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romanna [79]
3 years ago
15

A 0.50 kilogram ball is held at a height of 20 meters. What is the kinetic energy of the ball when it reaches halfway after bein

g released?
Physics
1 answer:
mars1129 [50]3 years ago
4 0
<span>(M G H)=(0.5 x 9.8 x 10) = 49 joules.</span>
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We can find the force by using the following formula;
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Fa = ma = 76 x 1.2 = 91.2
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So, the force is 836 N.
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A force of 80 N is exerted on an object on a frictionless surface for a distance of 4 meters. If the object has a mass of 10 kg,
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Your cousin Jannik skis down a blue square ski slope, with an initial speed of 3.6 m/s. He travels 15 m down the mountain side b
fenix001 [56]

Answer: The loss of energy due to friction is equal to 1,253 J.

Explanation:

The problem tells us that the skier has an initial speed of 3.6 m/s, which means that his initial kinetic energy is as follows:

K₁ = 1/2 m v₁² = 1/2 . 58.0 Kg. (3.6)² (m/s)² =  376 J

After coming to a  flat landing, his final speed is 7.8 m/s, so the final kinetic energy is as follows:

K₂ = 1/2 m v₂² = 1/2. 58.0 Kg. (7.8)² (m/s)² = 1,764 J

Now, when skying down the slope the increase in kinetic energy only can come from another type of energy, in this case, gravitational potential energy.

If we take the ground flat level as a Zero reference, the initial gravitational potential energy, can be written as follows, by definition:

U₁ = m.g. h (1)

Now, we don't know the value of the height h, but we know that the incline has a 18º angle above the horizontal, and that the distance travelled along the incline is 15 m.

By definition, the sinus of an angle, is equal to the proportion between the height and the hypotenuse , so we can write the following equation:

sin 18º = h / 15 m ⇒ h = 15 m. sin 18º = 4.6 m

Replacing in (1), we get:

U₁ = 58.0 Kg. 9.8 m/s². 4.6 m = 2,641 J

So, we can get the total initial mechanical energy, as follows:

E₁ = K₁ + U₁ = 376 J + 2,641 J = 3,017 J

After arriving to the flat zone, all potential energy has become in kinetic energy, even though not completely, due to the effect of friction.

This remaining kinetic energy can be written as follows:

E₂ = K₂ = 1,764 J

The difference E₂-E₁, is the loss of energy due to friction forces acting during the travel along the 15 m path, and is as follows:

ΔE= E₂ - E₁ = 1,764 J - 3,017 J = -1,253 J

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Ayuda porfavorrrrrrrrrrr
zhuklara [117]

what do you need help with can i help with something or do you what me to give you the answer ask me anething and i got youExplanation:

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