Tuesday, May 31, 2011
Asbestos
http://www.google.co.nz/imgres?imgurl=http://www.mesotheliomalegaladvice.org/wp-content/neg_images/ce945440b1069faa055caf4c1740b094.jpg&imgrefurl=http://www.mesotheliomalegaladvice.org/asbestos-pictures&usg=__l466o4RYYrptcuOON94GTnFkjYc=&h=940&w=944&sz=142&hl=en&start=0&sig2=7XDRpNLLYT_SxdxaBaFsTQ&zoom=1&tbnid=WnsU8scSRgHOmM:&tbnh=130&tbnw=131&ei=7pDlTaO7LI6UvAPn0pS-Bg&prev=/search%3Fq%3Dasbestos%26um%3D1%26hl%3Den%26client%3Dfirefox-a%26sa%3DN%26rls%3Dorg.mozilla:en-GB:official%26biw%3D1280%26bih%3D606%26tbm%3Disch&um=1&itbs=1&iact=hc&vpx=277&vpy=104&dur=566&hovh=224&hovw=225&tx=99&ty=125&page=1&ndsp=19&ved=1t:429,r:1,s:0&biw=1280&bih=606
Asbestos can be found in Old ceilings, boiler heat ducts, Water pipes and old heat appliances
Asbestos can be found in Old ceilings, boiler heat ducts, Water pipes and old heat appliances
Sunday, May 15, 2011
Health and safety presentation
Here is my presentation when uploaded into google docs the file dosnt come out as how i made it would you be able to give you my email so i can send the direct word file ?
Sunday, May 1, 2011
My Summary
A loose electron within a conducter will travel randomly to loose spaces in other atoms valence shells untill a electromotive force is applied this will cause the electrons movement to be directed towards the positive pole causing current
Tuesday, April 5, 2011
What i Learnt
this week i learnt about the physics revolving around mass acceleration and force i also learnt formulas that use involving change of temprture, gravity and the specific heat capacity of material . I learnt about the hazards associated with fires, there different ratings and how to counteract them
Finding an unkown angle
- Step 1 The two sides we know are Opposite (300) and Adjacent (400).
- Step 2 SOHCAHTOA tells us we must use Tangent.
- Step 3 Use your calculator to calculate Opposite/Adjacent = 300/400 = 0.75
- Step 4 Find the angle from your calculator using tan-1
Tan x° = opposite/adjacent = 300/400 = 0.75
tan-1 of 0.75 = 36.9
Solving unknown side
| Start with: | sin 39° = d/30 | |
| Swap: | d/30 = sin 39° | |
| Calculate sin 39°: | d/30 = 0.6293… | |
| Multiply both sides by 30: | d = 0.6293… x 30 = 18.88 to 2 decimal places. | |
The depth the anchor ring lies beneath the hole is 18.88 m
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