please help a tough question for physics?

Question

Fluids and dynamics?
1. Water flows through a 2 cm opening in the side of a tank. If the opening is 8 m below the surface of the water,what volume of a water would escape from the tank per minute?If a stream of water escapes from another hole 2 m above the base of the tank and strikes the ground at a horizontal distance of 25 m.How far below the surface of the water is the hole?

2. A sphere 1 m in diameter floats half submerged in a tank of oil( Sg = 0.80). a.) What is the upward force acting on the sphere? b.) What is the minimum weight of an anchor w/ a density of 20,000 N/m^3 that will be required to submerge the sphere completely? c.) What minimum weight of a body placed on the top of the sphere will be required to just submerge sphere completely?

Answer 1

1.
The velocity of the outflowing water in the opening is given by Torricelli's law:
v = √(2·g·h)

For the first opening:
v = √(2·9.81m/s·8m) = 12.53m/s
The volume flow is given by the product of area of the opening times fluid velocity:
V = A·v = π/4·d²·v = π/4·(0.02m²)·12.53m/s = 3.936·10⁻³m³/s = 0.2362m³/min
( I take the opening as circle of diameter 2cm, the question is a not clear about this)

For the second opening:
The trajectory of the water stream is the same as the trajectory of a projectile launched at the height y₀with an angle θ to the horizontal and the velocity v. The "range of the projectile" is given by:
d = v·cosθ/g · (v·sinθ + √((v·sinθ)² + 2·g·y₀)
For θ=0 this reduces to:
d = v·√(2·y₀/g)
The velocity at the opening is
v = d·√(g/(2·y₀))
Combine this with Torricelli's law to get the water surface:
h = v²/(2·g) = d²·g/(2·y₀)·1/2g= d²/(4·y₀) = (25m)²/(4·2m) = 78.125m

2.
a)
The lifting force on a floating body is equal to the gravity force of the displaced fluid volume.
Fa = V·ρ·g = π /12·d³·ρ·g = π /12·(1m)³·(0.8·1000kg/m³)· 9.81m/s² = 2054.6 N
b)
The lift force exerted on the completely submerged sphere is twice as high as the force Fa. To submerge the sphere completely the anchor must exert an additional force of magnitude Fa on the sphere. The force exerted by the anchor is its weight minus the buoyant force acting on the anchor:
Fa = W - V·ρ·g = W(1 - ρ·g / ρa)
where ρa is the density of the anchor in terms of weight per volume.
The weight of the anchor is:
W = Fa / (1 - ρ·g / ρa) = 2054.6N/(1-(0.8·1000·9.81)/ 20000) = 3381.5 N
c)
Same idea as in problem 2b). Since the is no buoyancy force acting on the body on the sphere its weight is:
W' = Fa = 2054.6N

Answer 2

se Bernoulli's principle
assume, 2cm means the radius of a circular opening. Further assume that that area of the top of the tank
is >> that the area of the opening.

we have the equation


p1+ρ1gy1+1/2ρ1v1^2
=p2+ρ2gy2+1/2ρ2v2^2
ρ1=ρ2

Now because of the area difference v1~0. and
because the tank is exposed to atmospheric
pressure p1=p2. So we have ρ1gy1=1/2ρ1v2^2

gy1=1/2v2^2

solving for v2

v2=sqrt(2gy1)
=sqrt(2*9.9*8)=12.5 m/s

dV/dt=A2v2

A2=πr^2=π*(0.02)^2=1.26E-3 m^2

dV/dt=1.26E-3 m^2*12.5 m/s=1.57 m^3/s

in this case we need to use equations of
projectile motion to determine the exit velocity
and then work backward to find the height of
the hole from the surface

we set up the projectile motion equations
ay, vy. ax, and vx aren"t a*y or v*x but a subscript
h is the height above the base=2
v is the exit velocity
ay=-g
vy=-gt
y=h-1/2gt^2

in the x direction
ax=0
vx=v
x=vt

combining the 6 equations we get the
parabolic projectile motion

y=h-1/2g(x/v)^2

when x=25 and y=0 (the water has hit the ground) we can solve for the initial velocity of the exit.

v=x*sqrt(g/(2h)
=25*sqrt(9.9/(2*2)
=39.3 m/s

now we can go back to Bernoulli and get the height
from the surface. We'll just reuse gy1=1/2v2^2 but this
time solving for y1

y1=1/2v2^2/g=1/2(39.3)^2/9.8
=78.9 m

2. Using Archimedes principle we now the buoyant force is equal the the weight of the volume displaced. for 1/2 a sphere the volume is 1/2*4/3*π*r^3 and the weight is the volume times the specific gravity times the density of water (1000 kg/m^3) so:
F=1/2*4/3*π*(1/2)^3*.8*1000
=209.4 N

Let's answer c first
instead of half the sphere it's the whole sphere so it's 2x the value above or 418.9 N

418.9 N has to also be the force fulling down but the anchor is in the liquid.

first determine the volume of the anchor. 418.9 N/20000 N/m^3=2.1E-2 m^3

The increase in force using Archimedes is:

1000 kg/m^3*.8*2.1E-2 m^3=16.8 N so, the anchor must weigh 418.9+16.8= 435.7 N