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

hi there!

your question asks at which segment does freezing occur.

since we're talking about water, the freezing point of water will be 32 degrees fahrenheit or below.

hope this ! ~