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3 years ago

Which one of the following energy conversion devices has the lowest energy efficiency?

Physics

1 answer:

Anna35 [415]3 years ago

3 0

Answer:

A) incandescent ligth bulb, its efficiency is about 10%

Explanation:

The incandescent bulb, that is, the well-known focus with its warm light, was one of the most useful inventions of the 19th century although its use is currently considered very inefficient. These lamps waste between 80 and 90 percent of the total electricity they consume by turning it into heat. The metal filament thus heated and which is the central part of the bulb, only converts the remaining energy into light. Its service life ranges from 750 to 1,000 hours.

This is why they are used in ovens for food preparation, because of the large amount of heat they generate.

The steam boiler in a power plant depends on the fuel that it is using, but a coal-fired power plant with modern technology its efficiency is about 40%

Electric motor are around 85-92%

In order to better understand the concept of efficiency it is as if we pay 100 dollars of gasoline for our weekly use, but of that 100 dollars the car only uses 10 dollars to do that activity the rest of the money the 90 dollars were lost because of the inefficiencies of the vehicle.

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Two small spheres with mass 10 gm and charge q, are suspended from a point by threads of length L=0.22m. What is the charge on e

Nataliya [291]

Answer:

$q=1.95*10^{-7}C$

Explanation:

According to the free-body diagram of the system, we have:

$\sum F_y: Tcos(15^\circ)-mg=0(1)\\\sum F_x: Tsin(15^\circ)-F_e=0(2)$

So, we can solve for T from (1):

$T=\frac{mg}{cos(15^\circ)}(3)$

Replacing (3) in (2):

$(\frac{mg}{cos(15^\circ)})sin(15^\circ)=F_e$

The electric force ( $F_e$ ) is given by the Coulomb's law. Recall that the charge q is the same in both spheres:

$(\frac{mg}{cos(15^\circ)})sin(15^\circ)=\frac{kq^2}{r^2}(4)$

According to pythagoras theorem, the distance of separation (r) of the spheres are given by:

$sin(15^\circ)=\frac{\frac{r}{2}}{L}\\r=2Lsin(15^\circ)(5)$

Finally, we replace (5) in (4) and solving for q:

$mgtan(15^\circ)=\frac{kq^2}{(2Lsin(15^\circ))^2}\\q=\sqrt{\frac{mgtan(15^\circ)(2Lsin(15^\circ))^2}{k}}\\q=\sqrt{\frac{10*10^{-3}kg(9.8\frac{m}{s^2})tan(15^\circ)(2(0.22m)sin(15^\circ))^2}{8.98*10^{9}\frac{N\cdot m^2}{C^2}}}\\q=1.95*10^{-7}C$

5 0

3 years ago

The four-firm concentration ratio for newstands is 17 and for electric lamp makers it is 89. The HHI for newstands is 105 and f

TEA [102]

Answer:

is; is not

Explanation:

The four-firm concentration ratio for newstands is 17 and for electric lamp makers it is 89. The HHI for newstands is 105 and for electric lamp makers it is 2 comma 850. The newstand market is an example of monopolistic competition. The electric lamp market is not an example of monopolistic competition

4 0

4 years ago

Two Earth satellites, A and B, each of mass m, are to be launched into circular orbits about Earth's center. Satellite A is to o

Pachacha [2.7K]

(a) 0.448

The gravitational potential energy of a satellite in orbit is given by:

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

where

G is the gravitational constant

M is the Earth's mass

m is the satellite's mass

r is the distance of the satellite from the Earth's centre, which is sum of the Earth's radius (R) and the altitude of the satellite (h):

r = R + h

We can therefore write the ratio between the potentially energy of satellite B to that of satellite A as

$\frac{U_B}{U_A}=\frac{-\frac{GMm}{R+h_B}}{-\frac{GMm}{R+h_A}}=\frac{R+h_A}{R+h_B}$

and so, substituting:

$R=6370 km\\h_A = 5970 km\\h_B = 21200 km$

We find

$\frac{U_B}{U_A}=\frac{6370 km+5970 km}{6370 km+21200 km}=0.448$

(b) 0.448

The kinetic energy of a satellite in orbit around the Earth is given by

$K=\frac{1}{2}\frac{GMm}{r}$

So, the ratio between the two kinetic energies is

$\frac{K_B}{K_A}=\frac{\frac{1}{2}\frac{GMm}{R+h_B}}{\frac{1}{2}\frac{GMm}{R+h_A}}=\frac{R+h_A}{R+h_B}$

Which is exactly identical to the ratio of the potential energies. Therefore, this ratio is also equal to 0.448.

