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The Label Coordinates VBA tool for placing Latitude/Longitude labels using MicroStation V8i was developed by Elivagner Barros de Oliveira at Bentley Systems Inc. ... ODOT has placed the command to invoke the Label Geographic Coordinates (v02) dialog on the Extra Tools tool box with a Lat/Long Labeler button.
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What maximum height will be reached by a stone thrown straight up with an initial speed of 35 m's?
s= vt+1/2gt^2 v2 = v02 + 2gs 0 = 352 - 2*9.8*s (taking down as the positive direction) 62.5 meters
How do you find the maximum range and height for angles 30 45 and 60?
1. You need to know the velocity of the projectile (V0) 2. The expressions for the range and height assume no air resistance (in vacuum) 3. The units must be consistent e.g. metres and g = 9.81 m/s2 Range in metres for 30 degree launch angle = sin 60 x V02 / 9.81 Range in metres for 45 degree launch angle = sin 90 x V02 / 9.81 Range in metres for 60 degree launch angle = sin 120 x V02 / 9.81 Max. height in metres for 30 degree launch angle = (V0 x sin 30)2 / 2g Max. height in metres for 45 degree launch angle = (V0 x sin 45)2 / 2g Max. height in metres for 60 degree launch angle = (V0 x sin 60)2 / 2g 2g is of course 9.81 x 2 = 19.62 m/s2 For interest, at 45 degree launch angle the max. height is 25% of the range.
Bernoulli's theorem proof?
Proof of Bernoulli's TheoremTo prove Bernoulli's theorem, we make the following assumptions:1. The liquid is incompressible.2. The liquid is non-viscous.3. The flow is steady and the velocity of the liquid is less than the critical velocity for the liquid.Imagine an incompressible and non-viscous liquid to be flowing through a pipe of varying cross-sectional area as shown in Fig. The liquid enters the pipe with a normal velocity v11 and at a height h1 above the reference level (earth's surface). It leaves the pipe with a normal velocity v2 at the narrow end B of cross-sectional area a2 and at a height h2 above the earth's surface. at its wide end A of cross-sectional area aIf r is the density of the incompressible liquid, then in accordance with the equation of continuity, the mass m of the liquid crossing any section of the pipe is given bya1 v1 = a2 v2 = m (say) or a1 v1 = a2 v2 = m/ ..... (ii)Let P1 and P2 be the values of the pressure due to the liquid at the ends A and B respectively. If the liquid moves from the end A to B under the action of pressure difference P1 - P2, then in accordance with energy conservation principle, the work done by the pressure energy of the liquid must appear as the increase in potential and kinetic energies of the liquid.The pressure energy exerts a force P1 a1 on the liquid at the end A. The liquid covers a distance v1 in one second at the end A and thereforeWork done per second on the liquid at the end A = 1 a1 v1The liquid reaches the end B against pressure P2 i.e. against a force P2 a2. At the end B, the liquid covers a distance v2 in one second and thereforeWork done per second by the liquid at the end B = 2 a2 v2Hence, net work done by the pressure energy in moving the liquid from the end A to B in one second = 1 a1 v1 - 2 a2 v2Using the equation (ii), we haveNet work done by the pressure energy per second = 1 m/ - p2 m/ = (1 - 2) m/ ... (iii)When the mass m of the liquid flows in one second from the end A to B, its height increases from h1 to h2.Therefore, increase in potential energy of the liquid per second= m g h2 - m g h1 = m g (h2 - h1) … (iv)Further, when the mass m of the liquid flows in one second from the end A to B, its velocity increases from v1 to v2.Therefore, increase in kinetic energy of the liquid per second = 1/2 mv22 - 1/2 mv12 = 1/2 m(v22 - v12) ... (v)According to work-energy conservation principle,Work done by the pressure energy per second = increase in potential energy per second + increase in kinetic energy per secondTherefore (P1 - P2) = m/? = mg (h2 - h1) + 1/2 m(v22 - v02) or P1/? - P2/? = gh2 - gh1 + 1/2 v22 - 1/2 v12 or P1/? + gh1 + 1/2 v12 = P2/? + gh2 + 1/2 v22 ..... (vi)or P/? + gh + 1/2 v2 = constantIt proves the Bernoulli's theorem. This is the most convenient form of Bernoulli's equation. However, it can be expressed in some other forms as explained below:Multiplying both sides of the equation (vi) by ?, we haveP1 + ?gh1 + 1/2 ?v12 = P2 + ?gh2 + 1/2 ?v22 ... (vii)In this equation, each term has got dimension of pressure.Again, dividing both sides of the equation (vi) by g, we haveP1/?g + h1 + v12/2g = P2/?g + h2 + v22/2g ..... (viii)In this equation, each term has got dimensions of length. The terms P/?g, h and v2/2g are called pressure head, elevation (or gravitational) head and velocity head respectively.When the liquid flows through a horizontal pipe (h1 = h2), then the equation (vi) becomesP1/ + 1/2 v12 + P2/ + 1/2 v22
