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Natasha2012
1 month ago
14

A steel cable lifting a heavy box stretches by ΔL . In order for the cable to stretch by only half of ΔL , by about what factor

must its diameter increase? 1.4 4.0 2.0 0.25 0.50
Physics
1 answer:
Keith_Richards [3.2K]1 month ago
6 0

Answer:

2.0

Explanation:

because I'm a geek and ik

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A player throws a football 50.0 m at 61.0° north of west. what is the westward component of the displacement of the football?
inna [3103]

Answer: 24.24 m

Explanation:

A player launches a football 50.0 m at an angle of 61° to the north of west. We will break this down into vertical and horizontal elements.

Horizontal component: 50 cos 61° = 24.24 m directed westward

Vertical component: 50 sin 61° = 43.73 m directed toward the north.

Refer to the diagram below.

Therefore, the westward displacement of the football corresponds to the horizontal component of the displacement, which is 24.24 m.

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Megan rode the bus to school, which is located 8 kilometers from her home. If Megan's frame of reference is her house, and it to
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Explanation: I don't know, sorry.
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Define kinetic energy and thermal energy. Describe what happens to each as the temperature of a substances increases.
Keith_Richards [3271]
Kinetic energy refers to the energy an object possesses while in motion, whereas thermal energy corresponds to heat energy. In situations where heat increases in materials, such as a solid turning into a liquid, the molecules start to move more rapidly, resulting in a rise in kinetic energy.
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1 month ago
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An end of a light wire rod is bent into a hoop of radius r. The straight part of the rod has length l; a ball of mass M is attac
ValentinkaMS [3465]

Answer:

arcsin(\frac{R\mu}{(R+l)\sqrt{\mu^2+1}})

Explanation:

By applying the Law of Sines,

sin \theta = \frac{sin \phi R}{ l + R}

Based on Newton's Law,

mg = N\sqrt{\mu^2+1}

And the final equation also derived from Newton's Law,

\mu N = mgsin\phi

Then by consolidating all the equations together,

\mu N = mgsin\phi = N\sqrt{\mu^2+1}sin\phi\\

sin\theta = \frac{\mu R}{ (l + R)\sqrt{\mu^2+1}}

Thus,

\theta = arcsin(\frac{R\mu}{(R+l)\sqrt{\mu^2+1}})

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