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3 years ago

Waves diffract the most when their wavelength is

Physics

1 answer:

djverab [1.8K]3 years ago

5 0

<h2>Answer: about the same size of the gap or slit</h2>

Diffraction happens when a wave (mechanical or electromagnetic wave, in fact, any wave) meets an obstacle or a slit .When this occurs, 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.

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.

Therefore:

<h2>Waves diffract the most when their wavelength is <u>about the same size of the gap </u></h2>

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A 3.0-kilogram mass is traveling in a circle of

dmitriy555 [2]

Answer:

$a=20\ m/s^2$

Explanation:

Given that,

The mass of an object, m = 3 kg

The radius of a circle, r = 0.2 m

The speed of the object, v = 2 m/s

We need to find the centripetal acceleration. Its formula is given by :

$a=\dfrac{v^2}{r}\\\\a=\dfrac{(2)^2}{0.2}\\\\a=20\ m/s^2$

So, the centripetal acceleration is $20\ m/s^2$ .

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3 years ago

What is the normal force acting on a 2.8 kg object that is being dragged across a floor

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Answer is in the file below

tinyurl.com/wpazsebu

7 0

3 years ago

According to newton's law of gravitation, what affects the force of attraction between two objects? distance between them angle

OlgaM077 [116]

The force of attraction between two objects Mass and distance.

<h3>What is newton's law of gravitation?</h3>

Every particle in the cosmos attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres, according to Newton's law of universal gravitation.

Inductive reasoning, as described by Isaac Newton, was used to deduce this general physical law from actual facts. It was created by Newton and is a component of classical mechanics. Philosophiae Naturalis Principia Mathematica, also known as "the Principia," was originally published on July 5, 1687. In April 1686, when Newton gave Book 1 of the unpublished book to the Royal Society, Robert Hooke said that Newton had learned the inverse square law from him.

According to the law, every point mass attracts every other point mass when a force applies along the line that intersects the two points, in today's parlance. The force is inversely equal to the square of the separation between the masses and directly proportional to their product.

$F = G\frac{m_{1} m_{2}}{r^{2} }$

to learn more about newton's law of gravitation go to - brainly.com/question/9373839

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3 0

2 years ago

I would like to know why this is the correct answer

Helen [10]

The acceleration of the object if the net force is decreased = 0.13 m/s²

<h3>Further explanation</h3>

Given

A net force of 0.8 N acting on a 1.5-kg mass.

The net force is decreased to 0.2 N

Required

The acceleration of the object if the net force is decreased

Solution

Newton's 2nd law :

$\tt \sum F=m.a$

The mass used in state 1 and 2 remains the same, at 1.5 kg

state 1

ΣF=0.8 N

m=1.5 kg

The acceleration, a:

$\tt a=\dfrac{\sum F}{m}\\\\a=\dfrac{0.8}{1.5}\\\\a=0.53`m/s^2$

state 2

ΣF=0.2 N

m=1.5 kg

The acceleration, a:

$\tt a=\dfrac{\sum F}{m}\\\\a=\dfrac{0.2}{1.5}\\\\a=0.13~m/s^2$

8 0

3 years ago

magine an astronaut on an extrasolar planet, standing on a sheer cliff 50.0 m high. She is so happy to be on a different planet,

Mama L [17]

Answer:

$\Delta t=(\frac{20}{g'}+\sqrt{\frac{400}{g'^2}+\frac{100}{g'} } )-(\frac{20}{g}+\sqrt{\frac{400}{g^2}+\frac{100}{g} } )$

Explanation:

Given:

height above which the rock is thrown up, $\Delta h=50\ m$

initial velocity of projection, $u=20\ m.s^{-1}$

let the gravity on the other planet be g'

The time taken by the rock to reach the top height on the exoplanet:

$v=u+g'.t'$

where:

$v=$ final velocity at the top height = 0 $m.s^{-1}$

$0=20-g'.t'$ (-ve sign to indicate that acceleration acts opposite to the velocity)

$t'=\frac{20}{g'}\ s$

The time taken by the rock to reach the top height on the earth:

$v=u+g.t$

$0=20-g.t$

$t=\frac{20}{g} \ s$

Height reached by the rock above the point of throwing on the exoplanet:

$v^2=u^2+2g'.h'$

where:

$v=$ final velocity at the top height = 0 $m.s^{-1}$

$0^2=20^2-2\times g'.h'$

$h'=\frac{200}{g'}\ m$

Height reached by the rock above the point of throwing on the earth:

$v^2=u^2+2g.h$

$0^2=20^2-2g.h$

$h=\frac{200}{g}\ m$

The time taken by the rock to fall from the highest point to the ground on the exoplanet:

$(50+h')=u.t_f'+\frac{1}{2} g'.t_f'^2$ (during falling it falls below the cliff)

here:

$u=$ initial velocity= 0 $m.s^{-1}$

$\frac{200}{g'}+50 =0+\frac{1}{2} g'.t_f'^2$

$t_f'^2=\frac{400}{g'^2}+\frac{100}{g'}$

$t_f'=\sqrt{\frac{400}{g'^2}+\frac{100}{g'} }$

Similarly on earth:

$t_f=\sqrt{\frac{400}{g^2}+\frac{100}{g} }$

Now the required time difference:

$\Delta t=(t'+t_f')-(t+t_f)$

$\Delta t=(\frac{20}{g'}+\sqrt{\frac{400}{g'^2}+\frac{100}{g'} } )-(\frac{20}{g}+\sqrt{\frac{400}{g^2}+\frac{100}{g} } )$

3 0

3 years ago