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3 years ago

How does the pupillary response prevent injury? What would happen without it?

Physics

2 answers:

fenix001 [56]3 years ago

8 0

Part I

In extension to measuring the amount of light that enters the eye, the pupillary light reflex presents a helpful diagnostic tool. It provides for examining the uprightness of the sensory and motor functions of the eye. if your pupils didn't shut suddenly when abruptly disclosed to light, you'll get a tremendous shot of light (simultaneously with UV rays and other possibly dangerous rays) to your retina, which can severely mess up your vision.

Part II

Without it, we would go blind. If there is not sufficient light and the pupils do not dilate, a small number of light will pass to the retina and the image will be damaged.

Aleksandr [31]3 years ago

6 0

Pupils dilate and constrict in order to allow an adequate amount of light to pass through the retina and vision. If there is not enough light and the pupils do not dilate, a small amount of light will pass to the retina and the vision will be damaged.

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A 3.30 x 10-2-kg block is resting on a horizontal frictionless surface and is attached to a horizontal spring whose spring const

avanturin [10]

Answer:

0.17547 m

Explanation:

m = Mass of block = $3.3\times 10^{-2}\ kg$

v = Velocity of block = 10.8 m/s

k = Spring constant = 125 N/m

A = Amplitude

The kinetic energy of the system is conserved

$\dfrac{1}{2}mv^2=\dfrac{1}{2}kA^2\\\Rightarrow \\\Rightarrow A=\sqrt{\dfrac{mv^2}{k}}\\\Rightarrow A=\sqrt{\dfrac{3.3\times 10^{-2}\times 10.8^2}{125}}\\\Rightarrow A=0.17547\ m$

The amplitude of the resulting simple harmonic motion is 0.17547 m

6 0

3 years ago

Suppose a car manufacturer tested its cars for front-en4 collisions by hauling them up on a crane and dropping then; from a cert

Brrunno [24]

Answer:

Generally from third equation of motion we have that

$v^2 = u^2 + 2a[s_i - s_f]$

Here v is the final speed of the car

u is the initial speed of the car which is zero

$s_i$ is the initial position of the car which is certain height H

$s_i$ is the final position of the car which is zero meters (i.e the ground)

a is the acceleration due to gravity which is g

$v^2 = 0 + 2g[H - 0]$

=> $v = \sqrt{ 2 g H}$

$H = 9.86 \ m$

Explanation:

Generally from third equation of motion we have that

$v^2 = u^2 + 2a[s_i - s_f]$

Here v is the final speed of the car

u is the initial speed of the car which is zero

$s_i$ is the initial position of the car which is certain height H

$s_i$ is the final position of the car which is zero meters (i.e the ground)

a is the acceleration due to gravity which is g

$v^2 = 0 + 2g[H - 0]$

=> $v = \sqrt{ 2 g H}$

When $v = 50 \ km/h = \frac{50 *1000}{3600} = 13.9 \ m/s$ we have that

$13.9 = \sqrt{ 2 g H}$

=> $H = \frac{13.9^2}{2 * 9.8}$

=> $H = 9.86 \ m$

6 0

3 years ago

A natural atom can be negatively charged by

svet-max [94.6K]

Answer:

By Gaining Electrons

Explanation:

A nuetral atom is negative when it gains electrons, and it can be positive when it loses electrons.

4 0

2 years ago

Read 2 more answers

Suppose that a car traveling to the west (-x direction) begins to slow down as it approaches a traffic light. Which statement co

dalvyx [7]

Answer:

Its acceleration is positive

Explanation:

As the car is moving in the negative x-direction than after applying brake then there will be a decrease in the acceleration but in the opposite direction.

As decreasing acceleration consider to be negative but the car is moving in negative direction which means increasing acceleration is negative by sign convention but by applying brake acceleration decrease but in opposite direction than it will give positive value of acceleration.

4 0

3 years ago

A 11 g plastic ball is moving to the left at 29 m/s. How much work must be done on the ball to cause it to move to the right at

Rasek [7]

Answer:

4.25 J

Explanation:

Given that

mass of plastic ball = 11 g

Mass of plastic ball = 0.011 kg

velocity of ball = 29 m/s

We know that from work power energy theorem

$W_{all}=Change\ in\ kinetic\ energy\ of\ system$

We know that kinetic energy of moving mass given as

$KE=\dfrac{1}{2}mv^2$

Now by pitting the values

$KE=\dfrac{1}{2}mv^2$

$KE=\dfrac{1}{2}\times 0.011\times 29^2$

KE= 4.25 J

So the work done on the ball is 4.25 J

8 0

3 years ago