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Gemiola [76]
3 years ago
5

A thin beam of light of wavelength 625 nm goes through a thin slit and falls on a screen 3.00 m past the slit. You observe that

the first completely dark fringes occur on the screen at distances of ±8.24 mm from the central bright fringe, and that the central bright fringe has an intensity of 2.00 W/m2 at its center. (a) How wide is the slit?
Physics
1 answer:
AleksAgata [21]3 years ago
8 0

Answer:

a = 2.275 10⁻⁴ m

Explanation:

This is a diffraction problem that is described by the equation

         a sin θ = m λ

         

The first dark minimum occurs for m = 1

         a = λ  / sin θ

The angle can be found by trigonometry,

       tan θ = y / x

       θ = tan⁻¹ y / x

Let's reduce the magnitudes to the SI system

        y = 8.24 mm = 8.24 10⁻³ m

        λ  = 625 nm = 625 10⁻⁹ m

       θ = tan⁻¹ 8.24 10⁻³ / 3.00

       θ = 0.002747 rad

We calculate

       a = 625 10⁻⁹ / sin 0.002747

       a = 2.275 10⁻⁴ m

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lilavasa [31]

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Now, if you have mass, then you have weight ... it's the answer you give when somebody asks "How much do you weigh ?". That's the force of gravity between you and the Earth, pulling you toward the center of the Earth. But it doesn't stop there. There's also a force of gravity pulling the Earth toward the center of you. The strength of it is EQUAL to your weight.

Your weight on Earth is EQUAL to the Earth's weight on YOU !

7 0
3 years ago
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An object is located in air, 25 cm from the vertex of the concave surface of a block of glass (as viewed from the air side of th
Marizza181 [45]

Your question is missing one part as "magnification"

i have completed the missing part below

Answer:

a. d_{i}=-0.0566cm

b. M=441.69

Explanation:

For this type of numerical we will use the following formulas

\frac{n1}{d_{o} }+\frac{n_{2} }{d_{i} }=\frac{n_{2}-n_{1}  }{R}......... Eq1

where,

n_{1}=refractive index of the medium surrounding refracting surface/object

i.e. n_{1}=n_{air}  

n_{2}= refractive index of the refracting surface/object

i.e. n_{2}=n_{glass}

d_{0}= distance of object from the vertex of the refracting surface

d_{i}=distance of image from the vertex of the refracting surface

R=radius of curvature of the refracting surface

M=\frac{d_{0} }{d_{i} } ........... Eq2

where,

M=magnification

Convention:

R>0\for\ the\ convex\ refractive\ surface\ of\ curvature\\\ R

Given:

n_{1}=n_{air}=1.0

n_{2}=n_{glass}=1.5

d_{0}=25cm

R=-11cm because refraction surface is concave

Required:

a. d_{i}=?

b. M=?

Solution:

a. putting values in eq1, we get

\frac{1.0}{25}+\frac{1.5}{d_{i} }=\frac{1.5-1.0}{-11}

\frac{1.5}{d_{i} }=-0.045-0.040

d_{i}=(-0.085)(\frac{1}{1.5} )

d_{i}=-0.0566cm

b. M=\frac{25}{-0.05665}

M=441.69

5 0
4 years ago
A 0.300 kg mass is attached to
scoundrel [369]

Answer:

0.144 J

Explanation:

From the question above,

Total elastic energy of the spring = Kinetic energy of the spring + potential energy of the spring

E' = E.k+Ep...................... Equation 1

Ek = E'-Ep......................... Equation 2

Where E' = total elastic energy, E.k = kinetic energy, E.p = potential  energy.

E'  = 1/2ke² = 1/2(26.6)(0.12²) = 0.19152 J.

Ep = 1/2ke² = 1/2(26.6)(0.06²)

Ep = 0.04788 J

Therefore,

Ek = 0.19152-0.04788

Ek = 0.144 J

8 0
3 years ago
Which statement best explains the speed of light waves as they travel from gas to solid?
igomit [66]

Answer:

Light waves slow down as they travel from gas to solid.

Explanation:

The speed of light is often quoted with reference to a vacuum. Light travels fastest in a vacuum and in a gas.

When light travels through other media such as solids or liquids, the speed of light is decreased due to absorption and scattering of photons by molecules as well as remissions.

Hence the speed of light decreases from gas to solid.

8 0
3 years ago
Suppose an elephant has a mass of 2800 kg. show answer No Attempt How fast, in meters per second, does the elephant need to move
Mumz [18]

Answer:

The speed of elephant must be 1.4 m/s.

Explanation:

Mass of the elephant, M = 2800 kg

Mass of sprinter, m = 77 kg

Speed of sprinter, v = 8.5 m/s

It is required to find the speed of elephant to have the same kinetic energy as that of sprinter. So,

K_e=K_s\\\\\dfrac{1}{2}MV^2=\dfrac{1}{2}mv^2\\\\MV^2=mv^2\\\\V=\sqrt{\dfrac{mv^2}{M}} \\\\V=\sqrt{\dfrac{77\times 8.5^2}{2800}} \\\\V=1.40\ m/s

So, the speed of elephant must be 1.4 m/s.

7 0
3 years ago
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