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3 years ago

8

It is possible to convert radiant energy into electrical energy using photovoltaic cells. Assuming equal efficiency of conversio

n, would infrared or ultraviolet radiation yield more electrical energy on a per-photon basis?

1 answer:

Finger [1]3 years ago

4 0

Answer:

Ultraviolet radiation would yield more electrical energy

Explanation:

The reason for the ultraviolet to generate more energy is that there would be getting more electrons per unit of time the photovoltaic cell, due to the higher frequency of the ultraviolet in comparison with the infrared radiation.

The infrared spectrum goes from 300 GHz (10^9 Hertz) to 400 THz or (10^12 Hertz).

The ultraviolet spectrum goes from 800 THz to 30.000 THz or (10^12 Hertz). This kind of radiation is responsible for skin burn from the sun and it´s the “ most usable” part from the sunlight in a photovoltaic cell.

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3.5 g of a hydrocarbon fuel is burned in a vessel that contains 250. grams of water initially at 25.00 C. After the combustion,

kherson [118]

Answer:

3.) 463 J/g

Explanation:

Heat given by the fuel is used to increase the temperature of the water

so it is given as

$Q = ms\Delta T$

here we will have

m = 250 g

$\Delta T = 26.55 - 25$

now we also know that

$s = 4186$

so we have

$Q = (0.250)(4186)(26.55 - 25)$

$Q = 1622.075 J$

so 3.5 g fuel gives above energy

so per gram of fuel will have total energy given as

$Q = \frac{1622.075}{3.5}$

$Q = 463.45 J/g$

6 0

3 years ago

Describe how Rutherford's experiments changed the accepted scientific model of the atom.

lorasvet [3.4K]

Rutherford's model of the atom (ESAAQ) Rutherford carried out some experiments which led to a change in ideas around the atom. His new model described the atom as a tiny, dense, positively charged core called a nucleus surrounded by lighter, negatively charged electrons.

4 0

3 years ago

A box weighing 52.4 N is sliding on a rough horizontal floor with a constant friction force of magnitude LaTeX: ff. The box's in

german

Answer:

The magnitude of the friction force exerted on the box is 2.614 newtons.

Explanation:

Since the box is sliding on a rough horizontal floor, then it is decelerated solely by friction force due to the contact of the box with floor. The free body diagram of the box is presented herein as attachment. The equation of equilbrium for the box is:

$\Sigma F = -f = m\cdot a$ (Eq. 1)

Where:

$f$ - Kinetic friction force, measured in newtons.

$m$ - Mass of the box, measured in kilograms.

$a$ - Acceleration experimented by the box, measured in meters per square second.

By applying definitions of weight ( $W = m\cdot g$ ) and uniform accelerated motion ( $v = v_{o}+a\cdot t$ ), we expand the previous expression:

$-f = \left(\frac{W}{g} \right)\cdot \left(\frac{v-v_{o}}{t}\right)$

And the magnitude of the friction force exerted on the box is calculated by this formula:

$f = -\left(\frac{W}{g} \right)\cdot \left(\frac{v-v_{o}}{t}\right)$ (Eq. 1b)

Where:

$W$ - Weight, measured in newtons.

$g$ - Gravitational acceleration, measured in meters per square second.

$v_{o}$ - Initial speed, measured in meters per second.

$v$ - Final speed, measured in meters per second.

$t$ - Time, measured in seconds.

If we know that $W = 52.4\,N$ , $g = 9.807\,\frac{m}{s^{2}}$ , $v_{o} = 1.37\,\frac{m}{s}$ , $v = 0\,\frac{m}{s}$ and $t = 2.8\,s$ , the magnitud of the kinetic friction force exerted on the box is:

$f = -\left(\frac{52.4\,N}{9.807\,\frac{m}{s^{2}} } \right)\cdot \left(\frac{0\,\frac{m}{s}-1.37\,\frac{m}{s} }{2.8\,s} \right)$

$f = 2.614\,N$

The magnitude of the friction force exerted on the box is 2.614 newtons.

5 0

3 years ago

In terms of volume,how do ml & cm3 relate to one another?

Keith_Richards [23]

1 milliliter = 1 cubic centimeter (cm^3)

3 0

3 years ago

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Converting 15 miles to kilometer

Dmitry [639]

15 miles to kilometers would be: 24.14 kilometers

8 0

3 years ago

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