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3 years ago

6

A reason that one typically does not notice a blind spot in the visual field is that

1 answer:

Sophie [7]3 years ago

3 0

Explanation:

Both of our eyes have a blind spot, in the retina of eye where there are no rods (light vision) or cons (color vision) cells. Its about a size of pinhead. The blind spot is place where the optic nerves exit the eye and connect to our brain. The problem is that we cannot notice this blind as the brain fills in information for us.

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A point charge of -2 µC is located at the origin. A second point charge of 6 µC is at x = 1 m, y = 0.5 m. Find the x and y coord

Soloha48 [4]

Answer:

x coordinate = -1.66 m

y coordinate is = -0.825m

Explanation:

Suppose z be the distance form the first charge and z + sqrt(1^2 +.5^2) be the distance from the second So z + sqrt(1+.25) = z + 1.12

We have k*2.0x10^-6/s^2 = k*6x10^-6/(s+1.12)^2

0.0356s^2 -0.019s-0.0897=0

s=1.876m

The angle of the line between the two charges is arctan(.5/1) = 26.6o

x coordinate = -1.876*cos(26.6) = -1.66m

y coordinate is -1.876*sin(26.6) = -0.825m

3 0

3 years ago

A _____________ is any visual, audio, audiovisual, or other sensory material used to enhance a verbal message.

Ber [7]

Answer:

Presentational aid

Explanation:

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2 years ago

When throwing curveballs in baseball, is it important to have air hit the ______ part of the ball.

Licemer1 [7]

Answer: A (smooth)

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2 years ago

Read 2 more answers

g An astronaut must journey to a distant planet, which is 189 light-years from Earth. What speed will be necessary if the astron

Butoxors [25]

Answer:

The value is $v = 2.999 *10^{8} \ m/s$

Explanation:

From the question we are told that

The time taken to travel to the planet from earth is $t = 189 \ light-years$

The time to be spent on the ship is $t_{s} = 12 \ years$

Generally speed can be obtained using the mathematical relation represented below

$t_s = 2 * t * \sqrt{1 - \frac{v^2}{c^2 } }$

The 2 in the equation show that the trip is a round trip i.e going and coming back

=> $12 = 2 * 189 * \sqrt{1 - \frac{v^2}{(3.0*10^{8})^2 } }$

=> $v = 2.999 *10^{8} \ m/s$

5 0

2 years ago

Consider a block on a spring oscillating on a frictionless surface. The amplitude of the oscillation is 11 cm, and the speed of

IRISSAK [1]

Answer:

The angular frequency of the block is ω = 5.64 rad/s

Explanation:

The speed of the block v = rω where r = amplitude of the oscillation and ω = angular frequency of the oscillation.

Now ω = v/r since v = speed of the block = 62 cm/s and r = the amplitude of the oscillation = 11 cm.

The angular frequency of the oscillation ω is

ω = v/r

ω = 62 cm/s ÷ 11 cm

ω = 5.64 rad/s

So, the angular frequency of the block is ω = 5.64 rad/s

6 0

3 years ago