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3 years ago

Why is an iron bolt attracted to a magnet?

Physics

2 answers:

icang [17]3 years ago

8 0

Answer:

A. When it is in a magnetic field, it becomes a temporary magnet.

Explanation:

An iron bolt is attracted to a magnet because when in a magnetic field, the iron becomes a temporary magnet.

This is because the iron aligns their electrons in the magnetic fields.

This causes that attraction between the magnet and the iron.
Metals like iron are said to be ferromagnetic
Unpaired electrons in iron spin in such a way that they align with the magnetic fields of the magnet.

PSYCHO15rus [73]3 years ago

3 0

Answer:

Explanation:

just took the test late but its right

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a child in a tree house uses a rope attached to a basket to lift a 26 nn dog upward through a distance of 5.0 mm into the house.

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Total work done is 0.13 Joules

The sum of the displacement and the component of the applied force of the object in the displacement direction is the work done by a force.

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1 year ago

A thin uniform rod of mass M and length L is bent at its center so that the two segments are now perpendicular to each other. Fi

Tatiana [17]

Answer:

(a) I_A=1/12ML²

(b) I_B=1/3ML²

Explanation:

We know that the moment of inertia of a rod of mass M and lenght L about its center is 1/12ML².

(a) If the rod is bent exactly at its center, the distance from every point of the rod to the axis doesn't change. Since the moment of inertia depends on the distance of every mass to this axis, the moment of inertia remains the same. In other words, I_A=1/12ML².

(b) The two ends and the point where the two segments meet form an isorrectangle triangle. So the distance between the ends d can be calculated using the Pythagorean Theorem:

$d=\sqrt{(\frac{1}{2}L) ^{2}+(\frac{1}{2}L) ^{2} } =\sqrt{\frac{1}{2}L^{2} } =\frac{1}{\sqrt{2} } L=\frac{\sqrt{2} }{2} L$

Next, the point where the two segments meet, the midpoint of the line connecting the two ends of the rod, and an end of the rod form another rectangle triangle, so we can calculate the distance between the two axis x using Pythagorean Theorem again:

$x=\sqrt{(\frac{1}{2}L)^{2}-(\frac{\sqrt{2}}{4}L) ^{2} } =\sqrt{\frac{1}{8} L^{2} } =\frac{1}{2\sqrt{2}} L=\frac{\sqrt{2}}{4} L$

Finally, using the Parallel Axis Theorem, we calculate I_B:

$I_B=I_A+Mx^{2} \\\\I_B=\frac{1}{12} ML^{2} +\frac{1}{4} ML^{2} =\frac{1}{3} ML^{2}$

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