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Sonja [21]
3 years ago
14

The spreading of waves behind an aperture ismore for long wavelengths and less for short wavelengths.less for long wavelengths a

nd more for short wavelengths.the same for long and short wavelengths.not discussed in this chapter.
Physics
1 answer:
kompoz [17]3 years ago
6 0
<h2>Answer: The spreading of waves behind an aperture ismore for long wavelengths and less for short wavelengths</h2>

Here we are talking about Diffraction and, in fact, waves diffract the most when their wavelength is about the same size of the gap or aperture.

Diffraction happens when a wave (mechanical or electromagnetic wave) meets an obstacle or a slit .When this occurs, <u>the wave bends around the corners of the obstacle or passes through the opening of the slit that acts as an obstacle, forming multiple patterns with the shape of the aperture of the slit. </u>

<u />

Note that the principal condition for the occurrence of this phenomena is that the obstacle must be comparable in size (similar size) to the size of the wavelength.

In other words, when the gap (or slit) size is larger than the wavelength, the wave passes through the gap and does not spread out much on the other side, but when the gap size is equal to the wavelength, maximum diffraction occurs and the waves spread out greatly.

This means the smaller the slit or obstacle (diffracting object), the wider the resulting diffraction pattern, and the greater the obstacle, the narrower de resulting patter.

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If your friend drops a chocolate bar to you from a height of 5.0 m above your hands,
Sladkaya [172]

Answer:

<h3>1.01 s</h3>

Explanation:

Using the equation of motion S = ut+1/2gt² to solve the problem where;

u is the initial velocity of the chocolate = 0m/s

t is the time taken

g is the acceleration due to gravity = 9.81m/s²

S is the height of fall = 5.0m

Substituting the given parameter into the formula to get the time t we have;

5 = 0(t)+1/2(9.81)t²

5 = 4.905t²

t² = 5/4.905

t² = 1.019

t = √1.019

t = 1.009 secs

<em>Hence it will take 1.01 secs for me to catch the chocolate bar</em>

6 0
3 years ago
A uniform, thin, solid door has height 2.30 m, width 0.875 m, and mass 24.0 kg. (a) find its moment of inertia for rotation on i
vekshin1
A uniform thin solid door has height 2.20 m, width .870 m, and mass 23.0 kg. Find its moment of inertia for rotation on its hinges. Is any piece of data unnecessary? So far, I don't understand how to calculate moments of inertia for things like this at all. I can do a system of particles, but when it comes to any ridgid objects, such as this door or rods or cylinders, I don't get it. So basically I have no idea where to even start with this. 

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3 years ago
According to the pressure-flow hypothesis, which of the following statements is not true?
adelina 88 [10]
"The movement of water into a nutrient-rich region of the phloem decreases the pressure in that region" is the statement that is not true according <span>to the pressure-flow hypothesis. The correct option among all the options that are given in the question is the fourth option or the last option. I hope it helps you.</span>
7 0
3 years ago
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A charged particle enters a uniform magnetic field B with a velocity v at right angles to the field. It moves in a circle with p
alukav5142 [94]

A) d. 10T

When a charged particle moves at right angle to a uniform magnetic field, it experiences a force whose magnitude os given by

F=qvB

where q is the charge of the particle, v is the velocity, B is the strength of the magnetic field.

This force acts as a centripetal force, keeping the particle in a circular motion - so we can write

qvB = \frac{mv^2}{r}

which can be rewritten as

v=\frac{qB}{mr}

The velocity can be rewritten as the ratio between the lenght of the circumference and the period of revolution (T):

\frac{2\pi r}{T}=\frac{qB}{mr}

So, we get:

T=\frac{2\pi m r^2}{qB}

We see that this the period of revolution is directly proportional to the mass of the particle: therefore, if the second particle is 10 times as massive, then its period will be 10 times longer.

B) a. f/10

The frequency of revolution of a particle in uniform circular motion is

f=\frac{1}{T}

where

f is the frequency

T is the period

We see that the frequency is inversely proportional to the period. Therefore, if the period of the more massive particle is 10 times that of the smaller particle:

T' = 10 T

Then its frequency of revolution will be:

f'=\frac{1}{T'}=\frac{1}{(10T)}=\frac{f}{10}

6 0
3 years ago
Encuentre en km.h la velocidad de un tigre que corre 550 km.h en 69min​
Len [333]

Answer:

v = 478.26 km/h

Explanation:

The question is "find in km.h the speed of a tiger that runs 550 km in 69min"

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v=\dfrac{550\ km}{1.15\ h}\\\\v=478.26\ km/h

So, the speed of the tiger is 478.26 km/h.

8 0
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