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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.
Orion Wisoky
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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
What is V02?
well it is connected to vo2 max which is the measurement of the lactic threshold when excercising.
What is V02 kinectics?
well it is connected to vo2 max which is the measurement of the lactic threshold when excercising.
What does vV02 mean?
i think its the same as a V02 max which is something i dont know about. good luck
Why does the stroke volume of the heart plateau at a certain VO2?
Stroke volume plateaus at a certain V02 because venous return becomes optimized when the redistribution of blood away from areas such as the gastrointestinal becomes optimized.
An object is dropped from a height of 120.0 m It's velocity upon hitting the ground is?
v2-v02=2gx,v0=0,g=9.806,x=120 m then v=48.5(m/s) in this calculation we have omit the air resistance
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
What is the speed of an object with known mass over a known distance?
There is not enough information here to answer the question, but if you know the force acting on the object, you can find acceleration and use the equation v2 = v02 + 2a*x where v is velocity, v0 is the initial velocity (often assumed to be 0), a is acceleration, and x is distance.
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.
What is the difference between vo2 max and vo2 peak?
The difference between VO2 max and VO2 peak is what the values represent. VO2 max represents the highest running oxygen amount achievable during a high intensity test. V02 peak represents the maximum value that can be reached without having it raised by a boost.
What does netball help?
Netball helps improve fitness and helps your teamswork skills. It improves ur cardio vasucalar system and can improve your v02 max,balance,agility,strength etc...
When you throw a ball in the air what type of energy does it gain as it rises?
When the ball leave your hand it goes up with an initial velocity v0 so that the action of your hand gives the ball the kinetic energy K=0.5 M v02 where M is the ball mass. When the ball goes up the kinetic energy decreases: a part is converted in potential energy while the balls is higher and higher and part is dissipates due to the attrition with air (that more precisely is due to air viscosity) and to the fact that air particles are put in motion by the arrival of the ball so that they gain kinetic energy at the expenses of the kinetic energy of the ball due to collision. Neglecting the last term that, due to the small density of air with respect to the ball is generally quite small, the viscosity dissipate energy mainly by generating heat. When the ball stops at maximum height the kinetic energy is reduced to zero (the velocity is zero) and all the energy is potential, while a certain amount of heat has been dissipated during the motion. If we call Q the quantity of generated heat and h the maximum height reached by the ball, for the energy conservation rule we have 0.5 M v02=Q+M g h If the air viscosity is negligible, or if the ball goes up in vacuum, Q is zero and we can deduce the maximum height the ball reaches h=(M v02)/( 2 M g)
What is the Kinetic Energy of 5kg of water just before it hits the turbin from the a dam 30m high?
(This assumes that the water is fallingverticallywith nohorizontalmovement.) The speed of the water at the top of the dam is 0. We can find the speed of the water at the bottom of the dam with the equation v2= v02+ 2ax, where v0is 0, a is 9.81 m/s2(accelerationdue to gravity), and x is 30.v2= 2(9.81)(30) = 588.6v = 24.26 m/sThekineticenergyis then (1/2)mv2= (1/2)(5)(24.26)2which is about 1472 Joules.
