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2 years ago

An electric iron is connected to the mains power supply of 220 V. When the electric iron is

Physics

1 answer:

Ilya [14]2 years ago

7 0

Answer:

Given: V = 220V, Pmin = 360W, Pmax = 840W

For minimum heating case:

We know that

Pmin = VI

360 = 220 X I

I = 1.63 amp

R = V/I

R = 220/1.63

R = 134.96ohms

For maximum heating case:

We know that

Pmax = VI

840 = 220 X I

I = 3.81 amp

R = V/I

R = 220/3.81

R = 57.74 ohms

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2 years ago

A uniform thin wire is bent into a quarter-circle of radius a = 20.0 cm, and placed in the first quadrant. Determine the coordin

Mashcka [7]

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The general equation to calculate the center of mass is:

$r_{cm}=1/M*\int\limits {r} \, dm$

Any differential of mass can be calculated as:

$dm = \lambda*a*d\theta$ Where "a" is the radius of the circle and λ is the linear density of the wire.

The linear density is given by:

$\lambda=M/L=M/(a*\pi/2)=\frac{2M}{a\pi}$

So, the differential of mass is:

$dm = \frac{2M}{a\pi}*a*d\theta$

$dm = \frac{2M}{\pi}*d\theta$

Now we proceed to calculate X and Y coordinates of the center of mass separately:

$X_{cm}=1/M*\int\limits^{\pi/2}_0 {a*cos\theta*2M/\pi} \, d\theta$

$Y_{cm}=1/M*\int\limits^{\pi/2}_0 {a*sin\theta*2M/\pi} \, d\theta$

Solving both integrals, we get:

$X_{cm}=2*a/\pi=12.73cm$

$Y_{cm}=2*a/\pi=12.73cm$

Therefore, the position of the center of mass is:

$r_{cm}=[12.73,12.73]cm$

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