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2 years ago

13

Give two examples of spatial interference which can be easily observed.

1 answer:

kifflom [539]2 years ago

3 0

Answer : space and kite

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Why are the element from period 2 grouped together

valentinak56 [21]

All of the elements in a period have the same number of atomic orbitals. For example, every element in the top row (the first period) has one orbital for its electrons. All of the elements in the second row (the second period) have two orbitals for their electrons. As you move down the table, every row adds an orbital.

4 0

3 years ago

Light of wavelength 597 nm falls on a double slit, and the first bright fringe of the interference pattern is seen at an angle o

Kazeer [188]

Answer:

2.2 µm

Explanation:

For constructive interference, the expression is:

$d\times sin\theta=m\times \lambda$

Where, m = 1, 2, .....

d is the distance between the slits.

Given wavelength = 597 nm

Angle, $\theta$ = 15.8°

First bright fringe means , m = 1

So,

$d\times sin\ 15.8^0=1\times \597\ nm$

$d\times 0.2723=1\times \597\ nm$

$d=2192.43481\ nm$

Also,

1 nm = 10⁻⁹ m

1 µm = 10⁻⁶ m

So,

1 nm = 10⁻³ nm

Thus,

<u>Distance between slits ≅ 2.2 µm</u>

8 0

3 years ago

Read 2 more answers

How do the most common forms of energy work?

DiKsa [7]

Answer:

activity 2

Explanation:

7 0

3 years ago

Charge: A piece of plastic has a net charge of +2.00 μC. How many more protons than electrons does this piece of plastic have? (

snow_lady [41]

Answer

given,

net charge = +2.00 μC

we know,

1 coulomb charge = 6.28 x 10¹⁸electrons

1 micro coulomb charge = 6.28 x 10¹⁸ x 10⁻⁶ electron

= 6.28 x 10¹² electrons

2.00 μC = 2 x 6.28 x 10¹² electrons

= 1.256 x 10¹³ electrons

since net charge is positive.

The number of protons should be 1.256 x 10¹³ more than electrons.

hence, +2.00 μC have 1.256 x 10¹³ more protons than electrons.

6 0

3 years ago

011 10.0 points

Sedbober [7]

<h2>The temperature of the air is 66.8° C</h2>

Explanation:

From the Newton's velocity of sound relationship , the velocity of sound is directly proportional to the square root of temperature .

In this case The velocity of sound = frequency x wavelength

= 798 x 0.48 = 383 m/sec

Suppose the temperature at this time = T K

Thus 383 ∝ $\sqrt{T}$ I

The velocity of sound is 329 m/s at 273 K ( given )

Thus 329 ∝ $\sqrt{273}$ II

Dividing I by II , we have

$\frac{383}{329}$ = $\sqrt{\frac{T}{273} }$

or $\frac{T}{273}$ = 1.25

and T = 339.8 K = 66.8° C

4 0

3 years ago