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FinnZ [79.3K]
3 years ago
8

What is the main function of the spongy bone'

Physics
2 answers:
storchak [24]3 years ago
7 0
Oooooo there's a spongy bone? that's cool! Lol okay okay, I will research it and help you out. 

Here's what I found:

Cancellous bone<span>, also known as </span>spongy<span> or </span>trabecular bone<span>, is one of the </span>two<span> types of </span>bone<span> tissue found in the human body. ... It is very porous and contains red </span>bone<span>marrow, where blood cells are made.</span>
Korolek [52]3 years ago
5 0
Provides structural support and facilitates movement of the joints and limbs <span />
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What is the percentage of the incoming solar radiation from the top of the atmosphere the gets reflected right back out into spa
san4es73 [151]

Answer: We can define the solar constant as a measure of the luminous flux density.

Explanation:

The solar constant or solar constant is the amount of energy radiated at the upper limit of the Earth's atmosphere per unit time perpendicular to the unit surface, at the Earth's mean distance from the sun. Amounts to 1367.7 W / m² ± 6 W / m². The sun's constant includes all kinds of electromagnetic radiation, not just visible light. The average value is 1,368 kW / m2 and changes slightly with solar cycles. The amount of these constant changes over one year and has different benefits.

4 0
3 years ago
Alex is standing still and throws a football with a speed of 10 m/s to his friend, who is also standing still. The two friends a
Phantasy [73]

The question is incomplete. It comes with a set of answer choices.


These are the answer choices:


Alex observes it as 10 m/s, and his friend observes it as less than 10 m/s.


Alex observes it as less than 10 m/s, and his friend observes it as 10 m/s.


Both Alex and his friend observe it as 10 m/s.


Both Alex and his friend observe it as less than 10 m/s.



Answer: Both Alex and his friend observe it as 10 m/s.


Justification:


1) The speed is relative to the frame of reference.


2) It is said that the both Alex and his friend are standing still.


3) Then, the speed they both see is the same, 10 m/s, respect the Earth (where they are standing still).


Of course, Alex is watching the ball moving away and his friend is seing it approaching, but it is not relevant for the question, as it deals with the speed which is only about magnitude, not direction.

7 0
3 years ago
Read 2 more answers
What is the function of a hypothesis in the scientific inquiry process
svp [43]
To make a educational guess based on the your observations
6 0
3 years ago
Velocity vector and acceleration vector in a uniform circular motion are related as.
mr_godi [17]

They are related as \bold{\underline {v}\,.\,\underline a }= \bold{0}

  • In a uniform circular motion, the magnitude of the speed does not change during the travel and only the instantaneous direction changes.
  • This speed is always directed along the tangent to the circle at a given point. (refer to the figure attached)
  • For any circular motion, the must-have acceleration is the centripetal acceleration that is directed towards the centre of the circular locus (if the motion has a tangential acceleration, it has a tangential acceleration additionally).
  • Therefore, both the directions of the tangential speed and the centripetal acceleration are orthogonal to each other (perpendicular: one is 90 degrees apart from the other).
  • In mathematics, 2 vectors (\underline p , \underline q) that are perpendicular to each other have a quality that their dot product (\underline p\,.\, \underline q) equal to zero vector (\bold 0) which is written as \undeline p\,.\, \underline q = \bold 0.
  • This quality can be considered when dealing with the velocity vector and the acceleration vector in a manner \underline v\,.\, \underline a =\bold 0.

#SPJ4

8 0
1 year ago
An insulating sphere is 8.00 cm in diameter and carries a 6.50 µC charge uniformly distributed throughout its interior volume.
Kobotan [32]

Explanation:

(a)   Formula to calculate the density is as follows.

            \rho = \frac{Q}{\frac{4}{3}\pi a^{3}}

                       = \frac{6.50 \times 10^{-6}}{\frac{4}{3} \times 3.14 \times (0.04)^{3}}

                     = 2.42 \times 10^{-2} C/m^{3}

Now, calculate the charge as follows.

            q_{in} = \rho(\frac{4}{3} \pi r^{3})

                      = 2.42 \times 10^{-2} C/m^{3} \times 4.1762 \times (0.01)^{3}

                      = 10.106 \times 10^{-8} C

or,                   = 101.06 nC

(b)  For r = 6.50 cm, the value of charge will be calculated as follows.

                q_{in} = \frac{Q}{\frac{4}{3}\pi a^{3}}

                          = \frac{6.50 \times 10^{-6}}{\frac{4}{3} \times 3.14 \times (0.065)^{3}}

                          = 7.454 \mu C

7 0
3 years ago
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