Homework 2

2-1

$F= pA$ where $A$ is the area. So $F = 2.90 \times 10^4 ~\rm N$.

2-2

Assuming a constant gravitational constant and that the atmospheric pressure is completely due to the weight of the atmosphere, we can obtain the total net force on the surface of the earth and divide it by the gravitational constant. $A = 4\pi R^2 = 1.60\times 10^{15} ~\rm m^2$, and $F= Mg = pA = 1.62\times 10^{20} ~\rm N$. Therefore, $M = 1.65 \times 10^{19}~\rm kg$.

2-3

According to the phase diagramme, the minimum pressure at which diamond can form at $T=1000~\rm C$ is $P=4 ~\rm GPa$. Assuming the uniform density of $\rho = 3.1~\rm g/cm^3$, $P= \rho gh$ is satisfied. Therefore, $h = P/\rho g = 1.32\times 10^5 ~\rm m$.

2-4

(a) The total weight of the water is $W= \rho g h A= 6.06\times 10^9 ~\rm N$. (b) The pressure is $P= \rho g h = 2.1\times 10^7 ~\rm Pa \approx 20~ atm.$ IMHO, you cannot survive in this condition with a bare body.

2-5

Since the pressure is $P= \rho gh$, we can set up an equation of compensation $\rho_1 g h_1 = \rho_2 g h_2$. In other words, $\rho_1 = 2.9~\rm g/cm^3$, $\rho_2 = 3.3 ~\rm g/cm^3$, $h_1 = D+6 ~\rm km$, and $h_2 = D$. Then $D = 43.5 ~\rm km$.

2-6

The pressure at equilibrium is $P= \rho_w g(h_1+h_2)=\rho_m g h_2$ where $\rho_w,~\rho_m$ are respectively the density of water and mercury. Then $h_1= h_2(\rho_m - \rho_w)/\rho_w = 12.6 ~\rm cm$.

2-7

When the balloon and the payload are floating in the air, the total net force on the balloon is zero. That means that the volume of the balloon is such that the air occupying the same volume would have the same total mass (weight) as the total mass (weight) of the balloon, helium, and the payload. Let $M$: mass of the payload and the balloon, $V$: volume of the balloon, $\rho_a$: density of the air, and $\rho_b$: density of helium. Then $M + \rho_bV = \rho_a V$. So $V= M/(\rho_a - \rho_b)=1.84\times 10^3 ~\rm m^3$.

2-8

$m$: mass of the aluminium. $\rho_a$ density of the aluminium. $\rho_w$: density of the water. (a) In the air, the tension is the same as the weight $mg = 9.8 ~\rm N$. (b) In the water, the buoyant force is $-\rho_w g m/\rho_a = -3.62 ~\rm N$. So the total tension to sustain the aluminium is $6.17~\rm N$.

2-9

$R = D/2$: radius of the pipe. $r=d/2$: radius of the torpedo. $v_1$: the speed of the water passing the torpedo. (a) From the equation of continuity of the flow, $A_1v_1 = A_2v_2$, or $\pi (R^2-r^2)v_1 = \pi R^2 v_2$. Then $v_2 =2.4 ~\rm m/s$. (b) From Bernoulli's equation, $P_1 + {1\over 2} \rho v_1^2 = P_2 {1\over 2} \rho v_2^2$. So $\Delta P = {1\over 2} \rho (v_1^2 -v_2^2) = 245 ~\rm Pa$.

2-10

Use the Bernoulli's equation to obtain the pressure difference and multiply the pressure by the area of the wing.