2 years ago

Direction of waves is parallel to distance of vibration in

1 answer:

3 0

I strongly go with A

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frozen [14]

Answer:

$\theta = 49.81^0$

Explanation:

Given that:

$\omega = 41.5 \ rev/min\\\\\omega = 41.5 *\frac{1}{60}* 2 \pi\\\\\omega = 4.45 \ rad/s\\\\\\diameter = 0.748 m$

If we let the piece of the close lose contact at ∠θ;

Then ; from force balance;

we have:

$\\\\mg sin \theta = \frac{mv^2}{r}\\\\sin \theta = \frac{2v^2}{dg}\\\\\theta = sin^{-1} (\frac{2v^2}{dg})$

where;

$v = \frac{\omega d}{2}\\\\v = \frac{4.45 *0.748}{2}\\\\v = 1.6643\\\\v^2 = 2.77$

Again:

$\theta = sin^{-1}(\frac{2v^2}{dg})\\\\\theta = sin^{-1}( \frac{2*2.77}{0.74*9.8})\\\\\theta = 49.81^0$

6 0

2 years ago

Anna007 [38]

Objects in contact with each other never get past the electron clouds of the atoms on their surfaces.

5 0

3 years ago

Nastasia [14]

Here you go fam .......

4 0

2 years ago

zaharov [31]

I don’t know sorry then ya soo

4 0

3 years ago

Amanda [17]

I got 5,225 by 50x4.18= 209(25)=5,225

6 0

2 years ago