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bagirrra123 [75]
3 years ago
7

______ creates heat in earths interior when dense substance sink into earths core.

Physics
1 answer:
anastassius [24]3 years ago
7 0

The answer is A. Friction

hope this helps :)

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A bag of rocks has a mass of 16.4 kg what is it weight here on the earth
Alex73 [517]
Answer
160,72N
Explanation
W=mg
=(16.4)(9.8)
6 0
2 years ago
If an object accelerates from rest, what will its velocity be after 1.3 s if it has a constant acceleration of 9.1 m/s^2?
HACTEHA [7]

\text{Given that,}\\\\\text{Initial velocity,} ~v_0 = 0~ \text{m~s}^{-1}\\\\\text{Time,  t = 1.3~sec}\\\\\text{Acceleration, a = 9.1 m s}^{-2}\\\\\\\\\text{Velocity,}\\\\v = v_0  +at\\\\\implies v = 0 + 9.1 \times 1.3 = 11.83~~ \text{m~s}^{-1}

5 0
2 years ago
Joe first focuses his attention (and his eyes) on the tree. The focal length of the cornea-lens system in his eye must be ______
kotykmax [81]

Answer: The focal length of the cornea-lens system in his eye must be LESS THAN the distance between the front and back of his eye.

Explanation:

The human eye the front part of the eye is the CORNEA. This is the tough white transparent part of the eye that helps in the refraction of light rays. While the backside of the eye is the RETINA. This is the part of the eye when images are focused.

When a normal eye is at rest, parallel rays from a distant object are focused on the retina. The ability of the eye - lens to focus points at different distances on the retina is known as accomodation. The adjustment of the eye lens to focus objects of varying distances is brought about by the ciliary muscles. The have the ability to change the shape of the eye which leads to change in focal length.

When a person with normal vision looks at a distant object at infinity, the lens brings parallel rays to focus on the retina. Thus, the furthest point which the eye can see distinctly is called the far point of the eye and it's infinity for a normal eye. But Joe was able to focus his eye on the tree, meaning that the tree was within his near point. This is the nearest point at which an object is clearly seen. Therefore, when the effective focal length of the cornea-lens system changes, it changes the location of the image of any object in one's field of view.

5 0
3 years ago
Determine the critical crack length for a through crack contained within a thick plate of 7150-T651 aluminum alloy that is in un
Mila [183]

Explanation:

Formula to determine the critical crack is as follows.

          K_{IC} = \gamma \sigma_{f} \sqrt{\pi \times a}

  \gamma = 1,     K_{IC} = 24.1

  [/tex]\sigma_{y}[/tex] = 570

and,   \sigma_{f} = 570 \times \frac{3}{4}

                       = 427.5

Hence, we will calculate the critical crack length as follows.

      a = \frac{1}{\pi} \times (\frac{K_{IC}}{\sigma_{f}})^{2}

        = \frac{1}{3.14} \times (\frac{24.1}{427.5})^{2}

       = 10.13 \times 10^{-4}

Therefore, largest size is as follows.

            Largest size = 2a

                                 = 2 \times 10.13 \times 10^{-4}

                                 = 20.26 \times 10^{-4}

Thus, we can conclude that the critical crack length for a through crack contained within the given plate is 20.26 \times 10^{-4}.

4 0
3 years ago
The wavelength of visible light is about 5500 times longer than the wavelength of x-rays. If you represent the wavelenth of an x
Pie

Answer:

458.33 ft

Explanation:

We are given that

Wavelength of of an x- ray photon=1 in

Wavelength of of an x- ray photon==\frac{1}{12}ft

1 in=\frac{1}{12}feet

We have to find the  length of line (in feet) drawn by you to represent the wavelength of visible light.

According to question

Wavelength of visible light=5500\times \frac{1}{12} ft

Wavelength of visible light=\frac{5500}{12}ft

Wavelength of visible light=458.33 ft

Hence, the 458.33 ft line must drawn  by you to represent the wavelength of visible light.

6 0
4 years ago
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