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3 years ago

What width slit will produce a central maximum that is 3.00 cm wide on a screen 1.80 m from the slit?

Physics

1 answer:

Digiron [165]3 years ago

3 0

The central maximum extends to the first minimum on either side.

So the first minimum occurs at 3cm / 2 = 1.5cm = 0.015m to the side from the center of the central maximum. 

Now using the formula,
λ = a(y/L)

Where a is the width of the slit, y = distance of first minimum from center, L = distance of screen from slit, and λ = wavelength of the light that strikes a single slit at normal incidence which is unknown here.

a = L x λ/y
a = 1.80 m x (λ / 0.015 m)
a = 120 λ

Use the value of λ for any light and you get the width of slit.

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You observe that a mass suspended by a spring takes 0.25 s to make a full oscillation. What is the frequency of this oscillation

Katarina [22]

Answer:

Frequency of oscillation, f = 4 Hz

time period, T = 0.25 s

Angular frequency, $\omega = 25.13 rad/s$

Given:

Time taken to make one oscillation, T = 0.25 s

Solution:

Frequency, f of oscillation is given as the reciprocal of time taken for one oscillation and is given by:

f = $\frac{1}{T}$

f = $\frac{1}{0.25}$

Frequency of oscillation, f = 4 Hz

The period of oscillation can be defined as the time taken by the suspended mass for completion of one oscillation.

Therefore, time period, T = 0.25 s

Angular frequency of oscillation is given by:

$\omega = 2\pi \times f$

$\omega = 2\pi \times 4$

$\omega = 25.13 rad/s$

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4 years ago

A forklift raises a 1,020 N crate 3.50 m up to a shelf. How much work is done by the forklift on the crate?

Anarel [89]

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8 0

3 years ago

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Calculate the binding energy per nucleon of the helium nucleus 52he. express your answer in millions of electron volts to four s

Vitek1552 [10]

The main formula is given by Eb/nucleon = Eb/ mass of nucleid
as for 52He, the mass is 5
so by applying Einstein's formula Eb=DmC², Eb=binding energy
52He-----------> 2 x 11p + 3 x10n is the equation bilan
so Dm=2 mp + (5-2)mn-mnucleus, mp=mass of proton=1.67 10^-27 kg
mn=mass of neutron=1.67 10^-27 kg
 m nucleus= 5
Dm= 2x1.67 10^-27 kg+ 3x1.67 10^-27 kg-5= - 4.9 J
Eb= - 4.9 J x c²= -4.9 x 9 .10^16= - 45 10^16 J
so the answer is Eb /nucleon = Eb/5= -9.10^16 J, but 1eV=1.6 . 10^-19 J
so -9.10^16 J/ 1.6 10^-19= -5.625 10^35 eV

the final answer is Eb /nucleon = -5.625 x10^35 eV

8 0

3 years ago

A disk with radius R and uniform positive charge density s lies horizontally on a tabletop. A small plastic sphere with mass M a

Yanka [14]

Answer:

a. F = Qs/2ε₀[1 - z/√(z² + R²)] b. h = (1 - 2mgε₀/Qs)R/√[1 - (1 - 2mgε₀/Qs)²]

Explanation:

a. What is the magnitude of the net upward force on the sphere as a function of the height z above the disk?

The electric field due to a charged disk with surface charge density s and radius R at a distance z above the center of the disk is given by

E = s/2ε₀[1 - z/√(z² + R²)]

So, the net force on the small plastic sphere of mass M and charge Q is

F = QE

F = Qs/2ε₀[1 - z/√(z² + R²)]

b. At what height h does the sphere hover?

The sphere hovers at height z = h when the electric force equals the weight of the sphere.

So, F = mg

Qs/2ε₀[1 - z/√(z² + R²)] = mg

when z = h, we have

Qs/2ε₀[1 - h/√(h² + R²)] = mg

[1 - h/√(h² + R²)] = 2mgε₀/Qs

h/√(h² + R²) = 1 - 2mgε₀/Qs

squaring both sides, we have

[h/√(h² + R²)]² = (1 - 2mgε₀/Qs)²

h²/(h² + R²) = (1 - 2mgε₀/Qs)²

cross-multiplying, we have

h² = (1 - 2mgε₀/Qs)²(h² + R²)

expanding the bracket, we have

h² = (1 - 2mgε₀/Qs)²h² + (1 - 2mgε₀/Qs)²R²

collecting like terms, we have

h² - (1 - 2mgε₀/Qs)²h² = (1 - 2mgε₀/Qs)²R²

Factorizing, we have

[1 - (1 - 2mgε₀/Qs)²]h² = (1 - 2mgε₀/Qs)²R²

So, h² = (1 - 2mgε₀/Qs)²R²/[1 - (1 - 2mgε₀/Qs)²]

taking square-root of both sides, we have

√h² = √[(1 - 2mgε₀/Qs)²R²/[1 - (1 - 2mgε₀/Qs)²]]

h = (1 - 2mgε₀/Qs)R/√[1 - (1 - 2mgε₀/Qs)²]

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3 years ago

Quartz has an index of refraction of 1.46. Diamond has in index of refraction of 2.42. In which material does a light ray enteri

Ksivusya [100]

Diamond slowdown light more than Quartz , because diamonds have a greater index of refraction. Light will bend when its move from one medium to another. The Index of Refraction of Material is found by comparing the speed of light in their respective mediums.

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3 years ago

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