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3 years ago

In binocular rivalry, you see one image in the left eye and an incompatible image in the right eye. what do you perceive?

1 answer:

8 0

What we perceive when we <span>see one image in the left eye and an incompatible image in the right eye using a binocular rivalry is that we usually see something halfway between one another. By definition, a binocular rivalry takes place when images gets alternated as presented in each eye.</span>

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What is the SI Unit for amplitude?

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This distance is known as the amplitude of the wave, and is the characteristic height of the wave, above or below the equilibrium position. Normally the symbol A is used to represent the amplitude of a wave. The SI unit of amplitude is the metre (m).

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3 years ago

Which of these values is equivalent to the change in momentum of

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If a ball is thrown straight up into the air with an initial velocity of 65 ft/s, its height in feet after t seconds is given by

fgiga [73]

Answer:

a) $v_{1}=\frac{(62.5-66)ft}{(2.5-2)s}=-7ft/s$

$v_{2}=\frac{(65.94-66)ft}{(2.1-2)s}=-0.6ft/s$

$v_{3}=\frac{(66.0084-66)ft}{(2.01-2)s}=0.84ft/s$

$v_{4}=\frac{(66.001-66)ft}{(2.001-2)s}=1ft/s$

b) $v=65-32(2)=1ft/s$

Explanation:

From the exercise we got the ball's equation of position:

$y=65t-16t^{2}$

a) To find the average velocity at the given time we need to use the following formula:

$v=\frac{y_{2}-y_{1} }{t_{2}-t_{1} }$

Being said that, we need to find the ball's position at t=2, t=2.5, t=2.1, t=2.01, t=2.001

$y_{t=2}=65(2)-16(2)^{2} =66ft$

$y_{t=2.5}=65(2.5)-16(2.5)^{2} =62.5ft$

$v_{1}=\frac{(62.5-66)ft}{(2.5-2)s}=-7ft/s$

$y_{t=2.1}=65(2.1)-16(2.1)^{2} =65.94ft$

$v_{2}=\frac{(65.94-66)ft}{(2.1-2)s}=-0.6ft/s$

$y_{t=2.01}=65(2.01)-16(2.01)^{2} =66.0084ft$

$v_{3}=\frac{(66.0084-66)ft}{(2.01-2)s}=0.84ft/s$

$y_{t=2.001}=65(2.001)-16(2.001)^{2} =66.001ft$

$v_{4}=\frac{(66.001-66)ft}{(2.001-2)s}=1ft/s$

b) To find the instantaneous velocity we need to derivate the equation

$v=\frac{df}{dt}=65-32t$

$v=65-32(2)=1ft/s$

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3 years ago

ONLINE CALCULATOR .A force of 187 pounds makes an angle of 73 degrees 36 ' with a second force. The resultant of the two forces

saul85 [17]

Answer:

The magnitudes of the second force is $Z = 129.9 N$

The magnitudes of the resultant force is $R = 256.047 N$

Explanation:

From the question we are told that

The force is $F = 187 \ lb$

The angle made with second force $\theta_o = 73 ^o 36' = 73 + \frac{36}{60} = 73.6^o$

The angle between the resultant force and the first force $\theta _1 = 29 ^o 1 ' = 29 + \frac{1}{60} = 29.0167^o$

For us to solve problem we are going to assume that

The magnitude of the second force is Z N

The magnitude of the resultant force is R N

According to Sine rule

$\frac{F}{sin (\theta _o - \theta_1 } = \frac{Z}{\theta _1}$

Substituting values

$\frac{187}{sin(73.3 - 29.01667)} =\frac{Z}{sin (29.01667)}$

$267.82 =\frac{Z}{0.4851}$

$Z = 129.9 N$

According to cosine rule

$R = \sqrt{F ^2 + Z^2 + 2(F) (Z) cos (\theta _o) }$

Substituting values

$R = \sqrt{187^2 + 129.9 ^2 + 2 (187 ) (129.9) cos (73.6)}$

$R = 256.047 N$

3 0

3 years ago