1/2mv^2 = Mgh, What Does V Equal To? | Yahoo Answers

v=\sqrt { 2gh } v = 2gh, where h = 2 m is the depth of the orifice from the free water surface. Determine the time for a particle of water leaving the orifice to reach point B and the horizontal distance x where it hits the surface.v = SQRT (2gh) = √ (2gh) Note: SQRT (2gh) and √ (2gh) means the square root of 2gh. Note that the mass m cancels out of the equation, meaning that all objects fall at the same rate. Thus, if h = 1 ft, and since g = 32 ft/s², then v² = 2 * 32 * 1 = 64 andSolve for h v = square root of 2gh. Rewrite the equation as . To remove the radical on the left side of the equation, square both sides of the equation. Simplify each side of the equation. Tap for more steps... Simplify the left side of the equation. Remove parentheses.V = sqrt (2gh) Simple math, BUT according to Bernouli's as pressure decreases velocity increases, and since Pressure = density (g) as y decreases as the water is drained from the barrel the Pressure decreases and the velocity should increase! Intuitively the answer makes sense but according to these equations it is the opposite?Is it just because of Torricelli's law where v = sqrt(2gh)? $\endgroup$ - Horse Oct 8 '16 at 18:39 $\begingroup$ Yes you are right! The velocity v2 at the increased hole stays the same according to Torricelli's law. $\endgroup$ - freecharly Oct 8 '16 at 18:44

Potential Energy by Ron Kurtus - Physics Lessons: School

Torricelli's law, also known as Torricelli's theorem, is a theorem in fluid dynamics relating the speed of fluid flowing from an orifice to the height of fluid above the opening. The law states that the speed v of efflux of a fluid through a sharp-edged hole at the bottom of a tank filled to a depth h is the same as the speed that a body (in this case a drop of water) would acquire in fallingCalculate final velocity as a function of initial velocity, acceleration and displacement using v^2 = u^2 + 2as. Solve for v, u, a or s; final velocity, initial velocity, acceleration ar displacement. Free online physics calculators for velocity and displacement.I'm going to try and give you some hints to help you find the answer to this. After all: you'll learn nothing from getting an answer spoon-fed. First of all: what is the source of this equation? You obviously didn't "invent" it yourself. The sourc...You may occasionally her someone speak of "escaping a planet's gravity." This is really an inaccurate statement, since every object in the Universe pulls on every other object, however small that pull might be.

Solve for h v = square root of 2gh | Mathway

Calculates the free fall energy and velocity without air resistance from the free fall distance.v=sqrt(2gh/r 1 2)r 2 With the initial linear velocity know the time before the projectile crashes into the earth and the range of the projectile can be determined. As the Trebuchet utilizes a sling the angle of release cannot be controlled.What is the importance of [tex]\upsilon[/tex] = [tex]\sqrt{2gh}[/tex] in physics? I've seen it in escape velocity problems, though with an R instead of h. I've seen it with conservation of momentum applications, and it's similar to PE = mgh. Yesterday my professor was discussing Bernoulli's equation and this equation came up again.Just kidding. Let's talk about v= sqrt (2gh). A wise man once said if something doesn't make sense, check your assumptions. Well, you assumed no air resistance and hence KE = PE And "v" used in the calculation is perhaps terminal velocity which is when weight = air resistanceGh = ½ v^2 **Multiply both sides by 2** 2gh = v^2 **Take square root of both sides** V = sqrt(2gh) C. Use your equation from part B to find the speed with which the hammer struck the ground. Givens G = 9.80 H = 3.67

Here's a conceptual instance that may lend a hand answer your query:

In an energy-conserving roller coaster, there are carts that began at leisure at top of a hill of peak H and have speed v once they achieve the bottom. In order to make the vehicles reach a speed 2v we need to INCREASE the peak of the hill to 4h.

Why?

Uo + Ko = Kf + Uf

It STARTS from rest so v=0, therefore Ko might be 0, and the carts get started at SOME top so there will be attainable firstly.

When it reaches the ground, the potential ultimate will probably be 0, and there shall be some Kf since it is clearly moving at some Velocity.

mgh + 0 = 1/2 mVf^2 + 0

Vf^2 = 2mgh / m = 2gh

Square root either side

Vf = sqrt (2gh)

To make the carts reach a pace 2v...we increase the height by way of 4)

Vf = sqrt (2g(4h))

Vf = sqrt (8gh)

Vf = 2 * sqrt (2gh)

We know that Vf = sqrt (2gh), so we substitute it in Vf = 2 * sqrt (2gh) to get:

Vf = 2 Vf

In your state of affairs:

"an object falling off a cliff"

Uo + Ko = Uf + Kf

mgh + 0 = 0 + 1/2mVf^2

Vf^2 = 2gh

Vf = sqrt (2gh)

If we would have liked to find the Vf midway of its height then after all it would be:

Vf = sqrt (2g(1/2h)) = sqrt (gh).

You are proper.

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