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3 years ago

14

A particle performing circular motion along the path of radius 2 metre with linear acceleration 5 metre per second square find i

ts angular acceleration

1 answer:

dangina [55]3 years ago

4 0

Answer:

What is the question though?

Explanation:

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The simple act of walking requires single leg

OlgaM077 [116]

you are so wise how do you do it?

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3 years ago

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Name TWO WEAKNESSES of the model pictured below

Nikolay [14]

Answer:

Here are a few:

1) The orbital radius of these planets is ridiculously small an in no way representative of their actual radii.

2) The planets will only line up like that once every 5200 years, making this very unrepresentative of their usual relations - although this does make their order in distance from the sun.

3) The nebulae, comet, lens flare, and other junk in the background is incorrect.

4) If this is meant as a representation of the planets, then Pluto should not be there as it is now considered a planetoid.

5) The planets are incorrectly scaled both to each other and to the sun.

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3 years ago

What is the pathway of sound through fluids starting at the oval window through to dissipation of the sound waves at the round w

ruslelena [56]

Perilymph of scala vestibule; endolymph of cochlear duct; perilymph of scala tympani

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3 years ago

A mass m at the end of a spring vibrates with a frequency

Wittaler [7]

Answer:

m = 0.59 kg.

Explanation:

First, we need to find the relation between the frequency and mass on a spring.

The Hooke's law states that

$F = -kx$

And Newton's Second Law also states that

$F = ma = m\frac{d^2x}{dt^2}$

Combining two equations yields

$a = -\frac{k}{m}x$

The term that determines the proportionality between acceleration and position is defined as angular frequency, ω.

$\omega = \sqrt{\frac{k}{m}}$

And given that ω = 2πf

the relation between frequency and mass becomes

$f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}$ .

Let's apply this to the variables in the question.

$0.88 = \frac{1}{2\pi}\sqrt{\frac{k}{m}}\\0.60 = \frac{1}{2\pi}\sqrt{\frac{k}{m+0.68}}\\\frac{0.88}{0.60} = \frac{\frac{1}{2\pi}\sqrt{\frac{k}{m}}}{\frac{1}{2\pi}\sqrt{\frac{k}{m+0.68}}}\\1.4667 = \frac{\sqrt{m+0.68}}{\sqrt{m}}\\2.15m = m + 0.68\\1.15m = 0.68\\m = 0.59~kg$

6 0

3 years ago

mote1985 [20]

Answer:

OC, OD, OA, OB

Explanation:

7 0

3 years ago