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3 years ago

What happens to the diffraction pattern when the number of lines per centimeter of a diffraction grating is increased?

1 answer:

3 0

The separation between two neighboring vital maxima increments. As the quantity lines per centimeter expand, the detachment d between neighboring openings ends up noticeably littler. The point that determines primary maxima of a diffraction grinding is given by transgression = m/d, where is the wavelength and m = 0, 1, 2, 3. For settled estimations of m and the wavelength, this connection demonstrates that as d abatements, increments. For a settled screen area, this implies the separation between two neighboring chief maxima additionally increments.

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1. Plot the following graphs:

VLD [36.1K]

Answer:

(a) The distance-time graph for an object with uniform speed is giving by a straight line sloped graph with a constant positive or negative gradient as shown in the attached diagram

(b) The distance-time graph for an object with non-uniform speed is giving by a curved line sloped graph with varying gradient as shown in the attached diagram

(c) The velocity-time graph for a car with uniform motion is giving by a horizontal line graph at the speed of constant motion with a zero gradient as shown in the attached diagram

(d) The velocity-time graph for a car moving with uniform acceleration is giving by a straight line sloped graph with a constant positive or negative gradient as shown in the attached diagram

(e) The velocity-time graph for a car moving with non-uniform acceleration is giving by a curved line sloped graph with varying gradient as shown in the attached diagram

(f) According to Newton's first law of motion, an object at rest will remain at rest with no motion unless acted by a force, an therefore, will have no motion with time

Explanation:

8 0

3 years ago

Which other subatomic particle has the same mass as a neutron

RideAnS [48]

Answer:

.................... protons :)

7 0

2 years ago

Read 2 more answers

The drawing shows two situations in which charges are placed on the x and y axes. They are all located at the same distance of 5

ra1l [238]

Answer:

For situation (a)

net charge E = E₊₂ + E₋₅ + E₋₃

E = K(q/d²)

where K = 8.99e9

d = 5.7cm = 5.7e-2m

Therefore,

E₊₂(x) = K(q/d²) = (8.99e9)× ((2.0e-6)÷(5.7e-2)) = 3.15e5(+x)

E₋₅(y) = K(q/d²) = (8.99e9)× ((5.0e-6)÷(5.7e-2)) = 7.88e5(+y)

E₋₃(x) = K(q/d²) = (8.99e9)× ((3.0e6)÷(5.7e-2)) = 4.73e5(+x)

thus

E = E₊₂ + E₋₅ + E₋₃

= 3.15e5(x) + 7.88e5(y) + 4.73e6(x)

= 7.88e6(x) + 7.88e6(y)

use Pythagorean theorem

I <em>E </em>I = $\sqrt{(7.89e5)^{2} + (7.89e5)^{2}}$ = 1.242e6 $\frac{N}{C}$

∅ = $tan^{-1}$ $(\frac{7.88e5}{7.88e5} )$ = $tan^{-1}(1)$ = 45°

Thus for (a) net magnitude = 1.115e6 $\frac{N}{C}$ @ 45° above +x axis

for situation (b)

net charge E = E₊₄ + E₊₁ + E₋₁ + E₊₆

E₊₄(x) = K(q/d²) = (8.99e9)× ((4.0e-6)÷(5.7e-2)) = 6.30e5(+x)

E₊₁(y) = K(q/d²) = (8.99e9)× ((1.0e-6)÷(5.7e-2)) = 1.58e5(-y)

E₋₁(x) = K(q/d²) = (8.99e9)× ((1.0e-6)÷(5.7e-2)) = 1.58e5(+x)

E₊₆(y) = K(q/d²) = (8.99e9)× ((6.0e-6)÷(5.7e-2)) = 9.46e5(+y)

thus,

E = E₊₄ + E₊₁ + E₋₁ + E₊₆

= 6.30e5(x) - 1.58e5(y) + 1.58e5(x) + 9.46e5(y)

= 7.88e5(x) + 7.88e5(y)

use Pythagorean theorem

I <em>E </em>I = $\sqrt{(7.88e5)^{2} + (7.88e5)^{2}}$ = 1.242e6 $\frac{N}{C}$

∅ = $tan^{-1}$ $(\frac{7.88e5}{7.88e5} )$ = $tan^{-1}(1)$ = 45°

Thus for (a) and (b) the net magnitude = 1.242e6 $\frac{N}{C}$ @ 45° above +x axis

Explanation:

I attached a sample image, i hope that corresponds to your question

5 0

2 years ago

In the nuclear fission process mass is converted into energy. Determine the total mass that must be converted to energy in one y

brilliants [131]

Answer:

1) Mass that needs to be converted at 100% efficiency is 0.3504 kg

2) Mass that needs to be converted at 30% efficiency is 1.168 kg

Explanation:

By the principle of mass energy equivalence we have

$E=mc^{2}$

where,

'E' is the energy produced

'm' is the mass consumed

'c' is the velocity of light in free space

Now the energy produced by the reactor in 1 year equals

$Energy=Power\times time\\\\\therefore Energy=1\times 10^{9}\times 365\times 24\times 3600\\\\Energy=31.536\times 10^{15}Joules$

Thus the mass that is covertred at 100% efficiency is

$mass=\frac{Energy}{c^{2}}\\\\mass=\frac{31.536\times 10^{15}}{(3\times 10^{8})^{2}}\\\\mass=\frac{31.536\times 10^{15}}{9\times 10^{16}}\\\\\therefore mass=0.3504kg$

Part 2)

At 30% efficiency the mass converted equals

$mass|_{30}=\frac{mass|_{100}}{0.3}\\\\mass|_{30}=\frac{0.3504}{0.3}\\\\mass|_{30}=1.168kg$

8 0

3 years ago

The temperature of a superconductor is gradually lowered. At the critical temperature, how does the resistivity of the supercond

san4es73 [151]

At critical temperature, the resistivity of the superconductor

B. It suddenly drops to zero

Explanation:

Materials can be classified into three different types depending on their resistance:

Conductors: these materials have generally low resistance and allow electricity to pass through easily. The resistance of a conductor increases linearly with the temperature
Insulators: these materials do not allow electricity to pass through - so they have very high resistance
Semi-conductors: these are materials that are insulators are room temperature, however they becomes conductors when heated. Therefore, the resistance of a semiconductor decreases when the temperature increases
Superconductors: these are special materials that are normally conductors; however, at very low temperatures (we are talking about temperature very near to 0 K), their resistance becomes suddenly zero.

Therefore, the correct answer is:

B. It suddenly drops to zero

Learn more about current and resistance:

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8 0

3 years ago