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The largest watermelon ever grown had a mass of 118 kg. Suppose this watermelon were exhibited on a platform 5.00 m above the ground. After the exhibition, the watermelon is allowed to slide along to the ground along a smooth ramp. How high above the ground is the watermelon at the moment its kinetic energy is 4.61 kJ?

Respuesta :

Answer: height = 3.98m

Explanation: by placing the watermelon at a height above the ground, it has a potential energy of the formulae

p = mgh

p = potential energy = 4.61kJ = 4610J

m = mass of watermelon = 118 kg

g = acceleration due gravity = 9.8 m/s²

4610 = 118 * 9.8 * h

h = 4610/ 118 * 9.8

h = 4610/ 1156.4

h = 3.98m

onyebuchinnaji

The height of the watermelon at the given kinetic energy is 3.99 m.

The given parameters:

  • mass of the watermelon, m = 118 kg
  • kinetic energy of the watermelon, K.E = 4.61 kJ = 4,610 J

The height above the ground from which the water melon is dropped is calculated as follows;

[tex]P.E = mgh\\\\h = \frac{P.E}{mg} \\\\h = \frac{4610}{118 \times 9.8} \\\\h = 3.99 \ m[/tex]

The height of the watermelon at the given kinetic energy is 3.99 m.

Learn more here:https://brainly.com/question/18160141

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