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well it is connected to vo2 max which is the measurement of the lactic threshold when excercising.
Christop Hodkiewicz
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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)
How do you work out velocity acceleration?
v2 = v02 + 2a(delta x)v = sqrt(v02 + 2a(delta x))This is based on the assumption that there is constant acceleration.Another way to find velocity is using a little Calculus, this method is better since it does not assume constant acceleration, therefore this would work even if there is a change in acceleration.Since v = dx/dt, you can differentiate position with respects to time to find instantaneous velocity, given that you do know the time.And dv/dt = a, therefore dv = a x dt, integrate both sides you get velocity. This approach again requires the knowledge of time.
If a car moving with a speed of 50km per hour can be stopped by brakes after at least 6 m what is the minimum stopping distance of the same car if it is moving at a speed of 100km per hour?
we have v2- v02 = 2 a delta x at the first situation we have v= 0 ( final speed) v0= 50 ( km/ h ) = 13.9 ( m/ s ) ( initial speed ) delta x ( stopping distance ) = 6 m so a ( acceleration ) = - 16.1 ( m / s2) ( minus acceleration because speed is decreasing ) in the next situation by supposing that we have the same negative acceleration we have v=0 v0= 100 ( km / h ) = 27. 8 ( m / s ) a = - 16.1 ( m/s2) so delta x will be = 24.0 m ( stopping distance with 100 km/h speed )
How far will a falling object move in 9 seconds?
We can find the velocity after 8 seconds using the equation v = v0 + at, sov = 0 + (9.8)(8), v = 78.4. We can then use the equation v2 = v02 + 2ax to find how far it will fall:78.42 = 02 + 2(9.8)xx = 78.42/(2*9.8) = 313.6The object will fall about 313.6 meters.(Your final answer may differ a little due to rounding differences.)
What is the velocity of a mass of 75 kg about to hit the ground after falling 80 meters?
Kinetic energy is defined as Ek= ½ m v². First, a free fall implies an acceleration of -9.81 m/s². If the initial height is 80 m, the final height is 0 m, the initial speed is 0 m/s, and the acceleration is -9.81 m/s², then the "free fall" time is about 4.039 s. The final speed will therefore be -39.62 m/s. The kinetic energy will be ½(75)(-39.62)² = 58,860 J. Second, 8 g is 0.008 kg. The kinetic energy of the bullet is ½(0.008)(1000)² = 4,000 J. The free falling person will have the greatest kinetic energy.
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
What is the coordenints V02?
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.
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.
