Can anyone pls help me out in dis i am struggling in dis!

In young’s double slit experiment, the measured fringe width is 0.5 mm for a Sodium light of 589 nm at a distance of 1.5 m. A br

Levart [38]

Answer:

(A). The order of the bright fringe is 6.

(B). The width of the bright fringe is 3.33 μm.

Explanation:

Given that,

Fringe width d = 0.5 mm

Wavelength = 589 nm

Distance of screen and slit D = 1.5 m

Distance of bright fringe y = 1 cm

(A) We need to calculate the order of the bright fringe

Using formula of wavelength

$\lambda=\dfrac{dy}{mD}$

$m=\dfrac{d y}{\lambda D}$

Put the value into the formula

$m=\dfrac{1\times10^{-2}\times0.5\times10^{-3}}{589\times10^{-9}\times1.5}$

$m=5.65 = 6$

(B). We need to calculate the width of the bright fringe

Using formula of width of fringe

$\beta=\dfrac{yd}{D}$

Put the value in to the formula

$\beta=\dfrac{1\times10^{-2}\times0.5\times10^{-3}}{1.5}$

$\beta=3.33\times10^{-6}\ m$

$\beta=3.33\ \mu m$

Hence, (A). The order of the bright fringe is 6.

(B). The width of the bright fringe is 3.33 μm.

3 0

3 years ago

A flat sheet of paper of area 0.365 m2 is oriented so that the normal to the sheet is at an angle of 60 ∘ to a uniform electric

andriy [413]

Answer:

A.) 3.65 N*m²/C B) No C) 0º D) 90º

Explanation:

A) The electric flux, when the electric field is uniform across a gausssian surface, can be calculated as the dot product of the electric field vector, and the vector representing the area of the surface (normal to the surface and directed outward it by convention), as follows:

Flux = E*A*cos φ

where E = 20 N/C, A = 0.365 m², φ = 60º.

Replacing by the values, we can get the value of the electric flux, as follows:

Flux = 20 N/C* 0.365 m²*0.5 = 3.65 N*m²/C

B) While the area remains constant, and doesn't change orientation, the value of the flux will be the same, regardless the shape of the sheet.

C) When the normal to the sheet and the electric field are parallel each other, the surface will intercept the maximum number of field lines, i.e. the flux will be directly E*A*cos 0º = E*A (maximum value possible).

D) When the electric field is tangent to the surface, this means that no field lines will be intercepted by the sheet, so the flux is zero.

In this case, φ = 90º, cos φ = 0

⇒ E*A*cos 90º = E*A*0 = 0

5 0

3 years ago

An alien spaceship traveling at 0.600 c toward the Earth launches a landing craft. The landing craft travels in the same directi

Arturiano [62]

The kinetic energy as measured in the Earth reference frame is 6.704*10^22 Joules.

To find the answer, we have to know about the Lorentz transformation.

<h3>What is its kinetic energy as measured in the Earth reference frame?</h3>

It is given that, an alien spaceship traveling at 0.600 c toward the Earth, in the same direction the landing craft travels with a speed of 0.800 c relative to the mother ship. We have to find the kinetic energy as measured in the Earth reference frame, if the landing craft has a mass of 4.00 × 10⁵ kg.

$V_x'=0.8c\\V=0.6c\\m=4*10^5kg$

Let us consider the earth as S frame and space craft as S' frame, then the expression for KE will be,

$KE=m_0c^2=\frac{mc^2}{1-(\frac{v_x^2}{c^2} )}$

So, to $V_x=(0.8+0.6)c-[\frac{0.6c*(0.8c)^2}{c^2}]=1.016$ find the KE, we have to find the value of speed of the approaching landing craft with respect to the earth frame.
We have an expression from Lorents transformation for relativistic law of addition of velocities as,

$V_x'=\frac{V_x-V}{1-\frac{VV_x}{c^2} } \\thus,\\V_x=V_x'(1-\frac{VV_x}{c^2} )+V$

Substituting values, we get,

$V_x=0.8c(1-\frac{0.8c*0.6c}{c^2} )+0.6c=(0.8c*0.52)+0.6c=1.016c$

Thus, the KE will be,

$KE=\frac{4*10^5*(3*10^8)^2}{\sqrt{1-\frac{(1.016c)^2}{c^2} } } =\frac{1.2*10^{22}}{0.179}=6.704*10^{22}J$

Thus, we can conclude that, the kinetic energy as measured in the Earth reference frame is 6.704*10^22 Joules.

Learn more about frame of reference here:

brainly.com/question/20897534

SPJ4

3 0

2 years ago

A 20-gram bullet traveling at 250 m/s strikes a block of wood that weighs 2 kg. With what velocity will the block and bullet mov

Shkiper50 [21]

Answer:

the velocity of the bullet-wood system after the collision is 2.48 m/s

Explanation:

Given;

mass of the bullet, m₀ = 20 g = 0.02 kg

velocity of the bullet, v₀ = 250 m/s

mass of the wood, m₁ = 2 kg

velocity of the wood, v₁ = 0

Let the velocity of the bullet-wood system after collision = v

Apply the principle of conservation of linear momentum to calculate the final velocity of the system;

Initial momentum = final momentum

m₀v₀ + m₁v₁ = v(m₀ + m₁)

0.02 x 250 + 2 x 0 = v(2 + 0.02)

5 + 0 = v(2.02)

5 = 2.02v

v = 5/2.02

v = 2.48 m/s

Therefore, the velocity of the bullet-wood system after the collision is 2.48 m/s

6 0

3 years ago

How does mass affect the change in velocity of an object? Immersive Reader

Bas_tet [7]

Answer:

I would say the 3rd one

Explanation:

5 0

3 years ago