VOLUMEcm3Vw=saturation=cmcm3VT-295 cm³water content=porosityvoid ratio=Total density (g/cm³)Dry density (g/cm³).AIRWATER 1 g/cm³3SOLIDSRD 2.70MASSMa-0 gramsMwgramsM = 415 gramsMT-490 grams

Answered: Image /qna-images/answer/8c0f8f95-7a8f-4fcc-ac2b-1321eb0f39af.jpg

VOLUME Va cm Vw= cm VT-295 cm3 water content= porosity void ratio= saturation Total density (g/cm³) Dry density (g/cm3). AIR WATER 1 g/cm³ SOLIDS RD 2.70 MASS Ma 0 grams Mw" grams M₂-415 grams My 490 grams

VOLUME Va cm Vw= cm VT-295 cm3 water content= porosity void ratio= saturation Total density (g/cm³) Dry density (g/cm3). AIR WATER 1 g/cm³ SOLIDS RD 2.70 MASS Ma 0 grams Mw" grams M₂-415 grams My 490 grams

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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Biological Processes

Biological solid waste includes bacterial and contaminated waste produced by living individuals. Its treatment is essential to prevent the spread of any air-borne or water-borne diseases or diseases transmitted from one organism to another in a food chain. There are many benefits of the biological solid waste treatment and some of them are: the minimized quantity of waste substance, waste control, eradicating any disease-causing pathogens, generation of biogas, composting, anaerobic digestion, bio drying, mechanical biological treatment, microbial fuel cell, phytoremediation, and biofiltration. The final products of the biological solid waste treatment may be recycled to obtain organic fertilizers or disposed of in SWDS (Solid Waste Disposal Site) after proper treatment.

Question

VOLUME
cm3
Vw=
saturation=
cm
cm3
VT-295 cm³
water content=
porosity
void ratio=
Total density (g/cm³)
Dry density (g/cm³).
AIR
WATER 1 g/cm³3
SOLIDS
RD 2.70
MASS
Ma-0 grams
Mw
grams
M = 415 grams
MT-490 grams

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