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eimsori [14]
1 year ago
10

Suppose an acorn with a mass of 3.17 g falls off a tree. At a particular moment during the fall, the acorn has a kinetic energy

of 0.119 J and a potential energy of 0.347 J with respect to the ground. How does the acorn's potential and kinetic energy change as it falls
Physics
1 answer:
denis-greek [22]1 year ago
6 0

Potential and kinetic energy both decrease with the acorn's falling potential and kinetic energy.

The acorn's potential energy is at its peak when it reaches the top of the tree, yet its kinetic energy is zero (i.e., it is not accelerating).

The height of the ball reduces along with the potential energy as the acorn tumbles down the tree, but the kinetic energy rises (energy due to motion)

The height will be 0 and the kinetic and potential energy will be zero at the ground. This demonstrates that as an item falls, both potential and kinetic energy are lost.

Learn more about Energy here

brainly.com/question/13881533

#SPJ4

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Answer:

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Answer:

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In a machine, work output is less than work input because some energy is converted into thermal energy. true or false.
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A train at a constant 60.0 km/h moves east for 40.0 min, then in a direction 50.0 degrees east of due north for 20.0 min, and th
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Answer:

Part a)

v = 7.57 km/h

Part b)

\theta = 67.5 degreeNorth of East

Explanation:

Speed of train towards East = 60 km/h

displacement towards East is given as

d_1 = 40 km

now it turns towards 50 degree East of North

so its distance is given as

d_2 = 20 km(sin50 \hat i + cos50\hat j)

d_2 = 15.3 \hat i + 12.8 \hat j

then finally it moves towards west for 50 min

d_3 = -50 \hat i

Now the total displacement of the train is given as

d = d_1 + d_2 + d_3

d = (40 + 15.3 - 50)\hat i + 12.8 \hat j

d = 5.3\hat i + 12.8 \hat j

now total time duration of the motion is given as

T = 40 min + 20 min + 50 min

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now average velocity is given as

v_{avg} = \frac{5.3\hat i + 12.8\hat j}{1.83}

v_{avg} = 2.89\hat i + 6.99\hat j

Part a)

magnitude of the average velocity is given as

v = \sqrt{v_x^2 + v_y^2}

v = \sqrt{2.89^2 + 6.99^2}

v = 7.57 km/h

Part b)

Direction of the velocity is given as

tan\theta = \frac{v_y}{v_x}

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6 0
3 years ago
Suppose that two objects attract each other with a gravitational force of 50N. If the mass
vagabundo [1.1K]

Answer:

112.5 N

Explanation:

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Let F be the new force of attraction

F/50 = ( G(3M)(3m)/(2r)^2 ) / (GMm/r^2)

[Elimiating G,M,m,r]

F = 112.5 N

7 0
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