(c) B

The total energy of a satellite is given by the sum of the potential energy and the kinetic energy:

$E=U+K=-\frac{GMm}{R+h}+\frac{1}{2}\frac{GMm}{R+h}=-\frac{1}{2}\frac{GMm}{R+h}$

For satellite A, we have

$E_A=-\frac{1}{2}\frac{GMm}{R+h_A}=-\frac{1}{2}\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}kg)(28.8 kg)}{6.37\cdot 10^6 m+5.97\cdot 10^6 m}=-4.65\cdot 10^8 J$

For satellite B, we have

$E_B=-\frac{1}{2}\frac{GMm}{R+h_B}=-\frac{1}{2}\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}kg)(28.8 kg)}{6.37\cdot 10^6 m+21.2\cdot 10^6 m}=-2.08\cdot 10^8 J$

So, satellite B has the greater total energy (since the energy is negative).

(d) $-2.57\cdot 10^8 J$

The difference between the energy of the two satellites is:

$E_B-E_A=-2.08\cdot 10^8 J-(-4.65\cdot 10^8 J)=-2.57\cdot 10^8 J$

4 0

3 years ago

I WILL MARK BRAINLIEST TO FIRST ANSWER IF CORRECT

charle [14.2K]

Answer:

Conventional current flows from cathode to anode. It is opposite to the direction of flow of electrons.

Explanation:

Hope this helped, Have a Great Day!!

3 0

3 years ago

Ejection of Electrons from Hydrogen by Incident Photons Light of wavelength 80 nm is incident on a sample of hydrogen gas, resul

timofeeve [1]

Answer:

a) $K_{max}$ = 1.9 eV = 3.04 10⁻¹⁹ J,b ) This means that some electrons are at the first excited level of the hydrogen atom, which is highly likely as the temperature rises.

Explanation:

a) To calculate the maximum kinetic energy of the expelled electrons let's use the relationships of the photoelectric effect

$K_{max}$ = h f - Φ

Where K is the kinetic energy, h the Planck constant that is worth 6.63 10⁻³⁴ Js, f the frequency and Φ the work function

The speed of light is related to wavelength and frequency

c = λ f

Let's analyze the work function, it is the energy needed to start an electron from a metal, in this case to start an electron from a hydrogen atom its fundamental energy is needed, so

Φ= E₀ = 13.6 eV

let's replace and calculate the energy of the incident photon

E = h c / λ

E = 6.63 10⁻³⁴ 3 10⁸/80 10⁻⁹

E = 2,486 10⁻¹⁸ J

Let's reduce to eV

E = 2,486 10⁻¹⁸ (1 eV / 1.6 10⁻¹⁹)

E = 15.5 eV

Now we can calculate the kinetic energy

$K_{max}$ = h c / f - fi

$K_{max}$ = 15.5 -13.6

$K_{max}$ = 1.9 eV

b) Extra energy = 10.2 eV

The total kinetic energy of electrons is

Total kinetic energy = 1.9 +10.2 = 12.1 eV

For the calculation we are assuming that all the electors are in the hydrogen base state, but for temperatures greater than 0K some electors may be in some excited state, so less energy is needed to tear them out of hydrogen atom.

Let's analyze this possibility

ΔE = E photon - Total kinetic energy electron

ΔE = 15.5 - 12.1

ΔE = 3.4 eV

If we use the Bohr ratio for the hydrogen atom

$E_{n}$ = 13.606 / n2

n = √ 13.606 / En

n = √ (13606 / 3.4)

n = 2

This means that some electrons are at the first excited level of the hydrogen atom, which is highly likely as the temperature rises.

8 0

3 years ago