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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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What forms of rocks are produced when rocks travel to fewer stages in the cycle?

Veronika [31]

Answer:

The two major sources of energy for the rock cycle are also shown; the sun provides energy for surface processes such as weathering, erosion, and transport, and the Earth's internal heat provides energy for processes like melting, and metamorphism

4 0

3 years ago

A resistor rated at 250 k Ω is connected across two D cell batteries (each 1.50 V) in series, with a total voltage of 3.00 V. Th

Andrej [43]

Given Information:

Resistance = R = 250 Ω

Voltage = V = 3 V

Tolerance = ±5%

Required Information:

Maximum current = Imax = ?

Minimum current = Imin = ?

Answer:

Imax = 12.6 uA

Imin = 11.4 uA

Explanation:

As we know from Ohm's law

I = V/R

The current will be maximum when R is minimum and the current will be minimum when R is maximum

Rmax = 250,000 + 0.05*250,000

Rmax = 250,000 + 12500

Rmax = 262500 Ω

Rmin = 250,000 - 0.05*250,000

Rmin = 250,000 - 12500

Rmin = 237500 Ω

Imax = V/Rmin

Imax = 3/237500

Imax = 12.6 uA

Imin = V/Rmax

Imin = 3/262500

Imin = 11.4 uA

6 0

3 years ago

If y = 0.02 sin (30x – 200t) (SI units), the frequency of the wave is

leva [86]

Answer:

31.831 Hz.

Explanation:

<u>Given:</u>

$\rm y = 0.02\sin(30x-200 t).$

The vertical displacement of a wave is given in generalized form as

$\rm y = A\sin(kx -\omega t).$

<em>where</em>,

A = amplitude of the displacement of the wave.
k = wave number of the wave = $\dfrac{2\pi }{\lambda}.$
$\lambda$ = wavelength of the wave.
x = horizontal displacement of the wave.
$\omega$ = angular frequency of the wave = $\rm 2\pi f$ .
f = frequency of the wave.
t = time at which the displacement is calculated.

On comparing the generalized equation with the given equation of the displacement of the wave, we get,

$\rm A=0.02.\\k=30.\\\omega =200.\\$

therefore,

$\rm 2\pi f=200\\\\\Rightarrow f = \dfrac{200}{2\pi}=31.831\ Hz.$

It is the required frequency of the wave.

3 0

3 years ago

tekilochka [14]

The answer is 25.37 hope this helps

5 0

3 years ago

One kind of baseball pitching machine works by rotating light and stiff rigid rod about a horizontal axis until the ball is movi

erastova [34]

Answer:

(a). The ball's centripetal acceleration is $16.17\times10^{2}\ m/s^2$

(b). The magnitude of the net force is 232.9 N.

Explanation:

Given that,

Mass of baseball = 144 g

Speed = 81 mph = 36.2 m/s

Distance = 81 cm

(a). We need top calculate the ball's centripetal acceleration just before it is released

Using formula of centripetal acceleration

$a=\dfrac{v^2}{r}$

Where, v = speed

r = radius

Put the value into the formula

$a=\dfrac{(36.2)^2}{81\times10^{-2}}$

$a=1617.82\ m/s^2$

$a=16.17\times10^{2}\ m/s^2$

(b). We need to calculate the magnitude of the net force that is acting on the ball just before it is released

Using formula of force

$F=\dfrac{mv^2}{r}$

Put the value into the formula

$F=\dfrac{144\times10^{-3}\times(36.2)^2}{81\times10^{-2}}$

$F=232.9\ N$

Hence, (a). The ball's centripetal acceleration is $16.17\times10^{2}\ m/s^2$

(b). The magnitude of the net force is 232.9 N.

4 0

3 years ago