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Musya8
1 month ago
13

A 1.0-kg block and a 2.0-kg block are pressed together on a horizontal frictionless surface with a compressed very light spring

between them. They are not attached to the spring. After they are released and have both moved free of the spring
the lighter block will have more kinetic energy than the heavier block.
the magnitude of the momentum of the heavier block will be greater than the magnitude of the momentum of the lighter block.
the heavier block will have more kinetic energy than the lighter block.
both blocks will both have the same amount of kinetic energy.
both blocks will have equal
Physics
2 answers:
Softa [3K]1 month ago
8 0
Once the blocks are released and have moved away from the spring, the lighter block will possess greater kinetic energy compared to the heavier block. Therefore, the correct choice from the provided options is the first one. I trust this helps you as intended.
Maru [3.3K]1 month ago
6 0
The correct answer is that the momentum of the heavier block will exceed that of the lighter one. It's important to note that momentum differs from energy and speed. It can be described as the product of an object's mass and its velocity. Consequently, a heavier object with greater mass will have increased momentum, in contrast to a lighter object which will not achieve the same momentum. Thus, when comparing both blocks, the heavier mass results in higher momentum, leading to the conclusion that the second option is correct.
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Response:

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Clarification:

Provided:

h(t) = 25 ft/sec

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At this point, we can determine the distance between the individual and the helicopter utilizing the Pythagorean theorem

D(t) = \sqrt{h^2 + x^2}

Now, let's calculate the derivative of distance in relation to time

\frac{dD}{dt} (t) = \frac{2h \cdot \frac{dh}{dt} +2x \cdot\frac{dx}{dt}} {2\sqrt{h^2 + x^2}}

By plugging in the values for h(t) and x(t) and simplifying, we arrive at,

\frac{dD}{dt}(t) = \frac{50t \cdot \frac{dh}{dt} + 20 \cdot \frac{dx}dt}{2\sqrt{625\cdot t^2 + 100 \cdot t^2}}

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