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max2010maxim [7]
4 years ago
14

Which example best represents translational kinetic energy?

Physics
2 answers:
Alexxandr [17]4 years ago
8 0

The correct answer is:

An apple falling from a tree is transitional kinetic energy

Explanation:

The energy of motion of a body, according to the work it would do if it were produced to rest. The translational kinetic energy depends on motion through space, and for a determined body of regular mass is equal to the output of half the mass times the intersection of the speed. and translational (the energy due to a motion from one position to another.

Lemur [1.5K]4 years ago
5 0

Answer: Option (e) is the correct answer.

Explanation:

There are three types of kinetic energy which are vibrational, rotational and translational kinetic energy.

A plucked string on a guitar  will produce sound and it represents vibrational kinetic energy.

The moving blades of a ceiling fan shows the rotation of blades and it represents rotational kinetic energy.

A spinning top also displays rotation therefore, it represents rotational kinetic energy.

A tuning fork struck on a rubber block will produce sound and it will vibrate. Therefore, it represents vibrational kinetic energy.

An apple falling from a tree shows that apple is linearly falls on the ground therefore, it represents translational kinetic energy.

Thus, we can conclude that an apple falling from a tree is the best example which represents translational kinetic energy.


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Final speed is 900.06 m/s at 0.2215^{\circ}  

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Mass of the first asteroid, m = 15\times 10^{3}\kg

Mass of the second asteroid, m' = 20\times 10^{3}\kg

Initial velocity of the first asteroid, v = 770 m/s

Initial velocity of the second asteroid, v' = 1020 m/s

Angle between the two initial velocities, \theta = 20^{\circ}

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Since, the velocities and hence momentum are vector quantities, then by the triangle law of vector addition of 2 vectors A and B, the resultant is given by:

\vec{R} = \sqrt{A^{2} + 2ABcos\theta + B^{2}}

Thus applying vector addition and momentum conservation, the final velocity is given by:

(m + m')v_{final} = \sqrt{(mv)^{2} + 2(mv)(m'v')cos20^{\circ} + (m'v')^{2}}                               (1)

Now,

(m +m')v_{final} = (35\times 10^{3})v_{final}

(mv)^{2} = (15\times 10^{3}\times 770)^{2} = 1.334\times 10^{14}

(m'v')^{2} = (20\times 10^{3}\times 1020)^{2} = 4.16\times 10^{14}

2(mv)(m'v')cos20^{\circ} = 2(15\times 10^{3}\times 770)(20\times 10^{3}\times 1020)cos20^{\circ} = 4.43\times 10^{14}

Now, substituting the suitable values in eqn (1), we get:

v_{final} = 900.06\ m/s

Now, the direction for the two vectors is given by:

\theta = sin^{- 1} \frac{m'v'sin20^{\circ}}{(m + m')v_{final}}

\theta = sin^{- 1} \frac{20\times 10^{3}\times 1020sin20^{\circ}}{(35\times 10^{3})\times 900.06} = 0.2215^{\circ}

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3 years ago
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