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Zigmanuir [339]
3 years ago
11

During heavy rain, a section of a mountainside measuring 2.5 km horizontally, 0.80 km up along the slope, and 2.0 m deep slips i

nto a valley in a mud slide.Assume that the mud ends up uniformly distributed over a surface area of the valley measuring 0.40 km 0.40 km and that mud has a density of 1900 kg/m3.What is the mass of the mud sitting above a 4.0 m2 area of the valley floor?
Physics
1 answer:
Crank3 years ago
4 0

Answer:

The mass of the mud is 3040000 kg.

Explanation:

Given that,

length = 2.5 km

Width = 0.80 km

Height = 2.0 m

Length of valley = 0.40 km

Width of valley = 0.40 km

Density = 1900 Kg/m³

Area = 4.0 m²

We need to calculate the mass of the mud

Using formula of density

\rho=\dfrac{m}{V}

m=\rho\times V

Where, V = volume of mud

\rho = density of mud

Put the value into the formula

m=1900\times4.0\times0.40\times10^{3}

m =3040000\ kg

Hence, The mass of the mud is 3040000 kg.

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Answer:

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6 0
3 years ago
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Novay_Z [31]

Answer:

E =  ρ ( R1²) / 2 ∈o R

Explanation:

Given data

two cylinders are parallel

distance = d

radial distance = R

d < (R2−R1)

to find out

Express answer in terms of the variables ρE, R1, R2, R3, d, R, and constants

solution

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so area is 2 \pi R × l

and we apply here gauss law that is

EA = Q(enclosed) / ∈o   ......1

so first we find  Q(enclosed) = ρ Volume

Q(enclosed) = ρ ( \pi R1² × l )

so put all value in equation 1

we get

EA = Q(enclosed) / ∈o

E(2 \pi R × l)  = ρ ( \pi R1² × l ) / ∈o

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Answer:

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a = 73.36 m/s².

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