Ask question

Ask question

All categories

3 years ago

14

For your science fair project you need to design a diffraction grating that will disperse the visible spectrum (400-700 nm) over

30.0∘ in first order.

1 answer:

elixir [45]3 years ago

3 0

Answer:

Explanation:

If d be the distance between two lines , for bright fringe in first order

d sinθ = λ

700 nm must be positioned at 30⁰

d sin30 = 700 x 10⁻⁹

d = 7 x 10⁻⁷ / .5

= 14 x 10⁻⁷

= 14 x 10⁻⁷

no of lines per m

1/ 14 x 10⁷

= 7.14 x 10⁵

You might be interested in

Six artificial satellites complete one circular orbit around a space station in the same amount of time. Each satellite has mass

oee [108]

Answer:

The ranking of the net force acting on different satellite from largest to smallest is ${F_E} > {F_F} > {F_A} = {F_B} = {F_D} > {F_C}$

Explanation:

In order to get a good understanding of this solution we need to understand that the main concepts used to solve this problem are centripetal force and velocity of satellite.

Initially, use the expression of the velocity of satellite and find out its dependence on the radius of orbit. Use the dependency in the centripetal force expression.

Finally, we find out the velocity of the six satellites and use that expression to find out the force experienced by the satellite. Find out the force in terms of mass (m) and radius of orbit (L) and at last compare the values of force experienced by six satellites.

Fundamentals

The centripetal force is necessary for the satellite to remain in an orbit. The centripetal force is the force that is directed towards the center of the curvature of the curved path. When a body moves in a circular path then the centripetal force acts on the body.

The expression of the centripetal force experienced by the satellite is given as follows:

${F_{\rm{c}}} = \frac{{m{v^2}}}{L}$

Here, m is the mass of satellite, v is the velocity, and L is the radius of orbit.

The velocity of the satellite with which the satellite is orbiting in circular path is given as follows:

$v = \frac{{2\pi L}}{T}$

Here, T is the time taken by the satellite.

The velocity of the satellite with which the satellite is orbiting in circular path is given as follows;

$v = \frac{{2\pi L}}{T}$

Since, all the satellites complete the circular orbit in the same amount of time. The factor of $\frac{{2\pi }}{T}$ is not affected the velocity value for the six satellites. Therefore, we can write the expression of v given as follows:

Substitute $v = \frac{{2\pi L}}{T}$ in the force expression ${F_{\rm{c}}} = \frac{{m{v^2}}}{L}$ as follows:

$\begin{array}{c}\\{F_c} = \frac{{m{{\left( {\frac{{2\pi L}}{T}} \right)}^2}}}{L}\\\\ = \frac{{4{\pi ^2}}}{{{T^2}}}mL\\\end{array}$

Since, all the satellites complete the circular orbit in the same amount of time. The factor of $\frac{{4{\pi ^2}}}{{{T^2}}}$ not affect the force value for six satellites.Therefore, we can write the expression of ${F_c}$ given as follows:

${F_c} = kmL$

Here, k refers to constant value and equal to $\frac{{4{\pi ^2}}}{{{T^2}}}$

${F_A} = k{m_A}{L_A}$

Substitute 200 kg for ${m_A}$ and 5000 m for LA in the expression ${F_A} = k{m_A}{L_A}$

$\begin{array}{c}\\{F_A} = k\left( {200{\rm{ kg}}} \right)\left( {5000{\rm{ m}}} \right)\\\\ = {10^6}k{\rm{ N}}\\\end{array}$

The force acting on satellite B from their rocket is given as follows: ${F_B} = k{m_B}{L_B}$

Substitute 400 kg for ${m_B}$ and 2500 m for in the expression ${F_B} = k{m_B}{L_B}$

$\begin{array}{c}\\{F_B} = k\left( {400{\rm{ kg}}} \right)\left( {2500{\rm{ m}}} \right)\\\\ = {10^6}k{\rm{ N}}\\\end{array}$

The force acting on satellite C from their rocket is given as follows: ${F_C} = k{m_C}{L_C}$

Substitute 100 kg for ${m_C}$ and 2500 m for in the above expression ${F_C} = k{m_C}{L_C}$

