A 4 kg box on the ground has a total of three horizontal forces on it as shown. The acceleration of the box is...

A thin film soap bubble (n=1.35) is floating in air. If the thickness of the bubble wall is 300nm, which of the following wavele

iren [92.7K]

Answer:

540 nm

Explanation:

According to the question,

The refractive index of the soap bubble, $n=1.35$ .

The thickness of the soap bubble wall is, $t=300 nm$ .

Now, for constructive interference of soap bubble.

$2nt=(m+\frac{1}{2})\lambda$ .

Now for first order m=1.

Therfore,

$\lambda =\frac{4}{3} tn$

Substitute all the variables in the above equation.

$\lambda =\frac{4}{3} (1.35)(300 nm)$ .

Therefore,

$\lambda =540 nm$ .

Therefore the visible light wavelength which is strongly reflected is 540 nm.

6 0

4 years ago

Where is the near point of an eye for which a contact lens with a power of +2.55 diopters is prescribed?Where is the far point o

tankabanditka [31]

Answer:

(a). The eye's near point is 68.98 cm from the eye.

(b). The eye's far point is 33.33 cm from the eye.

Explanation:

Given that,

Power = 2.55 D

Object distance = 25 cm for near point

Object distance = ∞ for far point

Suppose where is the far point of an eye for which a contact lens with a power of -3.00 D is prescribed for distant vision?

(a) We need to calculate the focal length

Using formula of power

$f =\dfrac{1}{P}$

Put the value into the formula

$f=\dfrac{100}{2.55}$

$f=39.21\ cm$

We need to calculate the image distance

Using formula of lens

$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$

Put the value into the formula

$\dfrac{1}{39.21}=\dfrac{1}{v}-\dfrac{1}{-25}$

$\dfrac{1}{39.21}-\dfrac{1}{25}=\dfrac{1}{v}$

$\dfrac{1}{v}=-\dfrac{1421}{98025}$

$v=-68.98\ cm$

The eye's near point is 68.98 cm from the eye.

(b). We need to calculate the focal length

Using formula of power

$f =\dfrac{1}{P}$

Put the value into the formula

$f=-\dfrac{100}{3.00}$

$f=-33.33\ cm$

We need to calculate the image distance

Using formula of lens

$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$

Put the value into the formula

$\dfrac{1}{-33.33}=\dfrac{1}{v}-\dfrac{1}{\infty}$

$-\dfrac{1}{33.33}+\dfrac{1}{\infty}=\dfrac{1}{v}$

$\dfrac{1}{v}=-\dfrac{1}{33.33}$

$v=-33.33\ cm$

The eye's far point is 33.33 cm from the eye.

Hence, (a). The eye's near point is 68.98 cm from the eye.

(b). The eye's far point is 33.33 cm from the eye.

5 0

4 years ago

If a ball is thrown vertically upward with a velocity of 160 ft/s, then its height after t seconds is s = 160t − 16t^2.

White raven [17]

Answer:

400ft. 32ft/s -32ft/s

Explanation:

In reality the gravitational acceleration is 9.81 so the quadratic coefficient of the function should be 9.81/2

Anyway for the sake of assumtion let us takes=160t-16t^2

ds/dt=160-32t=0

t=160/32= 5 seconds.

s=160*160/32-16*(160/32)^2= 400 mts

s=384 mts

160t-16t^2=384

i.e

16t^2-160t+384=0

t^2-10t+24=0

(t-6)(t-4)=0

t=[4,6]

we have to take t=4 because it is all the up i.e <5

velocity =v=ds/dt=160-32t

v=160-32*4=32 ft/sec still going up

for all the way down take t=6 whuch is >5

v=160-6*32=-32 ft/sec (falling down!!!)

6 0

3 years ago

Marcy pulls a backpack on wheels down the 100 m hall. The 60N force is applied at an angle of 30° above the horizontal. How much

shutvik [7]

Work= Fcos∆×S

W=60N×cos 30⁰×100

W=60×0.866×100=5196.1J

6 0

3 years ago

The Milky Way Galaxy is (a) another name for our solar system; (b) a small group of stars visible in our night sky; (c) a collec

gtnhenbr [62]

Answer:

C

Explanation:

please give brainliest

3 0

3 years ago