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How much potential energy does a ball have when it reaches the top of its ascent being 1kg with a velocity of 30msec?
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What happens when potential energy and kinetic energy of a ball as it rolls down a hill?
Although your question was silent on the matter, we assume that the ball is at rest at the top of the ramp. At that point, it has no kinetic energy, since it's not moving. Its energy is all potential (due to gravity). When you release the ball to roll down the ramp, potential energy is converted to kinetic energy. (Remember, energy can't be created or consumed; it merely changes its form.) You can calculate the ball's potential energy by multiplying its height, h, and its mass, m, by the acceleration of gravity, g. That is, Ep = hmg. Kinetic energy can be calculated by multiplying 0.5 by the object's mass and the square of its velocity, v2. That is, Ek = (1/2)mv2. So, if you equate the two equations (Ep = Ek) you get hmg = (1/2)mv2. That is a powerful formula that allows you to solve for v if you know h and vice versa. If the ball has significant size and mass and it rolls down the ramp the kinetic Energy gets a little more complicated. The (1/2)mv^2 refers to the linear velocity of the center of mass but you should also add to that the rotational energy of the ball about its center of mass. This formula is (1/2)Iw^2; where I is the ball's moment of inertia and w is the balls angular velocity
How fast must an object be thrown up if it is to stay in the air for 5 seconds before hitting the ground again?
An object can be thrown vertically upwards or at an angle to the ground, in both cases it is needed that time of flight be 5 seconds. This means it's time of ascent (going up) is 2.5 seconds and time of descent(coming down) is also 2.5 seconds. So, it reaches highest point 2.5 seconds after it is thrown. At highest point the vertical component of velocity1 of the object becomes zero for an instance.Now, kinematics equation can be used to solve this.v = uy - g*twhere v is final velocity at top =0. uy is initial vertical velocity1. g is accleration due to gravity(9.8ms-2). t is time of ascent.putting values we get.0 = uy - 9.8 * 2.5or uy = 24.5 ms-1So we need to throw an object with vertical velocity = 24.5 meters per second so that it remains in air for 5 seconds.1. If object is thrown at an angle then vertical component of velocity of projection is taken.If object is thrown vertically upwards then vertical component of velocity of projection issame as velocity.vertical component of velocity of projection(uy) = u*sin(θ), where u is velocity of projectionand θ is angle of projection with respect to horizontal.2. Accleration due to gravity is different for different places. 9.8ms-2 is an approximatevalue.3. Here, air resistance and wind speeds have been neglected as they make the calculation verytedious and they are always varying from time to time and place to place.
When a rock thrown straight up climbs for 3.00 s before falling. Neglecting air resistance with what velocity did the rock strike the ground?
You can use Newton's equations of motion: At the top of the climb its velocity u = 0 m/s Its acceleration is acceleration due to gravity a ≈ 9.8 m/s Time of descent t = time of ascent = 3.00 s (I'll assume positive is towards the ground) v = u + at ≈ 0 m/s + 9.8 m/s² × 3.00 s = 29.4 m/s HOWEVER, this is the velocity (towards the ground) reached when the rock has returned to height from which it was thrown (released) above the ground - unless the rock was "thrown" by an explosive force at ground level, the rock will not have reached the ground at this point: there is still the distance from which it was "thrown". Which means its final velocity at ground level can be found using: v² = u² + 2as v = velocity it hits the ground u ≈ 29.4 m/s (as found above) s = distance above ground from which the rock was "thrown" = height_of_throw m a = acceleration due to gravity ≈ 9.8 m/s → v² = u² + 2as → v ≈ √((29.4 m/s)² + 19.6 m/s² × height_of_throw m) = √(864.36 + 19.6 × height_of_throw) m/s
What happens to the pressure of the air outside of the balloon as it goes higher?
The pressure inside the balloon will be higher because the balloon will try to get smaller and thus the balloon will ascent due to the low density of the helium inside the balloon.
Would the springs inside a bathroom scale be more compressed or less compressed if you weighed yourself in an elevator that moved upward at constant velocity?
While you're moving up or down at constant speed, the situation is exactly the same as standing on the ground. The scale reads your weight, and the springs are in the same condition as they are when the scale is on the floor of your bathroom. -- During the brief few seconds when you're accelerating downward ... starting down from an upper floor or stopping your ascent at an upper floor ... your weight temporarily seems to be decreased, and the springs are less compressed. -- During the brief few seconds when you're accelerating upward ... starting up from a low floor or stopping your descent at a low floor ... your weight temporarily seems to be increased, and the springs are more compressed.
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