$\begin{array}{c}\\{F_C} = k\left( {100{\rm{ kg}}} \right)\left( {2500{\rm{ m}}} \right)\\\\ = 0.25 \times {10^6}k{\rm{ N}}\\\end{array}$

The force acting on satellite D from their rocket is given as follows: ${F_D} = k{m_D}{L_D}$

Substitute 100 kg for ${m_D}$ and 10000 m for ${L_D}$ in the expression ${F_D} = k{m_D}{L_D}$

$\begin{array}{c}\\{F_D} = k\left( {100{\rm{ kg}}} \right)\left( {10000{\rm{ m}}} \right)\\\\ = {10^6}k{\rm{ N}}\\\end{array}$

The force acting on satellite E from their rocket is given as follows: ${F_E} = k{m_E}{L_E}$

Substitute 800 kg for ${m_E}$ and 5000 m for ${L_E}$ in the expression ${F_E} = k{m_E}{L_E}$

$\begin{array}{c}\\{F_E} = k\left( {800{\rm{ kg}}} \right)\left( {5000{\rm{ m}}} \right)\\\\ = 4.0 \times {10^6}k{\rm{ N}}\\\end{array}$

The force acting on satellite F from their rocket is given as follows: ${F_F} = k{m_F}{L_F}$

Substitute 300 kg for ${m_F}$ and 7500 m for ${L_F}$ in the expression ${F_F} = k{m_F}{L_F}$

$\begin{array}{c}\\{F_F} = k\left( {300{\rm{ kg}}} \right)\left( {7500{\rm{ m}}} \right)\\\\ = 2.25 \times {10^6}k{\rm{ N}}\\\end{array}$

The value of forces obtained for the six-different satellite are as follows.

$\begin{array}{l}\\{F_A} = {10^6}k{\rm{ N}}\\\\{F_B} = {10^6}k{\rm{ N}}\\\\{F_C} = 0.25 \times {10^6}k{\rm{ N}}\\\\{F_D} = {10^6}k{\rm{ N}}\\\\{F_E} = 4.0 \times {10^6}k{\rm{ N}}\\\\{F_F} = 2.25 \times {10^6}k{\rm{ N}}\\\end{array}$

The ranking of the net force acting on different satellite from largest to smallest is ${F_E} > {F_F} > {F_A} = {F_B} = {F_D} > {F_C}$

7 0

4 years ago

If you can see the blinking insertion point in a field, you are in ________.

Sonja [21]

You would be in an editing mode when you can see a blinking insertion point on a field. In addition, in any word processing software, it would basically signify that you can already type on the document. Another name for the insertion point would be the I-beam as displayed on the screen.

7 0

3 years ago

Why do organisms need to breathe?

Korvikt [17]

Answer:

Most living things need oxygen helps organisms grow,reproduce,and turn food into energy.Oxygen gives our cells the ability to break down food in order to get the energy we need to survive.

4 0

3 years ago

If the bar magnet is flipped over and the south pole is brought near the hanging ball, the ball will be?

disa [49]

The ball may attracted to the magnet.

<h3>How can we understand that the hanging ball will be attracted to the magnet or not?</h3>

From the question, we understand that the ball is attracted by the north pole of the bar magnet, then the bar magnet flipped over and the south pole is brought near the hanging ball.
As we know, in this type of experiments of bar magnet most of the times the ball is made out of steel.
Steel is a magnetic material.
Magnetic materials gets attracted to the magnet at both the North and South pole.
This can be compared to how neutral objects also gets attracted to the positively and negatively charged rods through the Polarization force.

So, If the bar magnet is flipped over and the south pole is brought near the hanging ball, The ball will be attracted to the magnet.

Learn more about the bar magnet:

brainly.com/question/27943723

#SPJ4

7 0

2 years ago

Two small objects, with masses m and m, are originally a distance r apart, and the gravitational force on each one has magnitude

elena-s [515]

32f. That's because the force is directly proportional to the product of the masses and inversely proportional to the square of the distance. So you get 2•(1/1/4)^2=2•16=32

6 0

4 years ago

Read 2 more answers