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3 years ago

5

A wave has an amplitude of 0.2 cm.

1 answer:

Anettt [7]3 years ago

3 0

The distance between the resting point and maximum height of the wave is 0.2 cm.

The amplitude is measured from the resting point up to the highest point of the wave.

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You stare at a bright red screen for so long that your red cones become saturated and no longer function. The red screen is then

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Answer: you'll see cyan color on the screen

Explanation:

Saturating the red cone causes them to stop functioning, hence you can't perceive the red part of white light. White light is made up of three main colors which are blue, red and green. When one can no longer perceive the red part of light, one is left with the grean and blue part. The green and blue part of light will superimpose to give a cyan color.

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3 years ago

Read 2 more answers

A cube is 4.4 cm on a side, with one corner at the origin. Part 1 (a) What is the unit vector pointing from the origin to the di

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Answer:

(a) $\hat{A} = \frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}}$

(b) $\theta = 85.44^{\circ}$

Solution:

As per the question:

Side of the cube, a = 4.4 cm

Coordinates of the diagonally opposite corner, A = <4.4, 4.4, 4.4> cm

Now,

(a) To calculate the unit vector:

$\hat{A} = \frac{\vec{A}}{|A|}$

$\hat{A} = \frac{4.4\hat{i} + 4.4\hat{j} + 4.4\hat{k}}{\sqrt{()4.4}^{2} + (4.4)^{2} + (4.4)^{2}}$

$\hat{A} = \frac{4.4(\hat{i} + \hat{j} + \hat{k})}{4.4\sqrt{3}}$

$\hat{A} = \frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}}$

(b) To calculate the angle between the two vectors say A and A' is given by:

$\vec{A}\cdot \vec{A'} = \vec{A}\vec{A'}cos\theta$

$\theta = cos^{- 1}(\frac{\vec{A}\cdot \vec{A'}}{\vec{A}\vec{A'}})$ (1)

Now,

The coordinates of the diagonally opposite corner, A' is <0, 0, 1> cm

Thus

$\vec{A'} = 0\hat{i} + 0\hat{j} + 1\hat{k} = \hat{k}$

Now,

Using equation (1) :

$\theta = cos^{- 1}(\frac{(\frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}})\cdot \hat{k}}{|A||A'|})$

$|A||A'| = (\sqrt{4.4^{2} +4.4^{2} + 4.4^{2}})(\sqrt{0^{2} + 0^{2} + 0^{2}}) = 7.261$

Thus

$\theta = cos^{- 1}(\frac{\frac{1}{\sqrt{3}}}{7.261})$

$\theta = cos^{- 1}(0.07946) = 85.44^{\circ}$

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