1A
Hint 2 Subhint 3 = L^2*g*d*rho_2
Hint 2 = L^2*g*d*rho_2  +  L^2*g*(L-d)*rho_3
Hint 3 = rho_1*L^3*g
Final = L(rho_1 – rho_3) / (rho_2 – rho_3)

2A
Hint 1 = rho_b * (V_b/(100^3)) = .300kg
Final = Hint 1 * g = 2.94N

2B
Hint 2 = Hint 3 = 873 Pa
Final = 1.00N (same pressure before and after = same weight)

2C
Hint 2 = rho_fluid * (V_b/(100^3)) * g = .524
Hint 3 = (rho – rho_b)Vg + F_r
Final = Weight from Part A – Hint 2 = F_r = 2.42N

2D
Hint 2 = same as Hint 2 in 2C by Archimedes Principle
Final = orig fluid wt + (ball density – fluid density)(V_b/(100^3))(g) = 3.42N

3A
Hint 4 = Pressure of air in the building * area of tank bottom = ~164,000N
Hint 5 Subhint 1 = p_surf + pgH = 130,000 + (1000)(g)(14.4) = ~183,000N
Hint 5 = Hint5Subhint1 * Area of tank bottom = ~330,000N
Final = Hint 5 – Hint 4 = ~166,000N

4A
rho_water(V)(g)

4B
(rho_water(V)(g) – T) / g

4C
(1 – T/(rho_water*V*g))*100 <-- percentage so *100

5A
Hint 2 = A_2*v_hole
Hint 3 = sqrt(2gh + (v_surf)^2) <-- use the given equation, the rho_1 and rho_2 are equal, cancel out
Hint 4 Subhint 1 = v_hole*A_2 / A_1
Hint 4 = sqrt (2gh / (1 – (A_2/A_1)^2)))
Hint 5 = - sqrt (2gh / ((A_2/A_1)^2 – 1))
Final = 51.9s

6A
Hint 3 Subhint 1 = A*dr*rho
Hint 3 = rho*dr*A*omega^2*r
Hint 4 = F_c / A
Hint 5 = rho*omega^2*r*dr
Final = .5*rho*omega^2*(r^2 – (r_0)^2)  +  p_0

7A
Hint 1 = rho*g*(h-x)
Final = rho*g*(h-x)*L*dx

7B
Hint 3 = rho*g*(h-x)*x*L*dx
Final = rho*g*L*h^3 / 6

8A
1.50m/s <-- flow rate in m^3/s over cross section area, same for 8B

8B
6.00m/s

8C
From lecture, .5*rho*((velocity at small opening)^2 – (velocity at big opening)^2) where rho is density of water (1000kg/m^3) = 1.69E4

8D
Find pressure at each point, convert to mmHg then to cmHg and find the difference = 12.7cm (1 Pa = .0075006 mmHg)

9A
F / (rho_water*l*w*g)

9B
Water level SINKS

9C
Part A – (F/(rho*l*w*g)) <-- NOTE that the rho in 9A is DIFFERENT from the rho here.  There is a rho_water and a plain rho.

10A
Rho*g*h_1

10B
Hint 2 = rho*g*h_2
Hint 3 Subhint 1 = p_1 + .5*rho*((v_1)^2 – (v_2)^2)
Hint 3 Subhint 2 = A_1*v_1 / A_2
Hint 3 = p_1 + .5*rho*(v_1)^2 * (1 – (A_1/A_2)^2)
Final = sqrt (2g(h_2 – h_1) / (1 – gamma^2)

11A
Hint 2 = dV/dt (givens) = ~83,300
Hint 3 = 40,000
Final = -40,000

11B
Hint 2 = .000314
Hint 3 = ~1.27E8
Hint 4 = Hint 3
Hint 5 = 3.32E-3
Final = 4.23E5

11C
The pressure that the water exerts on the walls of the glass bottle is greater than the pressure required to break the glass.