What type of energy comes out<br> of a generator?

A refracting telescope has a 1.48 m diameter objective lens with focal length 15.4 m and an eyepiece with focal length 3.28 cm.

vivado [14]

Answer:

m = 469.51

Explanation:

Given that,

The diameter of a refracting telescope is 1.48 m

Focal length of the objective lens is 15.4 m

Focal length of an eyepiece is 3.28 cm

We need to find the angular magnification of the telescope. The ratio of focal length of objective lens to the focal length of the eye piece is called angular magnification of the telescope. So,

$m=\dfrac{f_o}{f_e}\\\\m=\dfrac{15.4}{3.28\times 10^{-2}}\\\\m=469.51$

So, the angular magnification of the telescope is 469.51.

7 0

4 years ago

Calculate the number of vacancies per cubic meter in iron at 850 °C. The energy for vacancy formation is 1.08 eV/atom. Furthermo

Fed [463]

Answer:

The number of vacancies per cubic meter is 1.18 X 10²⁴ m⁻³

Explanation:

$N_v = N*e[^{-\frac{Q_v}{KT}}] = \frac{N_A*\rho _F_e}{A_F_e}e[^-\frac{Q_v}{KT}}]$

where;

N $_A$ is the number of atoms in iron = 6.022 X 10²³ atoms/mol

ρFe is the density of iron = 7.65 g/cm3

AFe is the atomic weight of iron = 55.85 g/mol

Qv is the energy vacancy formation = 1.08 eV/atom

K is Boltzmann constant = 8.62 X 10⁻⁶ k⁻¹

T is the temperature = 850 °C = 1123 k

Substituting these values in the above equation, gives

$N_v = \frac{6.022 X 10^{23}*7.65}{55.85}e[^-\frac{1.08}{8.62 X10^{-5}*1123}}]\\\\N_v = 8.2486X10^{22}*e^{(-11.1567)}\\\\N_v = 8.2486X10^{22}*1.4279 X 10^{-5}\\\\N_v = 1.18 X 10^{18}cm^{-3} = 1.18 X 10^{24}m^{-3}$

Therefore, the number of vacancies per cubic meter is 1.18 X 10²⁴ m⁻³

3 0

4 years ago

The average distance between collisions for atoms in a real gas is known as the mean free path (see pp. 298-9 in McKay). Which o

dusya [7]

Answer:

300 nm

Explanation:

R = Gas constant = 8.314 J/molK

r = Atomic radii = $1\times 10^{-10}\ m$

d = Atomic diameter = $2r=2\times 10^{-10}\ m$

At STP

T = Temperature = 273.15 K

P = Pressure = 100 kPa

$N_A$ = Avogadro's number = $6.022\times 10^{23}$

The mean free path is given by

$\lambda=\frac{RT}{\sqrt2d^2N_AP}\\\Rightarrow \lambda=\frac{8.314\times 273.15}{\sqrt2 \pi \times (2\times 10^{-10})^2\times 6.022\times 10^{23}\times 100000}\\\Rightarrow \lambda=2.12165\times 10^{-7}\ m=212.165\times 10^{-9}\ m=212.165\ nm$

The answer that best represents the mean free path for gas molecules is 300 nm

4 0

4 years ago

Una cámara fotográfica analógica (no digital) tiene dos lentes intercambiables. Uno de foco 55mm y el otro de 200 mm. Toma una f

Sergeu [11.5K]

Answer:

f = 55mm, h ’= -9.89 cm

f = 200 mm, h ’= 42.5 cm

Explanation:

For this exercise let's start by finding the distance to the image, using the equation of the constructor

$\frac{1}{f} = \frac{1}{p} + \frac{1}{q}$

where f is the focal length, p and q are the distances to the object and image, respectively

lens with f₁ = 55mm = 0.55cm

$\frac{1}{q} = \frac{1}{f} - \frac{1}{p}$

$\frac{1}{q_1}$ = $\frac{1}{0.55} - \frac{1}{10}$

$\frac{1}{q_1}$ = 1.718

q₁ = 0.582 m

lens with f₂ = 200mm = 2m

$\frac{1}{q_2}$ = $\frac{1}{2} - \frac{1}{10}$

$\frac{1}{q_2}$ = 0.4

q₂ = 2.5 m

the magnification of a lens is given by

m = $\frac{h'}{h} = - \frac{q}{p}$

h ’= $- \frac{q}{p} \ h$

let's calculate for each lens

f = 55mm

h '= - 0.582 / 10 1.7

h ’= 0.0989 m

h ’= -9.89 cm

f = 200 mm

h '= - 2.5 / 10 1.7

h ’= -0.425 m

h ’= 42.5 cm

The negative sign indicates that the image is real and inverted

4 0

3 years ago

Alicia drew a pentagon with equal side length and equal angles.Then she added red lines of symmetry to her drawing. How many lin

IRISSAK [1]

The sum of angles in a regular pentagon is
180*(5 - 2) = 540°
Each internal angle is 540/5 = 108°.

Each vertex creates a line of symmetry to the midpoint of the opposite side,
as shown in the figure.

Answer: 5 lines of symmetry.

7 0

3 years ago

What type of energy comes out of a generator?