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nikdorinn [45]
3 years ago
14

A toaster filament dissipates 100W when plugged into a 130V outlet; assuming a constant filament resistance, identify the power

dissipated by the filament when the toaster is when plugged into a 120V outlet. A. 85W OB.92W C. 100W OD. 108W E. 120W
Physics
1 answer:
Mrac [35]3 years ago
4 0

Answer:

A. 85 W

Explanation:

case 1: toaster is connected to 130 V outlet

P = Power rating of toaster = 100 W

R = Resistance of the filament

V = Voltage of the source = 130 Volts

Power is given as

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

100 = \frac{130^{2}}{R}

R = 169 ohm

case 2: toaster is connected to 120 V outlet

P = Power dissipated

R = Resistance of the filament = 169 ohm

V = Voltage of the source = 120 Volts

Power dissipated  is given as

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

P = \frac{120^{2}}{169}

P = 85 W

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kykrilka [37]

The maximum speed of Tim is 16.95 m/s.

The given parameters:

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<h3>What is Newton's second law of motion?</h3>
  • Newton's second law of motion states that, the force applied to an object is directly proportional to the product of mass and acceleration of the object.

The net force on Tim is calculated by applying Newton's second law of motion as follows;

T - \mu _k F_n = ma\\\\T - \mu _k F_n = m\frac{v}{t} \\\\T - \mu_k mg = m\frac{v}{t} \\\\t(\frac{T - \mu_k mg}{m} )= v\\\\8 (\frac{220 \ -\  0.1 \times 71 \times 9.8}{71} ) =v \\\\ 16.95 \ m/s = v

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Learn more about net horizontal force here: brainly.com/question/21684583

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Answer:

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<em>We have to find the minimum co-efficient of kinetic friction between the surface and the block that will prevent the block from returning to its equilibrium with non-zero speed.  </em>

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From the FBD we can say that:

⇒ Normal force, N=Mg                                   <em>...equation(i)</em>

⇒ Elastic potential energy, PE = \frac{kx^2}{2}               <em>  ...equation (ii)</em>

⇒ Frictional force, f = \mu_kN                                <em> ...equation (iii)</em>

⇒ Plugging (i) in (iii).

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⇒ \frac{kx}{2Mg}=\mu_k                    <em>...dividing both sides wit Mg.</em>

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