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SSSSS [86.1K]
4 years ago
10

A toaster is plugged into an 120-volt outlet. If it has a resistance of 460 ohms, how much power does it use?

Physics
2 answers:
Setler79 [48]4 years ago
6 0

Answer:31.3W

Explanation: JUST A FORMULA

P = V*V/R = 120*120/460 = 31.3W

jeyben [28]4 years ago
4 0

the power which is used by the toaster when plugged into 120 V outlet having resistance of 460 ohm, is 31.3 W.

<u>Explanation:</u>

The toaster is given to be plugged into a 120 V outlet. The power it uses can be calculated using the following method,

Given that Voltage V=120 V

Resistance R=460 ohm

As we know that,

\text { Power }=\text { Voltage } \times \text { Current }

\text {And Current}=\frac{\text {Voltage}}{\text {Resistance}}

\text { therefore, Power }=\text { Voltage } \times \text { Current }=\text { Voltage } \times \frac{\text { voltage }}{\text { Resistance }}

\text { Power }=\frac{\text { Voltage }^{2}}{\text { Resistance }}

Substituting the values, we get,  

\text { Power }=\frac{120 \times 120}{460}=\frac{14400}{460}=31.3 \mathrm{Watts}

Therefore, the power which is used by the toaster when plugged into 120 V outlet having resistance of 460 ohm, is 31.3 W.

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The other name for 'net force' is 'unbalanced force'. What is the name of the force that could be applied to an object that woul
jeka94

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Friction

Explanation:

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3 years ago
A 5.0 g coin is placed 15 cm from the center of a turntable. The coin has static and kinetic coefficients of friction with the t
Kaylis [27]

Answer:

Angular speed will reach 6.833rad/s before the coin starts slipping

Explanation:

There is no question but I'll asume the common one: Calculate the speed of the turntable before the coin starts slipping.

With a sum of forces:

Ff = m*a

Ff=m*V^2/R

At this point, friction force is maximum, so:

\mu*N=m*V^2/R

\mu*m*g=m*V^2/R

Solving for V:

V=\sqrt{\mu*g*R}

V=1.025 m/s

The angular speed of the turntable will be:

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7 0
3 years ago
MODERN PHYSICS
frosja888 [35]

Answer:

D. n=6 to n=2

Explanation:

Given;

energy of emitted photon, E = 3.02 electron volts

The energy levels of a Hydrogen atom is given as;  E = -E₀ /n²

where;

E₀ is the energy level of an electron in ground state =  -13.6 eV

n is the energy level

From the equation above make n, the subject of the formula;

n² = -E₀ / E

n² = 13.6 eV / 3.02 eV

n² = 4.5

n = √4.5

n = 2

When electron moves from higher energy level to a lower energy level it emits photons;

E = E_0(\frac{1}{n_1^2}-\frac{1}{n_2^2} )\\\\\frac{1}{n_1^2}-\frac{1}{n_2^2} = \frac{E}{E_o} \\\\\frac{1}{4} -\frac{1}{n_2^2} = \frac{3.02}{13.6} \\\\\frac{1}{4} -\frac{1}{n_2^2} =0.222\\\\\frac{1}{n_2^2} = 0.25 - 0.22\\\\\frac{1}{n_2^2} = 0.03\\\\n_2^2 = 33.33\\\\n_2 = \sqrt{33.33} = 6

For a photon to be emitted, electron must move from higher energy level to a lower energy level. The higher energy level is 6 while the lower energy level is 2

Therefore,  The electron energy-level transition is from n = 6 to n = 2

3 0
3 years ago
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