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anyanavicka [17]
3 years ago
11

Bone with a latticework structure is called

Physics
1 answer:
malfutka [58]3 years ago
7 0
Spongy or the cancellous bone is the bone with lattice work structure
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In 1985, Said Aouita set the world record for the 1500m race in a time of 3 minute 29.46 second ? What was his average speed ?
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 but tpla ybut tpla y but tpl but tpla y but tpla y but tpla ya y but tpla y but tpla y
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3 years ago
What is 1 and 2 I need it asap
ExtremeBDS [4]
The correct answer for the first one is A) Breaking rocks into smaller pieces. The correct answer to the second one is A) It has more energy. Hope this helps.
5 0
3 years ago
I need help with this question how to solve it for Brass and Cooper
Ksenya-84 [330]

Take into account that density and relative density are given by:

\begin{gathered} \text{density}=\text{ mass/volume} \\ \text{relative density = density/density of water} \end{gathered}

Take into account that the volume associated to each of the given sustances in the table is determined by the Level Difference (because it is the change in the volume of the water of the recipient in which the substance is immersed).

The density of water in kg/m^3 is 1000 kg/m^3.

Due to the density must be given in kg/m^3, it is necessary to express the volumes of the table in m^3 and mass in kg, then, consider the following conversion factor:

1 m^3 = 1000000 ml

1 kg = 1000 g

Then, you obtain the following results:

Brass:

\begin{gathered} 53.2g\cdot\frac{1kg}{1000g}=0.0532kg \\ 6ml\cdot\frac{1m^3}{1000000ml}=0.000006m^3 \\ \text{density}=\frac{0.0532kg}{0.000006m^3}\approx8866.67\frac{kg}{m^3} \\ \text{relative density=}\frac{(\frac{8866.66kg}{m^3})}{(1000\frac{kg}{m^3})}\approx8.87 \end{gathered}

Cooper:

\begin{gathered} 57.4g=0.0574kg \\ 6ml=0.000006m^3 \\ \text{density}=\frac{0.0574kg}{0.000006m^3}\approx9566.67\frac{kg}{m^3} \\ \text{relative density=}\frac{\frac{9566.67kg}{m^3}}{1000kg}=9.57 \end{gathered}

3 0
1 year ago
Assignment
Igoryamba

Answer:

step bro was stuck on the elevator

Explanation:

5 0
2 years ago
Two sources of light of wavelength 700 nm are 9 m away from a pinhole of diameter 1.2 mm. How far apart must the sources be for
Lelechka [254]

Answer:

The distance is  D  =  0.000712 \ m

Explanation:

From the question we are told that

    The wavelength of  the  light source is  \lambda  =  700 \ nm = 700 *10^{-9} \  m

     The distance from a pin hole is  x  =  9\ m

       The  diameter of the pin  hole is  d =  1.2 \ mm  =  0.0012 \ m

     

Generally the distance which the light source need to be in order for their diffraction patterns to be resolved by Rayleigh's criterion is

mathematically represented as

              D  =  \frac{1.22 \lambda }{d }

substituting values

             D  =  \frac{1.22 * 700 *10^{-9} }{ 0.0012 }

             D  =  0.000712 \ m

5 0
3 years ago
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