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galben [10]
3 years ago
5

The heater element of a particular 120-V toaster is a 8.9-m length of nichrome wire, whose diameter is 0.86 mm. The resistivity

of nichrome at the operating temperature of the toaster is 1.3 × 10-6 Ω ∙ m. If the toaster is operated at a voltage of 120 V, how much power does it draw
Physics
1 answer:
AleksAgata [21]3 years ago
3 0

Answer:

Power, P = 722.96 watts

Explanation:

It is given that,

Voltage, V = 120 V

Length of nichrome wire, l = 8.9 m

Diameter of wire, d = 0.86 mm

Radius of wire, r = 0.43 mm = 0.00043 m

Resistivity of wire, \rho=1.3\times 10^{-6}\ \Omega-m

We need to find the power drawn by this heater. Power is given by :

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

And, R=\rho\dfrac{l}{A}

P=\dfrac{V^2\times A}{\rho\times l}

P=\dfrac{120^2\times \pi (0.00043)^2}{1.3\times 10^{-6}\times 8.9}

P = 722.96 watts

So, the power drawn by this heater element is 722.96 watts. Hence, this is the required solution.    

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Explanation:

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In order of the increasing frequency and the photon energy and the decreasing wavelength the spectrum are:  

radio waves, microwaves, infrared radiation, visible light, ultraviolet radiation, X-rays and gamma rays.

The energy of the radio waves photons is the lowest of all the other waves in the electromagnetic spectrum.

Also, E={h\times \nu}

Where,  

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Thus, energy is directly proportional to the frequency. The radio waves have the lowest frequency.

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Difference between effort and load​
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2 years ago
Hello, I want to ask. . anyone knows the answer.
stealth61 [152]
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5 0
2 years ago
A disk of radius 25 cm spinning at a rate of 30 rpm slows to a stop over 3 seconds. What is the angular acceleration? B. How man
Ne4ueva [31]

Answer:  

A. α = - 1.047 rad/s²  

B. θ = 14.1 rad  

C. θ = 2.24 rev  

Explanation:  

A.  

We can use the first equation of motion to find the acceleration:

\omega_f = \omega_i + \alpha t  

where,  

ωf = final angular speed = 0 rad/s  

ωi = initial angular speed = (30 rpm)(2π rad/1 rev)(1 min/60 s) = 3.14 rad/s  

t = time = 3 s  

α = angular acceleration = ?  

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<u>α = - 1.047 rad/s²</u>

B.  

We can use the second equation of motion to find the angular distance:

\theta = \omega_it +\frac{1}{2}\alpha t^2\\\theta = (3.14\ rad/s)(3\ s) + \frac{1}{2}(1.04\ rad/s^2)(3)^2  

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C.  

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<u>θ = 2.24 rev</u>

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