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3 years ago

ME Project- Decomposer

Physics

1 answer:

Alenkasestr [34]3 years ago

3 0

Answer:

Explanation:

Decomposer.

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1) Think of planting a tree. You dig the hole, put it in, and fill it back in. Next,

pogonyaev

Answer:

at the top of the tree I hope it will help you please follow me

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2 years ago

Jak możemy służyć innym ludziom. Pilne

Troyanec [42]

Could u put it in english u can use goggle translation if u do not know how to

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4 years ago

A bike and rider together have a mass of 60 kg. If the bike and rider have an acceleration of 2.0 m/s^2, what is the force on th

Mice21 [21]

The net force on the bike and the rider is 120 N

Explanation:

We can solve this problem by applying Newton's second law of motion, which states that:

F = ma

where

F is the net force exerted on an object

m is the mass of the object

a is its acceleration

For the bike and the rider in this problem, we have

m = 60 kg is their combined mass

$a=2.0 m/s^2$ is their acceleration

Therefore, the net force on them is

$F=(60)(2.0)=120 N$

Learn more about Newton's second law:

brainly.com/question/3820012

#LearnwithBrainly

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3 years ago

A student pushes on a crate with a force of 100 N directed to the right. What force does the crate exert on the student?

sattari [20]

Answer:

The Force exerted on the student by crate will be 100 N.

Explanation:

As per the given Question student is trying to push the crate in right direction with the force of 100N.

And we know that, Newton's third law states that every action has equal and opposite reaction.

So from the Newton's 3rd law of motion it is very clear the student must be experiencing the same amount of force which he is applying.

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3 years ago

Red light of wavelength 633 nmnm from a helium-neon laser passes through a slit 0.400 mmmm wide. The diffraction pattern is obse

erik [133]

Answer:

$y_1 = 0.004589 \ m$

$y_2 =0.009179 \ m$

Explanation:

From the question we are told that

The wavelength of the red light is $\lambda _r = 633 \ nm = 633 *10^{-9} \ m$

The width of the slit is $d = 0.40 mm = 0.40 *10^{-3} \ m$

The distance of the screen from the point of diffraction is $D = 2.9 \ m$

Generally the width of the central bright fringe is mathematically represented as

$y_1 = \frac{\lambda * D}{d}$

=> $y_1 = \frac{633 *10^{-9} * 2.90 }{0.40 *10^{-3}}$

=> $y_1 = 0.004589 \ m$

Generally the width of the first bright fringe on either side of the central one is mathematically represented as

$y_2 = 2 * y_1$

=> $y_2 = 2 * 0.004589$

=> $y_2 =0.009179 \ m$

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3 years ago