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3 years ago

15

What is four possible energy sources for a circuit

2 answers:

VashaNatasha [74]3 years ago

8 0

<span>electric, solar, wind, and geothermal.</span>

Dima020 [189]3 years ago

4 0

the four sources needed for a circuit are 1 wind

2voltage

3constant currents

4energy

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This sphere contains all of the frozen water on Earth

erica [24]

Cryosphere is the sphere that contains all of the frozen water on Earth. The correct option among all the options that are given in the question is the fourth option or the last option. Cryosphere is actually those parts of the surface of the earth where the water is in a frozen state, like in seas and oceans.

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2 years ago

A coating is being applied to reduce the reflectivity of a pane of glass to light with a wavelength of 522 nm incident near the

fredd [130]

Answer:

t = 94.91 nm

Explanation:

given,

wavelength of the light = 522 nm

refractive index of the material = 1.375

we know the equation

c = ν λ

where ν is the frequency of the wave

c is the speed of light

$\nu= \dfrac{c}{\nu\lambda}$

$\nu = \dfrac{3\times 10^8}{522 \times 10^{-9}}$

ν = 5.75 x 10¹⁴ Hz

the thickness of the coating will be calculated using

$t = \dfrac{\lambda}{4\mu_{material}}$

$t = \dfrac{522 \times 10^{-9}}{4\times 1.375}$

t = 94.91 nm

the thickness of the coating will be equal to t = 94.91 nm

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2 years ago

A tank has the shape of an inverted circular cone with height 16m and base radius 3m. The tank is filled with water to a height

rewona [7]

Answer:

$W=17085KJ$

Explanation:

From the question we are told that:

Height $H=16m$

Radius $R=3$

Height of water $H_w=9m$

Gravity $g=9.8m/s$

Density of water $\rho=1000kg/m^3$

Generally the equation for Volume of water is mathematically given by

$dv=\pi*r^2dy$

$dv=\frac{\piR^2}{H^2}(H-y)^2dy$

Where

y is a random height taken to define dv

Generally the equation for Work done to pump water is mathematically given by

$dw=(pdv)g (H-y)$

Substituting dv

$dw=(p(=\frac{\piR^2}{H^2}(H-y)^2dy))g (H-y)$

$dw=\frac{\rho*g*R^2}{H^2}(H-y)^3dy$

Therefore

$W=\int dw$

$W=\int(\frac{\rho*g*R^2}{H^2}(H-y)^3)dy$

$W=\rho*g*R^2}{H^2}\int((H-y)^3)dy)$

$W=\frac{1000*9.8*3.142*3^2}{9^2}[((9-y)^3)}^9_0$

$W=3420.84*0.25[2401-65536]$

$W=17084965.5J$

$W=17085KJ$

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2 years ago

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