During heavy rain, a section of a mountainside measuring 1.9 km horizontally (perpendicular to the slope), 0.55 km up along the slope, and 2.4 m deep slips into a valley in a mud slide. Assume that the mud ends up uniformly distributed over a surface area of the valley measuring 1.4 km x 1.4 km and that the mass of a cubic meter of mud is 1900 kg. What is the mass of the mud sitting above a 3.5 m2 area of the valley floor

the mass of the mud sitting above a 3.5 m2 area of the valley floor is 8509273.5 kg

Explanation:

Given the data in the question;

first we compute the total volume of mud

V = ( 1.9×1000)m × (0.55×1000)m × (2.4×1000)m

V = 2.508 × 10⁹ m³

same volume of mud spread uniformly over 1.4 × 1.4 km² area

Hence,

V = (1.4×1000)m × (1.4×1000)m × depth

2.508 × 10⁹ = 1400m × 1400m × depth

2.508 × 10⁹ m³ = 1960000m² × depth

depth = 2.508 × 10⁹ m³ / 1960000 m²

depth = 1279.59 m

so volume of the mud sitting above 3.5 m² area of the valley floor will be;

⇒ 3.5 m² × 1279.59 m

⇒ 4478.565 m³

so mass will be

m = 4478.565 × 1900 kg

m = 8509273.5 kg

Therefore, the mass of the mud sitting above a 3.5 m2 area of the valley floor is 8509273.5 kg

a and c

c is 100% but i am confused in a