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3 years ago

the maximum displacement of a particle, within a wave, above or below its equilibrium position, is called__________

Physics

2 answers:

Travka [436]3 years ago

7 0

It’s called the amplitude

Gre4nikov [31]3 years ago

3 0

Answer:

the amplitude.

Explanation:

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2 years ago

Which planet has the most extreme temperature variations

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A dog of mass 4 kg runs up a hill of height 8 m. How much gravitational potential energy does the dog gain?

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3 years ago

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A simple pendulum has length of 820mm. Calculate the frequency (g = 9.8 ms -2)

Vadim26 [7]

Answer:

$\huge\boxed{\sf f=0.55 \ Hz}$

Explanation:

Given Data:

Length = l = 820 mm = 0.82 m

Acceleration due to gravity = g = 9.8 ms⁻²

Required:

Frequency = f = ?

Formula:

$\displaystyle f =\frac{1}{2 \pi} \sqrt{\frac{g}{l} }$

Solution:

$\displaystyle f =\frac{1}{2 \pi} \sqrt{\frac{g}{l} } \\\\Put\ the\ givens\\\\f=\frac{1}{2 \pi} \sqrt{\frac{9.8}{0.82} }\\\\ f = 0.159 \times \sqrt{11.95} \\\\f=0.159 \times 3.457\\\\f=0.55 \ Hz\\\\\rule[225]{225}{2}$

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2 years ago

An important dimensionless parameter in certain types of fluid flow problems is the Froude number defined as V/√g.l where V is a

prohojiy [21]

Answer:

1.24611

Explanation:

V = Velocity = 10 ft/s

L = Length = 2 ft

g = Acceleration due to gravity = 32.2 ft/s²

Froude number is given by

$Fr=\dfrac{V}{\sqrt{gL}}\\\Rightarrow Fr=\dfrac{10}{\sqrt{32.2\times 2}}\\\Rightarrow Fr=1.24611$

Converting to SI units

$10\ ft/s=10\times \dfrac{1}{3.281}$

$32.2\ ft/s^2=32.2\times \dfrac{1}{3.281}$

$2\ ft=2\times \dfrac{1}{3.281}$

$Fr=\dfrac{V}{\sqrt{gL}}\\\Rightarrow Fr=\dfrac{10\times \dfrac{1}{3.281}}{\sqrt{32.2\times \dfrac{1}{3.281}\times 2\times \dfrac{1}{3.281}}}\\\Rightarrow Fr=1.24611$

The Froude number is 1.24611

The Froude number is equal. The Froude number is dimensionless as the units cancel each other. In order for this to happen the units used need to be consitent either imperial or SI.

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3 years ago