All categories

3 years ago

How is aperture measured? What do the measurements mean?

1 answer:

4 0

Aperture is measured in F-stops, in which the f-stops is the amount of light allowed to pass through the aperture, which simply put means that the smaller the aperture, the higher the f-stops. What it does is reduce the amount of light that reaches the film, so the higher the f-stops, the less light reaches the film.

You might be interested in

Which one i need help

Lostsunrise [7]

Answer:

Explanation:

7 0

3 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

3 years ago

Read 2 more answers

Describe the difference between a transverse wave and longitudinal wave. Include the parts that you would find on a transverse w

spayn [35]

A transverse wave is a moving wave in which the current is perpendicular to the direction of the wave or path of propagation. A longitudinal wave are waves in which the displacement of the median is in the direction of the propagation.
Example:
Transverse- pond ripple
Longitudinal- crest and troff

6 0

3 years ago

Read 2 more answers

A 12-g wad of sticky clay is hurled horizontally at a 100-g wooden block initially at rest on a horizontal surface. The clay sti

klasskru [66]

Answer:

91.5 m/s

Explanation:

m = mass of clay = 12 g = 0.012 kg

M = mass of wooden block = 100 g = 0.1 kg

d = distance traveled by the combination before coming to rest = 7.5 m

μ = Coefficient of friction = 0.65

V = speed of the combination of clay and lock just after collision

V' = final speed of the combination after coming to rest = 0 m/s

acceleration caused due to friction is given as

a = - μ g

a = - (0.65) (9.8)

a = - 6.37 m/s²

Using the kinematics equation

V'² = V² + 2 a d

0² = V² + 2(- 6.37) (7.5)

V = 9.8 m/s²

v = speed of clay just before collision

Using conservation of momentum

m v = (m + M) V

(0.012) v = (0.012 + 0.100) (9.8)

v = 91.5 m/s

6 0

3 years ago

Atomic hydrogen produces a well-known series of spectral lines in several regions of the electromagnetic spectrum. Each series f

navik [9.2K]

Answer:

n₁ = 3

Explanation:

The energy of the states in the hydrogen atom is explained by the Bohr model, the transitions heal when an electron passes from a state of higher energy to another of lower energy,

ΔE = $E_{nf}$ - E₀ = - k²e² / 2m (1 / $n_{f}$ ²2 - 1 / n₀²)

The energy of this transition is given by the Planck equation

E = h f = h c / λ

h c / λ = -k²e² / 2m (1 / no ²- 1 / no²)

1 / λ = Ry (1/ $n_{f}$ ² - 1 / n₀²)

Let's apply these equations to our case

λ = 821 nm = 821 10⁻⁹ m

E = h c / λ

E = 6.63 10⁻³⁴ 3 10⁸/821 10⁻⁹

E = 2.423 10⁻¹⁹ J

Now we can use the Bohr equation

Let's reduce to eV

E = 2,423 10⁻¹⁹ J (1eV / 1.6 10⁻¹⁹) = 1,514 eV

$E_{nf}$ - E₀ = -13.606 (1 / $n_{f}$ ² - 1 / n₀²) [eV]

Let's look for the energy of some levels

n $E_{n}$ (eV) $E_{nf}$ - E $E_{ni}$ (eV)

1 -13,606 E₂-E₁ = 10.20

2 -3.4015 E₃-E₂ = 1.89

3 -1.512 E₄- E₃ = 0.662

4 -0.850375

We see the lines of greatest energy for each possible series, the closest to our transition is n₁ = 3 in which a transition from infinity (n = inf) to this level has an energy of 1,512 eV that is very close to the given value

8 0

3 years ago