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dimulka [17.4K]

2 years ago

14

Las placas de un capacitor de placas paralelas están en forma horizontal. En el espacio se coloca una losa de material dieléctri

co con que llena la mitad inferior del espacio entre las placas. ¿Cuál es la nueva capacitancia resultante

1 answer:

snow_lady [41]2 years ago

5 0

I’m sorry I do not understand what this says

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The first artificial satellite to orbit the Earth was Sputnik I, launched October 4, 1957. The mass of Sputnik I was 83.5 kg, an

9966 [12]

Answer:

$-4.941*10^8J.$

Explanation:

To solve this exercise it is necessary to take into account the concepts related to gravitational potential energy, as well as the concept of perigee and apogee of a celestial body.

By conservation of energy we know that,

$\Delta U = \Delta_{perogee}-\Delta_{Apogee}$

Where,

$U= \frac{-GmM_e}{r}$

Replacing

$\Delta U = \frac{-GmM_e}{r_p}- \frac{-GmM_e}{r_a}$

$\Delta U = GmM_e (\frac{1}{r_A}-\frac{1}{r_p})$

Our values are given by,

$m = 85.5Kg$

$M_e = 5.97*10^{24}Kg$

$r_A = 7330Km$

$r_p = 6610Km$

$G = 6.67*10^{-11}Nm^2/Kg^2$

Replacing at the equation,

$\Delta U = (6.67*10^{-11})(85.5)(5.97*10^{24}) (\frac{1}{7330}-\frac{1}{6610})$

$\Delta U = -4.941*10^8J$

Therefore the Energy necessary for Sputnik I as it moved from apogee to perigee was $-4.941*10^8J.$

4 0

2 years ago

11. The primary job of the cell wall is to

hodyreva [135]

11. protect the cell and keep its shape.

12. chloroplast

3 0

2 years ago

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The 630-nm light from a helium-neon laser irradiates a grating. The light then falls on a screen where the first bright spot is

dimaraw [331]

Answer:

464.8 nm

Explanation:

The second wavelength of light can be calculated using the next equation:

$\lambda = \frac{x*d}{L}$

<u>Where:</u>

<em>λ : is the wavelength of light</em>

<em>x: is the distance from the central maximum</em>

<em>d: is the distance between the spots </em>

<em>L: is the lenght from the screen to the bright spot</em>

For the first wavelength of light we have:

$\lambda_{1} = \frac{x_{1}*d}{L}$

$630 \cdot 10^{-9} m = \frac{0.61 m*d}{L}$

$\frac{d}{L} = \frac{630 \cdot 10^{-9} m}{0.61 m} = 1.033 \cdot 10^{-6}$ (1)

For the second wavelength of light we have:

$\lambda_{2} = \frac{x_{2}*d}{L}$

$\lambda_{2} = 0.45 m*\frac{d}{L}$ (2)

By entering equation (1) into equation (2) we have:

$\lambda_{2} = 0.45 m* 1.033 \cdot 10^{-6} = 4.648 \cdot 10^{-7} m = 464.8 nm$

Therefore, the second wavelength is 464.8 nm

I hope it helps you!

3 0

2 years ago

you notice a solid air freshener seems to disappear over time. you think it must be changing into a gas. this is an example of

poizon [28]

This is an example of sublimation where a substance goes directly from solid to a gas, skipping the liquid stage.

3 0

2 years ago

A horizontal pipe 18.0 cm in diameter has a smooth reduction to a pipe 9.00 cm in diameter. If the pressure of the water in the

Rzqust [24]

Answer:

The rate of flow of water is 71.28 kg/s

Solution:

As per the question:

Diameter, d = 18.0 cm

Diameter, d' = 9.0 cm

Pressure in larger pipe, P = $9.40\times 10^{4}\ Pa$

Pressure in the smaller pipe, P' = $2.80\times 10^{4}\ Pa$

Now,

To calculate the rate of flow of water:

We know that:

Av = A'v'

where

A = Cross sectional area of larger pipe

A' = Cross sectional area of larger pipe

v = velocity of water in larger pipe

v' = velocity of water in larger pipe

Thus

$\pi \frac{d^{2}}{4}v = \pi \frac{d'^{2}}{4}v'$

$18^{2}v = 9^v'$

v' = 4v

Now,

By using Bernoulli's eqn:

$P + \frac{1}{2}\rho v^{2} + \rho gh = P' + \frac{1}{2}\rho v'^{2} + \rho gh'$

where

h = h'

$\rho = 10^{3}\ kg/m^{3}$

$9.40\times 10^{4} + \frac{1}{2}\rho v^{2} = 2.80\times 10^{4} + \frac{1}{2}\rho (4v)^{2}$

$6.6\times 10^{4} = \frac{1}{2}\rho 15v^{2}$

$v = \sqrt{\frac{2\times 6.6\times 10^{4}}{15\times 10^{3}}} = 8.8\ m/s$

Now, the rate of flow is given by:

$\frac{dm}{dt} = \frac{d}{dt}\rho Al = \rho Av$

$\frac{dm}{dt} = 10^{3}\times \pi (\frac{18}{2}\times 10^{- 2})^{2}\times 8.8 = 71.28\ kg/s$

3 0

2 years ago

Read 2 more answers