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3 years ago

An effective early childhood curriculum can be

1 answer:

6 0

An effective early childhood curriculum can be learning how to read and write. Without learning how to read, we wouldn’t be able to communicate through this app, or through online school. We wouldn’t know directions, and we wouldn’t be able to enjoy stories. If we didn’t know how to write, I would not be able to help you figure out the answer to this problem, and your teacher wouldn’t be able to grade anything. Think about it, you ir entire education is based of being able to read and write.

Hope this helps :)

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Which of the following could be used to convert chemical energy to heat energy?

Mandarinka [93]

Answer:

Explanation:

Inside a battery there are chemical reactions. When the battery is powering a flashlight, this chemical energy gives a heat one.

3 0

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:

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

$i_{2}=4/65\: A$

$i_{3}=11/65\: A$

$i_{4}=1/65\: A$

$i_{5}=3/65\: A$

Finalmente:

La corriente correspondiente a R8 y R9 es 0.23 A
La corriente correspondiente a R7 es 0.17 A.

Pudes aprender mas de mallas aquí:

https://brainly.lat/tarea/11593276

4 0

3 years ago

An oil layer that is 5.0 cm thick is spread smoothly and evenly over the surface of water on a windless day. What is the angle o

Fofino [41]

Answer:

32.1

Explanation:

NOTE: You did not state the angle of incidence, and thus, I will be using 45° as my angle of incidence, all you need to do is replace it with your own value if it's different.

To solve this question, we are going to be using Snell's Law.

Snell's law describes the relationship between the angles of incidence and refraction, when referring to light or other waves passing through a boundary between two different isotropic media, such as water, glass, or air.

Snell's law is mathematically given as

sin(A1)/sin(A2) = n2/n1, where

n1 = incidence index

n2 = refracted index

A1 = incidence angle

A2 = refracted angle

The refraction index of oil is 1.15, and that of water is 1.33, so

if we take oil first,

sin A2 = (n1.sinA1)/n2

sin A2 = (1 * sin 45)/1.15

sin A2 = 0.7071/1.15

sin A2 = 0.6149

A2 = sin^-1 0.6149

A2 = 37.9°

Then

sin A3 = (1.15 * sin 37.9) / 1.33

sin A3 = (0.6149 * 1.15) / 1.33

sin A3 = 0.7071 / 1.33

sin A3 = 0.5317

A3 = sin^-1 0.5317

A3 = 32.1

4 0

3 years ago

As an aid in working this problem, consult Interactive Solution 3.41. A soccer player kicks the ball toward a goal that is 20.0

alekssr [168]

Answer:

V=14.9 m/s

Explanation:

In order to solve this problem, we are going to use the formulas of parabolic motion.

The velocity X-component of the ball is given by:

$Vx=V*cos(\alpha)\\Vx=15.7*cos(31^o)=13.5m/s$

The motion on the X axis is a constant velocity motion so:

$t=\frac{d}{Vx}\\t=\frac{20.0}{13.5}=1.48s$

The whole trajectory of the ball takes 1.48 seconds

We know that:

$Vy=Voy+(a)*t\\Vy=15.7*sin(31^o)+(-9.8)*(1.48)=-6.42m/s$

Knowing the X and Y components of the velocity, we can calculate its magnitude by:

$V=\sqrt{Vx^2+Vy^2} \\V=\sqrt{(13.5)^2+(-6.42)^2}=14.9m/s$

6 0

4 years ago

You drop a rock off a bridge. The rock's height, h (in feet above the water), after t seconds is modeled by h = – 16 t2 + 541. W

marin [14]

<span>h ( t) = h(1 sec) = -16t^2 + 541

so h (2 sec) = -16*(2)^2 + 541 = -64 + 541 = <span>477 ft

Therefore, </span></span>the height of the rock after 2 seconds is 477 feet.

I hope my answer has come to your help. Thank you for posting your question here in Brainly.

5 0

4 years ago

An effective early childhood curriculum can be​

An effective early childhood curriculum can be