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pogonyaev
3 years ago
9

Snakes and lizards are rarely found near polar environment explain why

Physics
1 answer:
Novosadov [1.4K]3 years ago
4 0
Reptiles, such as snakes and lizards, are cold-blooded animals. This means that they do not have the ability to control their body heat. For this reason, they often lay out in the sun to warm up. If their enviornent is cold they obtain that body temperature. They are very slow in colder environments.In cold weather these animals have a very difficult time moving because their muscles are very cold. Without heat in their environment, they cannot warm them up.
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The gnaphosid spider Drassodes cupreus has evolved a pair of lensless eyes for detecting polarized light. Each eye is sensitive
BartSMP [9]

Answer:

Part A

The intensity is  I = 618 W/m^2  

Part B

The intensity is  I_1= 81.884 W/m^2

Explanation:

From the question we are told that

       The intensity of the light detected by first eye is I = 700 W/m^2

Now at initial state according the question the light  ray is perpendicular to the eye so it means that it is at 90° the eye

Now the first question is to obtain the intensity the first eye (the first in this case is the one focused on the light  )would detect when the head is rotated by 20°  its previous orientation

This  is mathematically evaluated  as

                   I = I_i cos^2 ( 20^o)

                    I = 700\  cos^2 (20)

                    I = 618 W/m^2  

Now the second  question is to obtain the intensity the first eye (the first eye  in this case is the one that is not  focused on  the light  )would detect when the head is rotated by 20°  its previous orientation

Now in this case the angle between the eye and the light is 90-20 = 70°

           So

               I_1 = 700 \  cos^2 (70)

                   I_1= 81.884 W/m^2

 

5 0
3 years ago
This question is related to inertia:
luda_lava [24]
The way I do it is suddenly, in the same sort of way that magicians try to pull a table cloth off a table when there's things on the table cloth.The sudden approach acts as an impulse of force and starts to accelerate the roll. But, the piece (assuming it has perforations) is off the roll before the roll can move, due to inertia. Then the roll will acclerate, move, slow down and stop. However, in accelerating, the roll will unravel. The bigger the impulse the more it will unravel.+++++++++++++++++++++++++++++++++++++++If on the other hand, the piece of paper is held firmly, and the roll is pulled, then the impulse is presumably given to the paper and the hand whose inertia is a lot more than that of the roll. So, I think I'd actually go for choice c)+++++++++++++++++++++++++++++++++++++This assumes that the roll is free to rotate.I think that a similar idea is behind the design and use of a "ballistic galvanometer". The charge is passed through the galvanometer quickly, as a current pulse. Then the needle starts to deflect, and the deflection is arranged to depend on the total charge that has passed through in the time of the current pulse.
3 0
3 years ago
The old rubber boot has two leaks. The top of the boot is 0.3 m higher than the leaks. What is the velocity of the water coming
Otrada [13]

Answer:

Leak 1 = 3.43 m/s

Leak 2 = 2.42 m/s

Explanation:

Given that the top of the boot is 0.3 m higher than the leaks. 

Let height H = 0.3m and the acceleration due to gravity g = 9.8 m/s^2

From the figure, the angle of the leak 1 will be approximately equal to 45 degrees. While the leak two can be at 90 degrees.

Using the third equation of motion under gravity, we can calculate the velocity of leak 1 and 2

Find the attached files for the solution and figure

7 0
3 years ago
A light ray passing from medium 1 to medium 2 is bent away from the perpendicular normal to the boundary surface. What happens t
SVETLANKA909090 [29]

Answer:

The answer is a for Plato users.

Explanation:

Since the angle of the refracted ray moves away from the normal, it must be traveling in a faster medium.

6 0
3 years ago
Read 2 more answers
Light is incident along the normal to face AB of a glass prism of refractive index 1.54. Find αmax, the largest value the angle
marusya05 [52]

To solve this problem it is necessary to use the concepts related to Snell's law.

Snell's law establishes that reflection is subject to

n_1sin\theta_1 = n_2sin\theta_2

Where,

\theta = Angle between the normal surface at the point of contact

n = Indices of refraction for corresponding media

The total internal reflection would then be given by

n_1 sin\theta_1 = n_2sin\theta_2

(1.54) sin\theta_1 = (1.33)sin(90)

sin\theta_1 = \frac{1.33}{1.54}

\theta = sin^{-1}(\frac{1.33}{1.54})

\theta = 59.72\°

Therefore the \alpha_{max} would be equal to

\alpha = 90\°-\theta

\alpha = 90-59.72

\alpha = 30.27\°

Therefore the largest value of the angle α is 30.27°

3 0
3 years ago
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