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A 2 kg object released from rest at the top of a tall cliff reaches a terminal speed of 37.5 m/s after it has fallen a height of 110 m . How much kinetic energy approximately did the air molecules gain from the falling object after it has fallen through this height?

Respuesta :

Answer:

3562.25 J

Explanation:

m = mass of the object = 2 kg

v = speed of the object after falling through height of 110 m , = 37.5 m/s

h = height through which the object fall = 110 m

KE = Kinetic energy gained by air molecules

Kinetic energy gained by air molecules is given as

KE = (0.5) m v² + mgh

KE = (0.5) (2) (37.5)² + (2) (9.8) (110)

KE = 3562.25 J

The air molecules have Kinetic energy of 3562.25 Joules.

How do you calculate the kinetic energy?

Given that, an object of mass 2 kg releases from the top with a speed of 37.5 m/s falls from the height of 110 m.

The kinetic energy of the air molecules will be the sum of the potential energy of the molecules and the kinetic energy of the object.

[tex]KE = \dfrac{1}{2} mv^2 + mgh[/tex]

Where, m is the mass of the object, v is the speed of the object, h is the height and g is the gravitational acceleration.

[tex]KE = \dfrac{1}{2}\times 2\times (37.5)^2 + 2\times 9.8\times110[/tex]

[tex]KE= 1406.25+2156\\[/tex]

[tex]KE=3562.25 \;\rm J[/tex]

Hence, the Kinetic Energy of the air molecules is 3562.25 Joules.

For more details about kinetic energy, follow the link given below.

https://brainly.com/question/999862.