All categories

3 years ago

For two traveling waves, if the crest of one wave coincides with a trough of another, what type of interference occurs?

1 answer:

3 0

When crest of one wave interferes with the trough of other wave, the amplitude of the resultant wave formed is less. Hence the type of interference is destructive interference.

You might be interested in

A car tire rotates with an average angular speed of 32 rad/s. In what time interval will the tire rotate 3.5 times? Answer in un

stiks02 [169]

Answer:

$\Delta t = 0.687\ s$

Explanation:

given,

Angular speed of the tire = 32 rad/s

Displacement of the wheel = 3.5 rev

Δ θ = 3.5 x 2 π

= 7 π rad

now,

Time interval of the car to rotate 7π rad

using equation

$\omega_{avg} =\dfrac{\Delta \theta}{\Delta t}$

$\Delta t=\dfrac{7\pi}{32}$

$\Delta t = 0.687\ s$

Time taken to rotate 3.5 times is equal to 0.687 s.

3 0

2 years ago

A whistle you use to call your hunting dog has a frequency of 21 kHz, but your dog is ignoring it. You suspect the whistle may n

uranmaximum [27]

The Doppler effect is the right concept to solve this problem. The Doppler effect is understood as the change in apparent frequency of a wave produced by the relative movement of the source with respect to its observer. Mathematically it can be described as,

$f = (1-\frac{v_0}{v})f_0$

Here,

$f_0$ = Frequency of the sound from the Whistle

f = Frequency of sound heard

v = Speed of the sound in the Air

Replacing we have that

$1- \frac{v_0}{343} = \frac{20kHz}{21kHz}$

$\frac{v_0}{343} = 1-\frac{20}{21}$

$\frac{v_0}{343} = \frac{1}{21}$

$v_0 = \frac{1}{21}(343)$

$v_0 = 16.33m/s$

Therefore the minimum speed to know if the whistle is working is 16.33m/s

3 0

3 years ago

A model airplane of mass 0.6 kg is attached to a horizontal string and flies in a horizontal circle of radius 6 m, making 1.6 re

nikitadnepr [17]

Answer:

$speed= 12.15\frac{m}{s}$

Explanation:

In this question we have given

mass of airplane=.6 Kg

Radius of horizontal circle,r=6m

Time taken to complete 1.6 revolution=5s

Therefore time taken to complete 1 revolution, t= $\frac{5}{1.6}$

t=3.1 s

We have to find the speed of airplane

We will first find the distance covered by airplane in tracing one circle which is equal to circumference of circle

We know that

Circumference,d = $2 \pi \times r$

or $d=2\times 3.14 \times 6$

$d=37.68 m$

We know that speed= $\frac{d}{t}$ ............(1)

Put value of d and t in equation 1

speed= $\frac{37.68}{3.1}$

$speed= 12.15\frac{m}{s}$

5 0

3 years ago

Its a motion phisic question but im a little confused. Help me fast please.

Alenkinab [10]

6: Short way: it cannot be 2.5, 3, or 5 because up to 5 seconds it only has positive velocity so it must be moving forwards.

Long Way: Velocity is in m / s, multiply that by time (s) to get m or displacement. From 0->5 you have a triangle under the curve, (1/2)(5)(20) = 50 meters displaced positive, you need to then look when velocity is under the curve and use a similar equation to solve for the area but make the answer negative. Find the point where it equals -50 and that is where it will have returned.

Answer to 6: B

7. I cannot see the problem enough to answer this. Just know if the line is above 0 velocity is positive so it is moving the direction it started, when it goes below 0 velocity is negative so it is moving opposite direction it started.

8. Accelration is change in velocity. Whatever the slope of the velocity graph is acceleration. At t=8 the slope is 0 because it is not going up or down.

Answer to 8: A

3 0

3 years ago

The objective lens and the eyepiece of a telescope are spaced 85 cm apart. If the eyepiece is123 D what is the total magnificati

dedylja [7]

Answer:

The magnification would be "103.55". A further explanation is given below.

Explanation:

The given values are:

Distance between lens and eyepiece,

L = 85 cm

Eyepiece is,