How much energy is required to move an electron through a potential difference of

In a Young's double-slit experiment, 610-nm-wavelength light is sent through the slits. The intensity at an angle of 2.95° from

Sliva [168]

Answer:

spacing between the slits is 405.32043 × $10^{-9}$ m

Explanation:

Given data

wavelength = 610 nm

angle = 2.95°

central bright fringe = 85%

to find out

spacing between the slits

solution

we know that spacing between slit is

I = 4 $I_{0}$ × cos²∅/2

so

I/4 $I_{0}$ = cos²∅/2

here I/4 $I_{0}$ is 85 % = 0.85

so

0.85 = cos²∅/2

cos∅/2 = √0.85

∅ = 2 × $cos^{-1}$ 0.921954

∅ = 45.56°

∅ = 45.56° ×π/180 = 0.7949 rad

and we know that here

∅ = 2π d sinθ / wavelength

so

d = ∅× wavelength / ( 2π sinθ )

put all value

d = 0.795 × 610× $10^{-9}$ / ( 2π sin2.95 )

d = 405.32043 × $10^{-9}$ m

spacing between the slits is 405.32043 × $10^{-9}$ m

7 0

3 years ago

In order to take advantage of wave power, water must always come in direct contact with the blade of a turbine to generate elect

alina1380 [7]

Answer:

false

Explanation:

4 0

3 years ago

Two stars in a faraway part of the Milky Way are orbiting each other as a binary star system. By

natita [175]

Answer: $M_{total}=$ 1.85

Explanation: Estimate the total mass of a binary system is done by a reformulation of <em>Kepler's Third Law</em>, which states that the square of the period of a planet's orbit is proportional to the cube of its semimajor axis, i.e.:

$a^{3}=(M_{1}+M_{2})P^{2}$

where

a is semimajor axis in astronomical units (AU);

P is period measured in years;

$M_{1}+M_{2}$ is total mass of the two-stars system;

For the two stars faraway in the Milky Way:

1 year is equivalent of 365 days, so period in years:

$P=\frac{594}{365}$

P = 1.63 years

Calculating total mass:

$a^{3}=(M_{total})P^{2}$

$M_{total}=\frac{a^{3}}{P^{2}}$

$M_{total}=\frac{1.7^{3}}{1.63^{2}}$

$M_{total}=$ 1.85

<u>The total mass of the two-object system is 1.85 mass units.</u>

3 0

3 years ago

Assume that the ammeter in the figure below is removed and the current that flows through the 4.0Ω path, I3, is unknown. Determi

qaws [65]

The circuit is in parallel connection

Equivalent resistance = 1/Req = 1/R1 + 1/R2 + 1/R3

From the information given,

R1 = 5

R2 = 2

R3 = 4

1/Req = 1/5 + 1/2 + 1/4 = (4 + 10 + 5)/20 = 19/20

Req = 20/19 = 1.053 ohms

I = V/R

Given that V = 12,

Current flow through circuit = 12/1.053 = 11.4 A

I1 + I2 + I3 = 11.4

I1 = 12/5 = 2.4 A

I2 = 12/2 = 6 A

I3 = 12/4 = 3A

8 0

1 year ago

In a collision experiment, the ratio of the velocity change between two carts of equal inertia is found to equal 1.

VashaNatasha [74]

Answer:

A) Therefore if I double the masses with are in the two terrine they are simplified and the radii of the speeds remain the same

B) If the masses are maintained and the speeds are doubled, the radius of the two speeds remains the same

Explanation:

A vehicle crash problem must be solved with the equation of the moment,

Initial instant Before crash

p₀ = m v₁ + mv₂

After the crash

$p_{f}$ = m $v_{1f}$ + m $v_{2f}$

p₀ = $p_{f}$

If the speed ratio before and after the crash is one

p₀ / $p_{f}$ = 1

We can assume that initially one of the cars was stopped

m v₁₀ = m $v_{2f}$

v₁₀ = $v_{2f}$

For the two speeds to be equal, the masses of the vehicles must be the same.

A) Therefore if I double the masses with are in the two terrine they are simplified and the radii of the speeds remain the same

B) If the masses are maintained and the speeds are doubled, the radius of the two speeds remains the same

5 0

3 years ago