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1 year ago

Constructive interference of two coherent waves will occur if the path difference is:___.

Physics

1 answer:

Olegator [25]1 year ago

8 0

Constructive interference of two coherent waves will occur if the path difference is λ/2.

<h3>Constructive interference:</h3>

When two waves are in phase and their maxima add, a process known as constructive interference occurs where the combined amplitude of the two waves equals the sum of their individual amplitudes.

The resultant wave is created by adding the amplitudes of two waves that are in phase and traveling in the same direction. The waves in this instance are said to have experienced beneficial interference. The upward displacement of the medium is higher than the displacement of the two interfering pulses because upward displacement occurs when the waves experience constructive interference. When the phase difference between the waves is an even multiple of (180°), constructive interference happens.

Learn more about constructive interference here:

brainly.com/question/17329186

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What is the relationship between area, force, and pressure? what is the relationship between area, force, and pressure? pressure

White raven [17]

Mathematically, relation between force, area and pressure is given by...
Pressure = force / area

hence, pressure is directly proportional to force but inversely proportional to area.

7 0

2 years ago

A parallel-plate capacitor with plates of area 360 cm2 is charged to a potential difference V and is then disconnected from the

Softa [21]

Answer:

$Q=3.9825\times 10^{-9} C$

Explanation:

We are given that a parallel- plate capacitor is charged to a potential difference V and then disconnected from the voltage source.

1 m =100 cm

Surface area =S= $\frac{360}{10000}=0.036 m^2$

$\Delta d=0.8 cm=0.008 m$

$\Delta V=100 V$

We have to find the charge Q on the positive plates of the capacitor.

V=Initial voltage between plates

d=Initial distance between plates

Initial Capacitance of capacitor

$C=\frac{\epsilon_0 S}{d}$

Capacitance of capacitor after moving plates

$C_1=\frac{\epsilon_0 S}{(d+\Delta d)}$

$V=\frac{Q}{C}$

Potential difference between plates after moving

$V=\frac{Q}{C_1}$

$V+\Delta V=\frac{Q}{C_1}$

$\frac{Qd}{\epsilon_0S}+100=\frac{Q(d+\Delta d)}{\epsilon_0S}$

$\frac{Q(d+\Delta d)}{\epsilon_0 S}-\frac{Qd}{\epsilon_0S}=100$

$\frac{Q\Delta d}{\epsilon_0 S}=100$

$\epsilon_0=8.85\times 10^{-12}$

$Q=\frac{100\times 8.85\times 10^{-12}\times 0.036}{0.008}$

$Q=3.9825\times 10^{-9} C$

Hence, the charge on positive plate of capacitor= $Q=3.9825\times 10^{-9} C$

6 0

3 years ago

Suppose you first walk 12.0 m in a direction 20? west of north and then 20.0 m in a direction 40.0? south of west. how far are y

Gnesinka [82]

The representation of this problem is shown in Figure 1. So our goal is to find the vector $\overrightarrow{R}$ . From the figure we know that:

$\left | \overrightarrow{A} \right |=12m \\ \\ \left | \overrightarrow{B} \right |=20m \\ \\ \theta_{A}=20^{\circ} \\ \\ \theta_{B}=40^{\circ}$

From geometry, we know that:

$\overrightarrow{R}=\overrightarrow{A}+\overrightarrow{B}$

Then using vector decomposition into components:

$For \ A: \\ \\ A_x=-\left | \overrightarrow{A} \right |sin\theta_A=-12sin(20^{\circ})=-4.10 \\ \\ A_y=\left | \overrightarrow{A} \right |cos\theta_A=12cos(20^{\circ})=11.27 \\ \\ \\ For \ B: \\ \\ B_x=-\left | \overrightarrow{B} \right |cos\theta_B=-20cos(40^{\circ})=-15.32 \\ \\ B_y=-\left | \overrightarrow{B} \right |sin\theta_B=-20sin(40^{\circ})=-12.85$

Therefore:

$R_x=A_x+B_x=-4.10-15.32=-19.42m \\ \\ R_y=A_y+B_y=11.27-12.85=-1.58m$

So if you want to find out <span>how far are you from your starting point you need to know the magnitude of the vector $\overrightarrow{R}$ , that is:
</span>
$\left | \overrightarrow{R} \right |=
\sqrt{R_x^2+R_y^2}=\sqrt{(-19.42)^2+(-1.58)^2}=\boxed{19.48m}$

Finally, let's find the <span>compass direction of a line connecting your starting point to your final position. What we are looking for here is an angle that is shown in Figure 2 which is an angle defined with respect to the positive x-axis. Therefore:

</span> $\theta_R=180^{\circ}+tan^{-1}(\frac{\left | R_y \right |}{\left | R_x \right |}) \\ \\ \theta_R=180^{\circ}+tan^{-1}(\frac{1.58}{19.42}) \\ \\ \theta_R=180^{\circ}+4.65^{\circ}=185.85^{\circ}$

6 0

3 years ago

If the light wave has a wavelength of 10m what would be its velocity

s2008m [1.1K]

If this case could ever happen, the speed would follow from this formula:

$v = f \cdot \lambda$

with f the frequency and lambda the wavelength. We are give a wavelength of 10m. The frequencies of the visible light can range between 400 to about 790 Terahertz, so let us pick a middle point of 600 THz ("green-ish") as a "representative."

$v = 600THz\cdot 10m = 6\cdot 10^{14} \frac{1}{s}\cdot 10 m = 6\cdot10^{15}\frac{m}{s}$

The speed of such a wave would have to be 6e+15 m/s (which would be 7 orders of magnitude higher than the universal speed of light constant)

7 0

3 years ago

Read 2 more answers

The 1350-kg car has a velocity of 24 km/h up the 8-percent grade when the driver applies more power for 18 s to bring the car up

Brrunno [24]

Answer:

Explanation:

We shall apply concept of impulse to solve the problem .

Impulse = force x time

impulse = change in momentum

force x time = change in momentum

initial speed u = 24 km/h = 6.67 m /s

final speed v = 65 km/h = 18.05 m /s

change in momentum = m v - mu

= m ( v-u )

= 1350 ( 18.05 - 6.67 )

= 15363 kg m/s

F x 18 = 15363

F = 853.5 N .

4 0

3 years ago