution:

dT/dx =T2-T1/L

&

q_x = -k*(dT/dx)

<u>Case (1)  </u>

dT/dx= (-20-50)/0.35==> -280 K/m

 q_x  =-50*(-280)*10^3==>14 kW

Case (2)

dT/dx= (-10+30)/0.35==> 80 K/m

 q_x  =-50*(80)*10^3==>-4 kW

Case (2)

dT/dx= (-10+30)/0.35==> 80 K/m

 q_x  =-50*(80)*10^3==>-4 kW

Case (3)

q_x  =-50*(160)*10^3==>-8 kW

T2=T1+dT/dx*L=70+160*0.25==> 110° C

Case (4)

q_x  =-50*(-80)*10^3==>4 kW

T1=T2-dT/dx*L=40+80*0.25==> 60° C

Case (5)

q_x  =-50*(200)*10^3==>-10 kW

T1=T2-dT/dx*L=30-200*0.25==> -20° C

note:

all graph are attached

6 0
3 years ago
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