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bija089 [108]
3 years ago
6

Since sinusoidal waves are cyclical, a particular phase difference between two waves is identical to that phase difference plus

a cycle. For example, if two waves have a phase difference of π4, the interference effects would be the same as if the two waves had a phase difference of π4+2π. The complete criterion for constructive interference between two waves is therefore written as follows: phase difference=0+2πnfor any integer n Write the full criterion for destructive interference between two waves
Physics
1 answer:
r-ruslan [8.4K]3 years ago
4 0

Answer:

nπ + π/2 for any integer n

Explanation:

Since destructive interference occurs every odd multiple of half wavelength, that is π/2, 3π/2, 5π/2 where the interference is half wavelength and in general, (n + 1/2)π where n is an integer. So, nπ + π/2 for any integer n

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Your friend has decided to make some money during the next State Fair by inventing a game of skill. In the game as she has devel
coldgirl [10]

Answer: from the vertical, one should aim  86.6°

Explanation:

height of the center of object = 7.0 m - 0.05 m = 6.95 m

now let the bullet hits centre at point A height x meters from the ground

also let t be the time taken for the bullet to hit the object

so distance travelled by the target will be

d = h - x = 6.95 - x

now using the equation of motion

d = 1/2gt²

so 1/2gt² = 6.95 - x

x = 6.95 - 1/2gt² .........let this be equ 1

let angle of fire be ∅

so v(cos∅) × t = 100

our velocity v is 1200 ft/sec = 365.76 m/s

365.76(cos∅) × t = 100 ........equ 2

also vertical position of the bullet after t is

y = y₀ + c(sin∅)t - 1/2gt²

y = 1 + 365.76(sin∅)t - 1/2gt² ----- equ 3

After time t. the vertical position x and y are same, else the bullet wouldn't have strike target at centre, so;

x = y

we substitute

equ 1 = equ 3

6.95 - 1/2gt² = 1 + 365.76(sin∅)t - 1/2gt²

6.95 - 1 = 365.76(sin∅)t - 1/2gt² +  1/2gt²

5.95 = 365.76(sin∅)t

t = 5.96 / 365.76(sin∅)

now input the above equ  into equ 2

365.76(cos∅) ×  5.96/365.76(sin∅) = 100

5.95(cos∅)/sin∅ = 100

tan∅ = 5.95/100 = 0.0595

∅ =  3.40°

therefore from the vertical, one should aim (90° - 3.40°) = 86.6°

4 0
3 years ago
Dos cargas iguales de 10^-7C estan separadas por una distancia de 2m. calcule la fuerza con que se repelen
m_a_m_a [10]

Answer:

F=2.25\times 10^{-5}\ N

Explanation:

According to question,

Charge 1 and charge 2 are 10^{-7}\ C

The distance between charges is 2 m

We need to find the force with which two positive charges repel. It is called electrostatic force of repulsion. It can be given by :

F=\dfrac{kq^2}{r^2}\\\\F=\dfrac{9\times 10^9\times (10^{-7})^2}{2^2}\\\\F=2.25\times 10^{-5}\ N

So, the electric force of repulsion is 2.25\times 10^{-5}\ N.

8 0
3 years ago
The flow of electrons through a circuit is measured in which of the following units? A. electrical pressure B. amperes C. volts
damaskus [11]

The total quantity of electrons that have flowed through a circuit is a
quantity of charge, measured in Coulombs, or in Ampere-seconds.

The <em><u>rate</u></em> of flow of electrons, or more accurately the rate of flow of
the charge on them, is electrical current.  Its unit is the Ampere. 
1 Ampere is 1 Coulomb of charge per second.


8 0
3 years ago
2) How could a motorcycle (a vehicle with less mass) make a van (a
jeyben [28]

it depends upon what state they are in like in motion or res

8 0
2 years ago
Calculate the electric field at the center of a square
pantera1 [17]

Answer:

E_y=1175510.2\ N.C^{-1}

The Magnitude of electric field is in the upward direction as shown directly towards the charge q_1.

Explanation:

Given:

  • side of a square, a=52.5\ cm
  • charge on one corner of the square, q_1=+45\times 10^{-6}\ C
  • charge on the remaining 3 corners of the square,q_2=q_3=q_4=-27\times 10^{-6}\ C

<u>Distance of the center from each corners</u>=\frac{1}{2} \times diagonals

diagonal=\sqrt{52.5^2+52.5^2}

diagonal=74.25\cm=0.7425\ m

∴Distance of center from corners, b=0.3712\ m

Now, electric field due to charges is given as:

E=\frac{1}{4\pi\epsilon_0}\times \frac{q}{b^2}

<u>For charge q_1 we have the field lines emerging out of the charge since it is positively charged:</u>

E_1=9\times 10^9\times \frac{45\times 10^{-6}}{0.3712^2}

  • E_1=2938775.5\ N.C^{-1}

<u>Force by each of the charges at the remaining corners:</u>

E_2=E_3=E_4=9\times 10^9\times \frac{27\times 10^{-6}}{0.3712^2}

  • E_2=E_3=E_4=1763265.3\ N.C^{-1}

<u> Now, net electric field in the vertical direction:</u>

E_y=E_1-E_4

E_y=1175510.2\ N.C^{-1}

<u>Now, net electric field in the horizontal direction:</u>

E_y=E_2-E_3

E_y=0\ N.C^{-1}

So the Magnitude of electric field is in the upward direction as shown directly towards the charge q_1.

8 0
3 years ago
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