Question

Question 4 10 pts Brian is on Mars and decides to dump his physics textbook into this never before seen hydrocarbon lake. The atmospheric pressure in Mars is 1/4 that of Earth's and the gravitational acceleration is 3m/s^2. The hydrocarbon liquid has a density of 1300 kg/m^3. If the book is experiencing a force of 426 N underneath the lake, how deep is the book inside the lake? The book has a surface area of 6.4x10^-3 m^2. Round answer to 3 sig figs. Useful constants 1000kg 9.8m 9 G = 6.67-10-11 Mcarth 5.97. 1024 kg, Rearth 6371km, Pwater = Mgun = 2. 1030 kg, Cwater = 4,184 Next → Previous Submit Qui No new data to save. Last checked at 11:32am

Answer 1

Answer:

Given, atmospheric pressure on the Mars is 1/4 that of Earth's. Earth's atomspheric pressure is 1 atm = 101325 N/m², that means for Mars, P_atm = 1/4(101325 N/m²) = 25331.25 N/m²

Acceleration due to gravity is g = 3 m/s²

Density of hydrocarbon liquid is $\rho$ = 1300 kg/m³

The force experienced by the book at the underneath of the lake is F = 426 N

Surface area of the book is A = 6.4 x 10^-3 m²

The pressure on the book at the bottom of the lake is P = F/A = (426 N) / (6.4 x 10^-3 m²) = 66562.50 N/m²

We need to use Bernoulli's formula for the problem to get the depth of the book.

P + ρgh + 1/2 ρv² = constant.

Where P is absolute pressure, ρ is the density of the fluid, v is the velocity of the fluid, g is theaceleration due to gravity and h is the height.

We need to take the two points, one is at the surface of the lake and the second point is at the bottom of the lake or at the position of the book in the lake.

Applying Bernoulli's formula at both points.

P₁ + ρgh₁ + 1/2 ρv₁² = P₂ + ρgh₂ + 1/2 ρv₂²

Here, the lake is static, so v₁ = v₂ = 0

P₁ + ρgh₁ = P₂ + ρgh₂

At the bottom of the lake, the height is zero,i.e., h₂ = 0 .

Thus, P₂ = P₁ + ρgh₁    ------------- (1)

Here, P₂ = P = 66562.50 N/m² and P₁ = P_atm = 25331.25 N/m²

From this expression, we can understand that the pressure increases with the depth of the lake.

Eq.(1) becomes, h₁ = (P₂ - P₁) / ρg

Substituting all the values, then we get,

h₁ = (66562.50 N/m² - 25331.25 N/m²) / (1300 kg/m³) (3 m/s²) = 10.572 m

Hence, this is the depth of the physics textbook inside the hydrocarbon lake.