physics water tank?

Question

Figure P9.81 shows a water tank with a valve at the bottom. If this valve is opened, what is the maximum height attained by the water stream coming out of the right side of the tank? Assume that h = 9.0 m, L = 1.50 m, and = 35° and that the cross-sectional area at A is very large compared with that at B.
m (above the level where the water emerges)

http://smg.photobucket.com/albums/v296/operaprincess315/?action=view¤t=p9_81.gif

Answer 1

The velocity of the water leaving the spout is the same as if it had fallen through a height equal to the depth of the spout.

The opening is h - l sin (theta) below the surface, so the velocity of the water can be found by solving

1/2 v^2 = g (h - l sin (theta))

The rest is kinematics; you have your initial velocity and angle and you can easily calculate the maximum height.

Answer 2

artwork may well be the substitute in skill capability required to improve the water to its very final top. substitute in skill capability is the burden of the water cases the substitute in that is top (making use of z for the vertical distance). The water is being pumped from floor point which may well be z = 0. the in simple terms right top is variable reckoning on what top in the sector the water finally ends up at. making use of a radius R (= 50 yet i will use R for now) and an perspective A between the radius and the horizontal, distance from the backside of the tank will become R + R*sin(A). this provides 0 on the backside (A = -pi/2) and a pair of*R on the real (A = pi/2). you're able to think of of the tank as being divided into horizontal tiers the place each and every of the water in any given point has a similar skill capability. So we choose the burden interior each and every point and that's in simple terms the element of quantity shaped like a disk cases the density of the water. Use a disk shaped element of quantity parallel to the floor so as that dV = pi*R^2*dz z = R*cos(A) dV = pi*R^2*cos^2(A)dz dW = d*pi*R^2*cos^2(A)dz ... d = density of water = sixty two.4 lbs/ft^3 placed dz in terms of dA z = R*sin(A) so dz = R*cos(A)dA dW = d*pi*R^3*cos^3(A)dA substitute in skill capability = weight cases substitute in top dPE = dW*z dPE = z*d*pi*R^3*cos^3(A) dA all of us understand the top in the tank and that's R + R sin(A) all of us understand the backside of the tank is 200 ft So z = 200 + R + Rsin(A) Now for the genuine numbers: z = 200 + 50 + 50sin(A) z = 250 + 50 sin(A) dPE = (250 + 50 sin(A))*d*pi*R^3*cos^3(A) dz dPE = (250 + 50 sin(A))*d*pi*R^3*cos^3(A) dz dPE = (d*pi*R^3)[250cos^3(A) + 50sin(A)cos^3(A)]dA Now combine this: PE = (d*pi*R^3){250*[sin(A) - sin^3(A)/3] - 50cos^4(A)/4} And now evaluate from A = -pi/2 to A = pi/2 The cos term would be 0 and the sines would be a million and -a million PE = 2*(d*pi*R^3){250*[2/3]} PE = 500*(2/3)*pi*sixty two.4*50^3 PE = 2600000000*pi PE = 8168140899 ft-lb notice: Ted S made a mistake while he claimed that x^2 + y^2 = 50^2. it is barely real on the middle of the sector. you could certainly see this by making use of thinking bearing directly to the real or the backside. needless to say x^2 + y^2 = 0 at the two those factors. that is going to be x^2 + y^2 = [R*cos(A)]^2. by making use of making use of what he did he's changing the tank from a sphere to a cylinder.