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2 years ago

5

Light of wavelength 480 nm illuminates a double slit, and the interference pattern is observed on a screen. At the position of t

he m = 4 bright fringe, how much farther is it to the more distant slit than to the nearer slit?

1 answer:

vova2212 [387]2 years ago

4 0

Answer:

BC =980nm

Explanation:

Given wavelength =490nm

m=2

Required to find (OB - OC) = BC (more distant slit - nearest slit ) = dsin theta

We have dsin theta = m*wavelength

or BC = m* wavelength = 2*490nm

= 980nm = 98.000A

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3. A 2kg wooden block whose initial speed is 3 m/s slides on a smooth floor for 2 meters before it comes to a

serious [3.7K]

Answer:

Calculating Coefficient of friction is 0.229.

Force is 4.5 N that keep the block moving at a constant speed.

Explanation:

We know that speed expression is as $\mathrm{V}^{2}=\mathrm{V}_{\mathrm{i}}^{2}+2 . \mathrm{a} . \Delta \mathrm{s}$ .

Where, ${V}_{i}$ is initial speed, V is final speed, ∆s displacement and a acceleration.

Given that,

${V}_{i}$ =3 m/s, V = 0 m/s, and ∆s = 2 m

Substitute the values in the above formula,

$0=3^{2}-2 \times 2 \times a$

0 = 9 - 4a

4a = 9

$a=2.25 \mathrm{m} / \mathrm{s}^{2}$

$a=2.25 \mathrm{m} / \mathrm{s}^{2}$ is the acceleration.

Calculating Coefficient of friction:

$\mathrm{F}=\mathrm{m} \times \mathrm{a}$

$\mathrm{F}=\mu \times \mathrm{m} \times \mathrm{g}$

Compare the above equation

$\mu \times m \times g=m \times a$

Cancel "m" common term in both L.H.S and R.H.S

$\text { Equation becomes, } \mu \times g=a$

$\text { Coefficient of friction } \mu=\frac{a}{g}$

$\mathrm{g} \text { on earth surface }=9.8 \mathrm{m} / \mathrm{s}^{2}$

$\mu=\frac{2.25}{9.8}$

$\mu=0.229$

Hence coefficient of friction is 0.229.

calculating force:

$\text { We know that } \mathrm{F}=\mathrm{m} \times \mathrm{a}$

$\mathrm{F}=2 \times 2.25 \quad(\mathrm{m}=2 \mathrm{kg} \text { given })$

F = 4.5 N

Therefore, the force would be <u>4.5 N</u> to keep the block moving at a constant speed across the floor.

7 0

2 years ago

What is the maximum magnitude of charge that can be placed on each plate if the electric field in the region between the plates

34kurt

The given question is incomplete. The complete question is as follows.

A parallel-plate capacitor has capacitance $C_{0}$ = 8.50 pF when there is air between the plates. The separation between the plates is 1.00 mm.

What is the maximum magnitude of charge that can be placed on each plate if the electric field in the region between the plates is not to exceed $3.00 \times 10^{4}$ V/m?

Explanation:

It is known that relation between electric field and the voltage is as follows.

V = Ed

Now,

Q = CV

or, Q = $C \times Ed$

Therefore, substitute the values into the above formula as follows.

Q = $C \times Ed$

= $8.50 pF \times (\frac{10^{-12} F}{1 pF})(3 \times 10^{4} m/s)(1 mm)(\frac{10^{-3} m}{1 mm})$

= $2.55 \times 10^{-10} C$

Hence, we can conclude that the maximum magnitude of charge that can be placed on each given plate is $2.55 \times 10^{-10} C$ .

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3 years ago

List 5 possible effects of not adhering to standards of measurement

murzikaleks [220]

Answer:

factual evidence of customer-service levels.

better understanding of cross-functional performance.

enhanced alignment of operations with strategy.

evidence-based determination of process improvement priorities.

detection of performance trends.

better understanding of the capability range of a process.

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3 years ago

A conducting loop has an area of 0.065 m2 and is positioned such that a uniform magnetic field is perpendicular to the plane of

aev [14]

Answer:

initial magnetic field 1.306 T

Explanation:

We have given area of the conducting loop $A=0.065m^2$

Emf induced = 1.2 volt

Initial magnetic field B = 0.3 T

Time dt = 0.087 sec

We know that induced emf is given by $e=\frac{d\Phi }{dt}=-A\frac{db}{dt}$

$1.2=0.065\times \frac{db}{0.087}$

$db=1.606T$

So initial magnetic field = 1.606-0.3= 1.306 T

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