astarloa

How can I find the volume of water lost from the cup?

"The actual goblet has two parts; a little base just one, the place that the advantage can be straight, and a large second i, where the sharpness is definitely diagonal. The lower portion incorporates a height associated with 0.5cm as well as a diam involving five.3cm. I've truly ray-calculated of which several times, it truly is undoubtedly 5 various.several, definitely not five. Inside top area the actual dimension in the bottoom of computer is definitely 5 various.3cm and also at the most notable this diam is definitely 7cm. The particular angle distance is definitely 8cm. Observe that this is actually the length of the slant; it's not a new directory description. With this particular selective information, precisely how might My partner and i discover the volume associated with water system that's been shed if your water table started at a vertical stature of seven.5cm, in addition to dropped by way of directory stature of just one.5cm? (vertical this time, not slant) Posting the formula (beyond just the answer, so i could make a point I am just doing your computations appropriate) can be considerably appreciated, as I should repeat this deliberation many different amounts of water-decrease."
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ringostarr
"The water in the bottom cylinder is always there, so ignore the cylinder. The top of the cone has a radius of 3.5cm and the bottom, a radius of 2.65cm. Extend a vertical line from the edge of bottom of the cone. This will form a right triangle with a hypotenuse of 8cm and a width of 0.85cm (3.5 - 2.65). The height of this triangle is thus 7.955cm. At a height of 1cm (1.5 minus the ignored 0.5 cylinder), the width of the triangle would be .85/7.955 or .11cm. At a height of 7cm, the width would be .75cm. You now have all the measurements of a frustrum. R = 2.65+.75 = 3.4cm, r = 2.65+.11=2.76cm, and h=7-1=6 so V=({pi}h/3)×(R² + Rr + r²) = 179.458cm³"
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sathyai
"Any size cup? "
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