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Leya [2.2K]
3 years ago
12

Narrow, bright fringes are observed on a screen behind a diffraction grating. The entire experiment is then immersed in water. D

o the fringes on the screen get closer together, get farther apart, remain the same, or disappear? Explain.
Physics
1 answer:
Aleks04 [339]3 years ago
8 0

Answer:

 n (a sin θ) =  m λ₀

n> 1, therefore the fringes move away from each other

Explanation:

The diffraction experiment the constructive interference fringes is described by

          a sin θ = m λ₀

in this equation it is assumed that the experiment emptied the air n = 1

When the same experiment is performed in water, the wavelength changes

           λₙ = λ₀ / n

execution for constructive interference

            a sin θ = m λₙ

we substitute

           a sin θ = m λ / n

           n (a sin θ) =  m λ₀

the refractive index of water is n = 1.33, so for the same wavelength the separation of the spectrum is multiplied by n> 1, therefore the fringes move away from each other

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ankoles [38]

Answer:

It will take 8.80 sec to fall from the building

Explanation:

We have given height pf the state building h = 380 m

Initial velocity will be 0 m /sec

So u = 0 m/sec

Acceleration due to gravity g=9.8m/sec^2

We have to find the fall time

According to second equation of motion h=ut+\frac{1}{2}gt^2

So 380=0\times t+\frac{1}{2}\times 9.8\times t^2

t^2=77.55

t = 8.80 sec

3 0
3 years ago
If two children, with masses of 16 kg and 25 kg, sit in seats opposite one another in a rotating merry-go-round with an arm leng
Jlenok [28]

Answer:

92.25 kgm^2

Explanation:

Assume both children bodies are point particles. The total moment of inertia about the rotation axis of 2 points particles of mass 16 kg and 25 kg at 1.5 m arm length is

\sum I = m_1r_1^2 + m_2r_2^2

where m_1 = 16 kg, m_2 = 25 kg are the masses of 2 children r_1 = R-2 = 1.5m are their distance to the center of rotation

\sum I = 16*1.5^2 + 25*1.5^2 = 1.5^2(16 + 25) = 92.25 kgm^2

4 0
3 years ago
30. Light has been described as a ___ and a ___
andrezito [222]

Answer:

Explanation: light has been described as a particle and a wave

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3 years ago
During a snowball fight two balls with masses of 0.4 kg and 0.6 kg, respectively, are thrown in such a manner that they meet hea
valkas [14]
Considering conservation of momentum;
m1v1 + m2v2 = m3v3

In which,
m1 = mass of snowball 1 = 0.4 kg
v1 = velocity of snowball 1 = 15 m/s
m2 = mass of snowball 2 = 0.6 kg
v2 = velocity of snow ball 2 = 15 m/s
m3 = combined mass = 1 kg
v3 = velocity after comination
Therefore;
0.4*15 + 0.6*15 = 1*v3
v3 = 6+9 = 15 m/s
KE = 1/2mv^2

Then,
KE1 = 1/2*0.4*15^2 = 45 J
KE2 = 1/2*0.6*15^2 = 67.5 J
KE3 = 1/2*1*15^2 = 112.5 J

Therefore, KE3 (kinetic energy after collision) = K1+K2 {kinetic energy before collision). And thus it is 100%.
3 0
3 years ago
A potential difference of 1.20 V will be applied to a 33.0 m length of 18-gauge copper wire (diameter = 0.0400 in.). Calculate (
Nostrana [21]

Answer:A) Current = 1.739A, B)current density, J = 2.147x10^6 A/m2

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)rate of thermal energy, P  =2.086W

Explanation:

Resistance  = R =   ρL/A

But the cross-section area of the wire. is given as

Diameter / 2 = 0.04/2 =0.02in to m = 0.02 / 39.37= 0.000508

A = πr^2 = π x  0.000508^2 = 8.10 x 10^-7

since resistivity of copper,ρ= 17x10-9 ohm.m

so resistance is   R =   ρL/A

17x10-9  x 33 / 8.1x10-7

= 0.69 ohm.

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B)current density, J = Current /Area

= 1.739/8.1x10-7

= 2.147x10^6 A/m2

c)magnitude of electric field , E =  Current density x resistivity =J ρ

E = 2.147 x 10^6  x 17  x 10^-9

E = 0.036 N/C

D)rate of thermal energy, P  = I² R =1.739² X 0.69

=2.086W

6 0
3 years ago
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