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Alecsey [184]
2 years ago
5

Peak wavelength of light coming from a star with a temp of 4300k

Physics
1 answer:
PSYCHO15rus [73]2 years ago
3 0

from Wein's law

\frac{k}{ \lambda_{max} }  = T \\ \lambda_{max} =  \frac{k}{T}  \\ \lambda_{max} =  \frac{3.898 \times  {10}^{ - 3} }{4300}   \\  = 9.065 \times  {10}^{ - 7}

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Drawing a ray diagram for an object far from convex lens requires how many rays?
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Answer

4

Explanation:

four rays can also explain the ray diagram if the object is at infinity

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2 years ago
Make a claim for the behavior of charged objects with other charged objects. Make a claim for the behavior of charged objects wi
emmainna [20.7K]

charged objects will either attract or repel other charged objects

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Your little sister is building a radio from scratch. Plans call for a 500 μH inductor wound on a cardboard tube. She brings you
gayaneshka [121]

Answer:

N = 195 turns

Explanation:

The inductance of the inductor, L = 500 μH = 500 * 10⁻⁶H

The length of the tube, l = 12 cm = 0.12 m

The diameter of the tube, d = 4 cm = 0.04 m

Radius, r = 0.04/2 = 0.02 m

Area of the tube, A = πr² = 0.02²π = 0.0004π m²

\mu_{0} = 4\pi * 10^{-7}

The inductance of a solenoid is given by:

L = \frac{\mu_{0}N^{2} A }{l}

500 * 10^{-6} = \frac{4\pi *10^{-7}  N^{2} *4\pi  *10^{-4}  }{0.12}\\500 * 10^{-6} = 0.00000001316N^{2} \\N^{2} = \frac{500 * 10^{-6}}{0.00000001316}\\N^{2} = 37995.44\\N = \sqrt{37995.44} \\N = 194.92 turns

8 0
2 years ago
Baseball Digest is but one of several publications on the sport of baseball. True or False?
Gnoma [55]

Answertrue

Explanation:

3 0
3 years ago
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Una barra de plata de 335.2 g con una temperatura de 100 ºC se introduce un calorímetro de aluminio de 60 g de masa que contiene
sdas [7]

Respuesta:

0,0560 cal / gºC.

Explicación:

Cantidad de calor; (Q)

Q = mcΔt; Δt = t2 - t1

m = masa, c = capacidad calorífica específica; Δt = cambio de temperatura

c de agua = 1 cal / gºC

c de aluminio = 0,22 cal / gºC

QTotal = Q de agua + Q de aluminio

Q de agua = 450 * 1 * (26 - 23) = 1350 cal

Q de aluminio = 60 * 0.22 * (26 - 23) = 39.6 cal

QTotal = 1350 + 39,6 = 1389,6 cal

Calor perdido = calor ganado

QTotal = calor perdido

- 1389,6 = 335,2 * c * (26 - 100)

-1389,6 = −24804,8 * c

c = 1389,6 / 24804,8

c = 0,056021 cal / gºC.

Capacidad calorífica específica de la plata = 0,0560 cal / gºC.

8 0
3 years ago
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