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3 years ago

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A soap bubble appears red (λ = 633nm) at the point on its front surface nearest to the viewer. Assuming n = 1.35, what is the sm

allest film thickness the film could have?

1 answer:

alexira [117]3 years ago

7 0

Answer:

The smallest film thickness is 117 nm.

Explanation:

Light interference on thin films can be constructive or destructive. Constructive interference is dependent on the film thickness and the refractive index of the medium.

For the first interference (surface nearest to viewer), the minimum thickness can be expressed as:

$2t_{min} = \frac{wavelenth}{2n}$

where n is the refractive index of the bubble film.

Therefore,

$2t_{min} = \frac{633x10^{-9} }{(2)(1.35)}$

$2t_{min} =2.344x10^{-7}$

∴ $t_{min} =\frac{2.344x10^{-7} }{2}$

$t_{min} = 1.17x10^{-7} m = 117 nm.$

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A spring stretches 3.5 cm when a 9 g object is hung from it. The object is replaced with a block of mass 26 g which oscillates i

evablogger [386]

Answer:

The period of motion of new mass T = 0.637 sec

Explanation:

Given data

Mass of object (m) = 9 gm = 0.009 kg

Δx = 3.5 cm = 0.035 m

We know that spring force is given by

F = m g

F = 0.009 × 9.81 = 0.08829 N

Spring constant

$k = \frac{F}{x}$

$k = \frac{0.08829}{0.035}$

k = 2.522 $\frac{N}{m}$

New mass $(m_1)$ = 26 gm = 0.026 kg

Now the period of motion is given by

$T = 2 \pi \sqrt{\frac{m}{k} }$

$T = 2 \pi \sqrt{\frac{0.026}{2.522} }$

T = 0.637 sec

This is the period of motion of new mass.

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3 years ago

If T, = 40 N, find T, and the mass of the weight (W).

seraphim [82]

Answer:4kg

Explanation:

acceleration due to gravity(g)=10m/s^2

Weight(w)=40N

Weight=mass x g

40=mass x 10

Divide both sides by 10

Mass =40/10

Mass=4kg

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3 years ago

If a planet has the same mass as the earth, but has twice the radius, how does the surface gravity, g, compare to g on the surfa

shepuryov [24]

Answer:

The surface gravity g of the planet is 1/4 of the surface gravity on earth.

Explanation:

Surface gravity is given by the following formula:

$g=G\frac{m}{r^{2}}$

So the gravity of both the earth and the planet is written in terms of their own radius, so we get:

$g_{E}=G\frac{m}{r_{E}^{2}}$

$g_{P}=G\frac{m}{r_{P}^{2}}$

The problem tells us the radius of the planet is twice that of the radius on earth, so:

$r_{P}=2r_{E}$

If we substituted that into the gravity of the planet equation we would end up with the following formula:

$g_{P}=G\frac{m}{(2r_{E})^{2}}$

Which yields:

$g_{P}=G\frac{m}{4r_{E}^{2}}$

So we can now compare the two gravities:

$\frac{g_{P}}{g_{E}}=\frac{G\frac{m}{4r_{E}^{2}}}{G\frac{m}{r_{E}^{2}}}$

When simplifying the ratio we end up with:

$\frac{g_{P}}{g_{E}}=\frac{1}{4}$

So the gravity acceleration on the surface of the planet is 1/4 of that on the surface of Earth.

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3 years ago

A cylindrical can holds 1 liters (1000 cm^3) of liquid. How should the height and radius be chosen to minimize the amount of mat

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I was struck by this many years ago while in graduate school. ... People with mental illness tend to die young , but so do their families. ... there is no medical assessment that can diagnose the absence of illness.

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4 years ago

The quantity of charge q (in coulombs) that has passed through a surface of area 2.00 cm2 varies with time

Makovka662 [10]

Explanation:

We have,

Surface area, $A=2\ cm^2=0.0002\ m^2$

The current varies wrt time t as :

$q(t) = 4t^3 + 5t + 6$

(a) At t = 2 seconds, electrical charge is given by :

$q(t) = 4t^3 + 5t + 6\\\\q(2) = 4(2)^3 + 5(2) + 6\\\\q=48\ C$

(b) Current is given by :

$I=\dfrac{dq}{dt}\\\\I=\dfrac{d(4t^3 + 5t + 6)}{dt}\\\\I=12t^2+5$

Instantaneous current at t = 1 s is,

$I=12(1)^2+5=17\ A$

(c) Current is, $I=12t^2+5$

Current density is given by electric current per unit area.

$J=\dfrac{I}{A}\\\\J=\dfrac{(12t^2+5)}{0.0002}\\\\J=5000(12t^2+5)\ A/m^2$

Therefore, it is the required explanation.

7 0

3 years ago