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3 years ago

What are things made by nature that are used by people?

Physics

2 answers:

icang [17]3 years ago

6 0

Wood, stone, iron, copper, steel, wildlife, and Us.

aleksandrvk [35]3 years ago

3 0

Wood is made by nature, and grown as trees!
We as humans cut them down, build some pretty cool things with them, and plant new trees!

Hope this helps! :)

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What is the basic difference between ultraviolet, visible, and infrared radiation?

N76 [4]

To me the basic difference is their frequency and wavelength.

For frequency :
Infrared < visible light < ultraviolet

For wavelength :
Infrared > visible light > ultraviolet

3 0

4 years ago

What type of change occurs when water change from a solid to a iquid?

Oksanka [162]

The type of change is melting

6 0

3 years ago

3. Does the Sun's light that reaches Earth produce a true continuous rainbow? If not, what colors

dangina [55]

Answer:

when there is no water in the atmosphere or when no drops are formed there is no rainbow..

he hydrogen transition lines would be missing in the visible range, the RED region

Explanation:

The sunlight that reaches the Earth is scattered by the atmosphere and radiated, as the atmosphere has mainly nitrogen, the radiated light is blue, this light does not produce a rainbow.

The sunlight must reach a drop of water that disperses the colors according to the refractive index of the water and in this way produces a rainbow, so when there is no water in the atmosphere or when no drops are formed there is no rainbow..

If the atmosphere were composed of Hydrogen, the hydrogen transition lines would be missing in the visible range, this corresponds to the so-called Balmer series and they are in the RED light region.

8 0

4 years ago

LOOKING FOR ALL YOU SCIENTISTS!!

tensa zangetsu [6.8K]

Answer:

Make sure you're getting paid. Unpaid interships are a complete and total waste.

Explanation:

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3 years ago

Read 2 more answers

Determine la resistencia equivalente de la "escalera" de

defon

Las respuestas a cada inciso son:

a) La resistencia equivalente de la "escalera" de resistores iguales de 125 Ω que se muestra en la figura adjunta es:

$R_{t} = 341.7 \: \Omega$

b) La corriente a través de cada uno de los tres resistores de la izquierda si se conecta una batería de 50.0 V entre los puntos A y B es:

Correspondiente a R8 y R9 es 0.23 A
Correspondiente a R7 es 0.17 A.

a) En la imagen adjuntada correspondiente a la Figura 26-40, podemos observar que las resistencias 1, 2 y 3 están en serie, por lo tanto la ressitencia equivalente entre estas 3 es:

$R' = R_{1} + R_{2} + R_{3}$

De aquí en adelante tendremos presente que las todas las resistencias son iguales entre sí y por ende igual a 125 Ω. Las notaciones del 1 al 9 son para poder mostrar la resolución del problema.

Entonces:

$R' = 3R$

Ahora, esta resistencia está en paralelo con la resistencia R₄, por lo tanto la resistencia equivalente entre estas dos es:

$\frac{1}{R''} = \frac{1}{R'} + \frac{1}{R_{4}} = \frac{1}{3R} + \frac{1}{R} = \frac{4R}{3R^{2}}$

$R'' = \frac{3}{4}R$

Luego, esta resistencia está en serie con las resistencias R₅ y R₆, por lo tanto:

$R''' = R'' + R_{5} + R_{6} = \frac{3}{4}R + 2R = \frac{11}{4}R$

Esta resistencia está ahora en paralelo con R₇, entonces:

$\frac{1}{R''''} = \frac{1}{R'''} + \frac{1}{R_{7}} = \frac{4}{11R} + \frac{1}{R} = \frac{15R}{11R^{2}}$

$R'''' = \frac{11}{15}R$

Finalmente, esta resistencis está en serie con las resistencias R₈ y R₉, por lo tanto la resistencia total es:

$R_{t} = R'''' + R_{8} + R_{9} = \frac{11}{15}R + 2R = \frac{41}{15}R = \frac{41}{15}*125 \: \Omega = 341.7 \: \Omega$

b) Para este inciso debemos usar la Ley de Kirchhoff, pues tenemos tres mallas. Supondremos que las corriente de cada malla fluiran en sentido horario, por lo tanto las ecuaciones de para cada malla serán:

Malla 1

Segun la ley de Ohm tenemos:

$V-i_{1}R-i_{3}R=0$ (1)

Malla 2

$-i_{2}R-i_{5}R-i_{2}R+i_{3}R=0$ (2)

Malla 3

$-i_{4}R-i_{4}R-i_{4}R+i_{5}R=0$ (3)

Recordemos tambien que:

$i_{1}=i_{2}+i_{3}$ (4)

$i_{2}=i_{4}+i_{5}$ (5)

Lo que debemos hacer ahora es resolver el sistema de ecuaciones y encontrar los valore de las corrientes. Por lo tanto, los valores de las corrientes serán:

$i_{1}=3/13\: A$