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STatiana [176]
3 years ago
8

3. Why is the term cold blooded a misconception? Explain​

Physics
1 answer:
jok3333 [9.3K]3 years ago
7 0

Answer:

The term “cold-blooded” implies that these animals are in a never-ending struggle to stay warm. That really isn't correct. A cold-blooded animal can warm up their blood by being in the sun for hours.

You might be interested in
What is the function of eye lens of the human eye<br>​
Artyom0805 [142]

Answer:

Lens of the human Eye is a important and one of that most complex sense organ.

Explanation:

Lens of the human eye it helps that in visualizing light and color perception and objects , glance on the human eye structure and function.

Sense organs are much pretty similar to the camera they  help us see the objects clear.

A human eye is the 2.3 cm in diameter and all filled some fluid, and there are following parts in eye:-  cornea,  Retina , Lens ,Pupil , optic nerves.

cornea :- the cornea is the first transparent part of is called cornea, enters the light through the cornea.

Retina :-it is light sensitive layer that consists of nerve cells,then transmitted to the brain through nerves.

Lens:- behind the pupil there is a transparent structure called lens,it shape focus light on the retina.

Pupil:-it control the value of light that enters the human eye.

Optic nerves is are the two types :- (1) cones (2) Rods .

7 0
3 years ago
Do all planetary systems look the same as our own?
Svet_ta [14]

Answer:

No

Explanation:

A planetary system consists of at least one star and non stellar objects revolving around it.

Our solar system has one star around which there are 8 planets. However there are star systems with more than one star. These systems are called binary systems. The size of stars also vary. They also vary by orbital configuration i.e, the planets have higher eccentricity than our solar system's. The planetary systems are also classified on the basis of the number of planets in them.

So, all planetary systems do not look the same as our own.

4 0
3 years ago
If a true heading of 135° results in a ground track of 130° and a true airspeed of 135 knots results in a groundspeed of 140 kno
vladimir2022 [97]

Answer:

245.1° and 13 knots

Explanation:

The given parameters are;

The true heading = 135°

The resultant ground track = 130°

The true airspeed = 135 knots

The ground speed = 140 knots

Given that the true airspeed the ground speed and the wind direction and magnitude form a triangle, we have;

From cosine rule, we have;

a² = b² + c² - 2×b×c×cos(A)

Where

a = The magnitude of the wind speed in knot

b = The true airspeed = 135 knots

c = The ground speed = 140 knots

A = The angle in between the true heading and the resultant ground track heading = 5°

Which gives;

a² = 135² + 140² - 2×135×140×cos(5 degrees) = 168.84 knots²

a = √168.84 = 12.9934 ≈ 13 knots

We have;

135 × sin(135 degrees) - 140× sin(130 degrees) = -11.7868

135 × cos(135 degrees) - 140× cos(130 degrees) = -5.469

Tan(θ) = -11.8/-5.5 = 2.155

θ = tan⁻¹(2.155) = 65.108°

Given that the wind is moving in opposite direction (slowing down the airplane, we add 180°, to get

Therefore, the angle direction = 180 + 65.108 = 245.1

Therefore, we have;

245.1° and 13 knots

3 0
3 years ago
Which area of earth is most similar to the suns convention zone
julsineya [31]
The area of the Earth that is most similar to the Sun's convection zone would be the mantle. The convection zone of the sun is its outermost layer where heat transfer by convection happens which is similar to the Earth's mantle. It would be the crust because it is the outer most layer.
3 0
2 years ago
What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1
notka56 [123]

Complete Question

Part of the question is shown on the first uploaded image

The rest of the question

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1 and q2 at x3 = -1.220 m ? Your answer may be positive or negative, depending on the direction of the force. Express your answer numerically in newtons to three significant figures.

Answer:

The net force exerted on the third charge is  F_{net}=  3.22*10^{-5} \ J

Explanation:

From the question we are told that

    The third charge is  q_3 =  55 nC =  55 *10^{-9} C

    The position of the third charge is  x = -1.220 \ m

     The first charge is q_1 =  -16 nC  =  -16 *10^{-9} \ C

     The position of the first charge is x_1 =  -1.650m

      The second charge is  q_2 =  32 nC  =  32 *10^{-9} C

      The position of the second charge is  x_2 =   0  \ m  

The distance between the first and the third charge is

      d_{1-3} =  -1.650 -(-1.220)

     d_{1-3} = -0.43 \ m

The force exerted on the third charge by the first is  

     F_{1-3} =  \frac{k  q_1 q_3}{d_{1-3}^2}

Where k is the coulomb's constant with a value  9*10^{9} \ kg\cdot m^3\cdot s^{-4}\cdot A^2.

substituting values

      F_{1-3} =  \frac{9*10^{9}* 16 *10^{-9} * (55*10^{-9})}{(-0.43)^2}

       F_{1-3} = 4.28 *10^{-5} \ N

 The distance between the second and the third charge is      

  d_{2-3} =  0- (-1.22)

   d_{2-3} =1.220 \ m

The force exerted on the third charge by the first is mathematically evaluated as

       F_{2-3} =  \frac{k  q_2 q_3}{d_{2-3}^2}

substituting values

       F_{2-3} =  \frac{9*10^{9} * (32*10^{-9}) *(55*10^{-9})}{(1.220)^2}

       F_{2-3} =  1.06*10^{-5} N

The net force is

      F_{net} =  F_{1-3} -F_{2-3}

substituting values

    F_{net} = 4.28 *10^{-5} - 1.06*10^{-5}

    F_{net}=  3.22*10^{-5} \ J

6 0
2 years ago
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