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3 years ago

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B) Calculate the equivalent capacitance of the network shown below between the points A nd 'B', Given: C1 = C2 = 12uF.C3 = 7uF,

CA = C5 = C6 =151F C6 =15uF

1 answer:

Akimi4 [234]3 years ago

8 0

As No diagram attached I am taking all are connected in series

We know

$\boxed{\sf C_{eq}=C_1+C_2\dots}$

$\\ \sf \longmapsto C_{eq}=12+12+7+15+15+15$

$\\ \sf \longmapsto C_{eq}=24+7+45$

$\\ \sf \longmapsto C_{eq}=76\mu F$

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The temperature and pressure at the surface of Mars during a Martian spring day were determined to be -41 oC and 900 Pa, respect

Fittoniya [83]

Answer:

Part a)

$\rho = 0.0205 kg/m^3$

Part b)

$\rho = 1.22 kg/m^3$

So density of atmosphere at Martian Surface is very less than the density at Earth.

Explanation:

Part a)

As per ideal gas equation we know that

$PM = \rho RT$

here we know that Martian atmosphere is equivalent to that of carbon

so we will have

$P = 900 Pa$

$T = 273 - 41 = 232 K$

now we will have

$(900)(0.044) = \rho (8.31)(232)$

$\rho = 0.0205 kg/m^3$

Part b)

Now for the earth surface the density of air is given for

$P = 101.6 kPa$

$T = 18 ^oC$

so we will have

$PM = \rho RT$

$(101.6\times 10^3)(0.029) = \rho(8.31)(273 + 18)$

$\rho = 1.22 kg/m^3$

So density of atmosphere at Martian Surface is very less than the density at Earth.

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3 years ago

A ball starts from rest and rolls down a hill at a constant acceleration of 5 m/s2. If it travels for 8 m how fast is it going i

makkiz [27]

Hi there!

We can use the following kinematic equation:

$v_f^2 = v_i^2 + 2ad$

vf = final velocity (? m/s)
vi = intial velocity (0 m/s)

a = acceleration (5 m/s²)

d = displacement (8 m)

Plug in the givens and solve.

$v_f^2 = 0 + 2(5)(8)\\\\v_f = \sqrt{80} = \boxed{8.944 \frac{m}{s}}$

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3 years ago

One similarity between work and power is that in order to calculate both you must know

svetoff [14.1K]

D.) In order to calculate both of them, we must know the "FORCE" on the system.

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3 years ago

Read 2 more answers

Explain the difference between balanced forces and action and reaction forces.

Monica [59]

Action-reaction forces<span> act on different objects; </span>balanced forces<span> act on the same object. </span>Balanced forces<span> can result in acceleration, </span>action-reaction forces<span> cannot. ... Newton's Third Law of Motion does not apply to </span>balanced forces<span>.</span>

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3 years ago

Para el siguiente conjunto de medidas, calcule EL ERROR RELATIVO PORCENTUAL: 1.34 m, 1.35 m, 1.37 m y 1.36 m

ivanzaharov [21]

Answer:

Ver explicacion abajo

Explanation:

En este caso para poder calcular el error relativo porcentual, es necesario calcular primero el error absoluto, que se calcula de la siguiente forma:

Error absoluto = Resultado exacto - aproximación

Sin embargo, no tenemos el resultado exacto de las medidas, pero podriamos conocerlo tomando el promedio de estas medidas y este es el que tomaremos como el verdadero resultado de las medidas:

Promedio de medidas = 1.34 + 1.35 + 1.37 + 1.36 / 4

Promedio de medidas = 1.355 m

Ya que tenemos el promedio, podemos calcular el error absoluto de cada medida y luego el error relativo porcentual:

Ea1 = 1.355 - 1.34 = 0.015

Ea2 = 1.355 - 1.35 = 0.005

Ea3 = 1.37 - 1.355 = 0.015

Ea4 = 1.36 - 1.355 = 0.005

Ya que tenemos los 4 errores absolutos, es posible calcular el porcentual:

%error relativo = (Error absoluto / resultado exacto) * 100

Aplicando la expresión con cada uno de los valores tenemos:

%Er1 = (0.015/1.34) * 100 = 1.12%

%Er2 = (0.005/1.35) * 100 = 0.37%

%Er3 = (0.015/1.37) * 100 = 1.09%

%Er4 = (0.005/1.36) * 100 = 0.37%

Espero que te sirva.

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3 years ago