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Mariana [72]
3 years ago
14

An electric device uses 650 watts of power. If the voltage is 120 volts, what is the resistance?

Physics
2 answers:
grigory [225]3 years ago
8 0

Answer: The correct answer is 22.15 ohm.

Explanation:

The expression for the power in terms of resistance and potential is as follows;

P=\frac{V^{2} }{R}

Here, V is the potential, P is the power and R is the resistance.

It is given in the problem that an electric device uses 650 watts of power. The voltage of this device is 120 V.

Calculate the resistance of the electric device by rearranging the above expression.

R=\frac{V^{2} }{P}

Put V= 120 V and P= 650 W.

R=\frac{(120)^{2} }{650}

R=22.15 ohm

Therefore, the value of the resistance is 22.15 ohm.

PSYCHO15rus [73]3 years ago
5 0
There are several handy-dandy formulas for electrical power.
Here they are:

--  Power  =  (voltage) x (current)
--  Power  =  (current²) x (resistance)
--  Power  =  (voltage²) x (resistance)

We just have to pick the one that's most convenient.

We know the power and the voltage, and we have to find the resistance.
So we would be smart to use a formula that has power, voltage, and
resistance in it.  The last formula on my list is exactly what we want.

--  Power  =  (voltage²) / (resistance)

     650 w  =  (120 v)² / (resistance)

     (650 w) x (resistance)  =  (120 v)²

           resistance  =  (120 v)² / (650 w)  =  22.2 ohms  .
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Complete Question:

In the "before" part of Fig. 9-60, car A (mass 1100 kg) is stopped at a traffic light when it is rear-ended by car B (mass 1400 kg). Both cars then slide with locked wheels until the frictional force from the slick road (with a low ?k of 0.15) stops them, at distances dA = 6.1 m and dB = 4.4 m. What are the speeds of (a) car A and (b) car B at the start of the sliding, just after the collision? (c) Assuming that linear momentum is conserved during the collision, find the speed of car B just before the collision.

Answer:

a) Speed of car A at the start of sliding = 4.23 m/s

b) speed of car B at the start of sliding = 3.957 m/s

c) Speed of car B before the collision = 7.28 m/s

Explanation:

NB: The figure is not provided but all the parameters needed to solve the question have been given.

Let the frictional force acting on car A, f_{ra} = \mu mg\\............(1)

Since frictional force is a type of force, we are safe to say f_{ra} = ma.......(2)

Equating (1) and (2)

ma = \mu mg\\a = \mu g\\\mu = 0.15\\a = 0.15 * 9.8 = 1.47 m/s^{2}

a) Speed of A at the start of the sliding

d_{A} = 6.1 m\\Speed of A at the start of sliding, v_{A} = \sqrt{2ad_{A} }\\ v_{A} = \sqrt{2*1.47*6.1 } \\v_{A} = \sqrt{17.934 } \\v_{A} = 4.23 m/s

b) Speed of B at the start of the sliding

d_{A} = 4.4 m\\Speed of A at the start of sliding, v_{B} = \sqrt{2ad_{B} }\\ v_{B} = \sqrt{2*1.47*4.4 } \\v_{B} = \sqrt{12.936 } \\v_{B} = 3.957 m/s

Let the speed of car B before collision = v_{B1}

Momentum of car B before collision = m_{B} v_{B1}

Momentum after collision = m_{A} v_{A} + m_{B} v_{B2}

Applying the law of conservation of momentum:

m_{B} v_{B1}  = m_{A} v_{A} +m_{B} v_{B2}

m_{A} = 1100 kg\\m_{B} = 1400 kg

(1400*v_{B1} ) = (1100 * 4.23) + ( 1400 * 3.957)\\(1400*v_{B1} ) = 10192.8\\v_{B1} = 10192.8/1400\\v_{B1 = 7.28 m/s

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D = distance between the cars at the start of time = 680 km

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A fisherman notices that his boat is moving up and down periodically without any horizontal motion, owing to waves on the surfac
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