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3 years ago

How many cycles/vibrations are there in 1 wavelength?

1 answer:

3 0

Te question is missing some information. However, I'll try to help you out with the steps.

We have the relation between velocity, wavelength and frequency expressed as follows:
c = λμ
where:
c is the speed of the wave
λ is the wavelength of the wave
μ is the frequency of the wave

Substituting in the above equation, we can get the frequency of the wave which is the number of vibrations per units of time

We can then get the periodic time which is 1/frequency

Hope this helps :)

You might be interested in

A package of mass m is released from rest at a warehouse loading dock and slides down a 3.0-m-high frictionless chute to a waiti

LuckyWell [14K]

Answer:

The speed of the package of mass m right before the collision $= 7.668\ ms^-1$

Their common speed after the collision $= 2.56\ ms^-1$

Height achieved by the package of mass m when it rebounds $= 0.33\ m$

Explanation:

Have a look to the diagrams attached below.

a.To find the speed of the package of mass m right before collision we have to use law of conservation of energy.

$K_{initial} + U_{initial} = K_{final}+U_{final}$

where $K$ is Kinetic energy and $U$ is Potential energy.

$K= \frac{mv^2}{2}$ and $U= mgh$

Considering the fact $K_{initial} = 0\ and U_{final} =0$ we will plug out he values of the given terms.

So $V_{1}{(initial)} =\sqrt{2gh} = \sqrt{2\times9.8\times3} = 7.668\ ms^-1$

Keypoints:

Sum of energies and momentum are conserved in all collisions.
Sum of KE and PE is also known as Mechanical energy.
Only KE is conserved for elastic collision.
for elastic collison we have $e=1$ that is co-efficient of restitution.

<u>KE = Kinetic Energy and PE = Potential Energy</u>

b.Now when the package stick together there momentum is conserved.

Using law of conservation of momentum.

$m_1V_1(i) = (m_1+m_2)V_f$ where $V_1{i} =7.668\ ms^-1$ .

Plugging the values we have

$m\times 7.668 = (3m)\times V_{f}$

Cancelling m from both sides and dividing 3 on both sides.

$V_f = 2.56\ ms^-1$

Law of conservation of energy will be followed over here.

c.Now the collision is perfectly elastic $e=1$

We have to find the value of $V_{f}$ for m mass.

As here $V_{f}=-2.56\ ms^-1$ we can use that if both are moving in right ward with $2.56$ then there is a $-2.56$ velocity when they have to move leftward.

The best option is to use the formulas given in third slide to calculate final velocity of object $1$ .

$V_{1f} = \frac{m_1-m_2}{m_1+m_2} \times V_{1i}= \frac{m-2m}{3m} \times7.668=\frac{-7.668}{3} = -2.56\ ms^-1$

Now using law of conservation of energy.

$K_{initial} + U_{initial} = K_{final}+U_{final}$

$\frac{m\times V(f1)^2}{2} + 0 = 0 +mgh$

$\frac{v(f1)^2}{2g} = h$

$h= \frac{(-2.56)^2}{9.8\times 3} =0.33\ m$

The linear momentum is conserved before and after this perfectly elastic collision.

So for part a we have the speed $=7.668\ ms^-1$ for part b we have their common speed $=2.56\ ms^-1$ and for part c we have the rebound height $=0.33\ m$ .

3 0

3 years ago

If the wavelength is changed to λ/2, does the central spot remain bright, does the central spot become dark, or do the fringes d

omeli [17]

Answer:The central spot becomes Dark

Explanation: it become dark because as the wavelength reduces,the velocity in the detector decreases, this time by 90degrees

4 0

3 years ago

How does an increase in temperature generally affect the rate of a reaction?

Zigmanuir [339]

Most reactions are exothemic. If the forward reaction of an equilibrium reaction is exothemic then the reverse reaction must be endothermic.

If a system in equilibrium is heated, it will move in exothermic direction to give out heat energy.

7 0

3 years ago

The pattern of weather in an area overtime is called

zzz [600]

The pattern of the temperature in a region day by day is called weather. The year by year temperature is a called climate.

6 0

3 years ago

At some airports there are speed ramps to help passengers get from one place to another. A speed ramp is a moving conveyor belt

Harrizon [31]

Answer:

It will take you 30.8 s to travel the 120 m of the ramp.

Explanation:

Hi there!

The equation for the position of an object moving in a straight line is:

x = x0 + v * t

Where:

x = position at time t

x0 = initial position

v = velocity

t = time

In this case, we will consider the start of the ramp as the origin of our reference system so that x0 = 0.

Now, let´s calculate the speed of the person walking on the ground:

x = v * t

120 m = v * 72 s

v = 120 m / 72 s

v = 1.7 m/s

If you walk on the ramp with that speed, your total speed will be your walking speed plus the speed of the ramp because both are in the same direction. Then, using the equation for the position:

x = v * t

In this case, v = speed of the ramp + walking speed

v = 2.2 m/s + 1.7 m/s = 3.9 m/s

120 m = 3.9 m/s * t

t = 120 m / 3.9 m/s = 30.8 s

It will take you 30.8 s to travel the 120 m

8 0

3 years ago