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inn [45]
3 years ago
12

The heater element of a 120 V toaster is a 5.4 m length of nichrome wire, whose diameter is 0.48 mm. The resistivity of nichrome

at the operating temperature of the toaster is 1.3 × 10-6 Ω·m. The toaster is operated at a voltage of 120 V. The power drawn by the toaster is closest to:
(A) 370 W
(B) 360 W
(C) 380 W
(D) 400 W
(E) 410 W
Physics
1 answer:
Ksenya-84 [330]3 years ago
7 0

Answer:

The power drawn by the toaster is closest to:

(A) 370 W

Explanation:

First we calculate the resistance of the nichrome wire (R).

R=\frac{pL}{A} =\frac{pL}{\pi r^{2} } \\

Where radious (r), resistance coefficient (p), and Length (L)

r=\frac{0.48}{2} 10^{-3} m\\\\L=5.4 m\\p=1.3*10^{-3}\varOmega  \\\\\\Replacing:\\\\R=\frac{1.3(10)^{-6} *5.4}{\pi* 0.24^{2} (10)^{-6}} =38.7940

After replace the value in the ohm law power formula to obtain the power consumed:

P=\frac{V^2}{R} =\frac{120^2}{38.7940} =371.191 Watts

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Answer:

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A 900 kg vehicle moves around a curve with an incline of 20\circ∘ at a speed of 12.5 m/s. If the curve has a radius of 50 meters
valentina_108 [34]

Answer:

The normal force experienced by the car is approximately 8223.2 N

Explanation:

The question relates to banking of road where the centripetal force for the circular motion of the vehicle is provided by the horizontal component of the normal reaction

The mass of the vehicle that moves around the curve, m = 900 kg

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The speed with which the vehicle moves around the curve, v = 12.5 m/s

The radius of the curve, R = 50 meters

We have;

N \cdot sin(\theta) = \dfrac{m \cdot v^2}{R}

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R = The radius of the curve around which the vehicle moves = 50 m

\therefore N = \dfrac{m \cdot v^2}{R \cdot sin(\theta)} = \dfrac{900 \times (12.5)^2}{50 \times sin(20^{\circ})}  = 8223.1998754586828969046217875927

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