During the exchange of air in your lungs oxygen is being taken into your bloodstream while this substance is being released

A grating has 470 lines/mm. how many orders of the visible wavelength 538 nm can it produce in addition to the m = 0 order?

Natali [406]

Three complete orders on each side of the m=0 order can be produced in addition to the m=0 order.

The ruling separation is d=1/(470mm-1)

$= 2.1 \times 10 {}^{ - 3}mm$

Diffraction lines occurs at an angle θ such that dsin=mλ,when λ is the wavelength and m is an integer.

Notice that for a given order,the line associated with a long wavelength is produced at a greater angle than the line associated with shorter wavelength.

we take λ to be the longest wavelength in the visible spectrum (538nm) and find the greatest integer value of m such that θ is less than 90°.

That is,find the greater integer value of m for which mλ<d.

since,d/λ

$= 538 \times 10 {}^{ - 9} m/2.1 \times 10 {}^{ - 6}$

There are three complete orders on each side of the m=0 order.

The second and third orders overlap.

learn more about diffraction from here: brainly.com/question/28168352

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4 0

1 year ago

What is the momentum of a 0.000015 kg mosquito flying straight at you with a velocity of 5.20 m/s?

Setler [38]

Answer:

Momentum, $p=7.8\times 10^{-5}\ N-m$

Explanation:

Given that,

Mass of a mosquito, m = 0.000015 kg

Velocity of the mosquito, v = 5.2 m/s

We need to find the momentum of the mosquito. The momentum of an object is given by :

p = mv

Put all the values in the above formula.

$p=0.000015\ kg\times 5.2\ m/s\\\\p=0.000078\ N-m\\\\\text{or}\\\\p=7.8\times 10^{-5}\ N-m$

So, the momentum of the mosquito is $7.8\times 10^{-5}\ N-m$ .

3 0

2 years ago

Which of the following never cause a change in the motion of an object? A. Net forces B. Unbalanced forces O C. Balanced forces

Karolina [17]

Answer:

balanced force never changes its motion

5 0

2 years ago

Write the device modeling equation for the first and second law of thermodynamics

anyanavicka [17]

1st the total energy of an isolated system is constant; energy can be transformed from one form to another, but can be neither created nor destroyed. ▲U=Q-W

2nd the total entropy can never decrease over time for an isolated system, that is, a system in which neither energy nor matter can enter nor leave.
DS (Greater than or equal to) 0

8 0

3 years ago

The barricade at the end of a subway line has a large spring designed to compress 2.00 m when stopping a 1.10 ✕ 105 kg train mov

Mrac [35]

Answer:

(a) k = 1684.38 N/m = 1.684 KN/m

(b) Vi = 0.105 m/s

(c) F = 1010.62 N = 1.01 KN

Explanation:

(a)

First, we find the deceleration of the car. For that purpose we use 3rd equation of motion:

2as = Vf² - Vi²

a = (Vf² - Vi²)/2s

where,

a = deceleration = ?

Vf = final velocity = 0 m/s (since, train finally stops)

Vi = Initial Velocity = 0.35 m/s

s = distance covered by train before stopping = 2 m

Therefore,

a = [(0 m/s)² - (0.35 m/s)²]/(2)(2 m)

a = 0.0306 m/s²

Now, we calculate the force applied on spring by train:

F = ma

F = (1.1 x 10⁵ kg)(0.0306 m/s²)

F = 3368.75 N

Now, for force constant, we use Hooke's Law:

F = kΔx

where,

k = Force Constant = ?

Δx = Compression = 2 m

Therefore.

3368.75 N = k(2 m)

k = (3368.75 N)/(2 m)

k = 1684.38 N/m = 1.684 KN/m

(c)

Applying Hooke's Law with:

Δx = 0.6 m

F = (1684.38 N/m)(0.6 m)

F = 1010.62 N = 1.01 KN

(b)

Now, the acceleration required for this force is:

F = ma

1010.62 N = (1.1 kg)a

a = 1010.62 N/1.1 x 10⁵ kg

a = 0.0092 m/s²

Now, we find initial velocity of train by using 3rd equation of motion:

2as = Vf² - Vi²

a = (Vf² - Vi²)/2s

where,

a = deceleration = -0.0092 m/s² (negative sign due to deceleration)

Vf = final velocity = 0 m/s (since, train finally stops)

Vi = Initial Velocity = ?

s = distance covered by train before stopping = 0.6 m

Therefore,

-0.0092 m/s² = [(0 m/s)² - Vi²]/(2)(0.6 m)

Vi = √(0.0092 m/s²)(1.2 m)

Vi = 0.105 m/s

4 0

3 years ago