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2 years ago

P2O5 is a covalent compound used to purify sugar. What is the name of this compound?

Physics

2 answers:

AfilCa [17]2 years ago

8 0

Answer:

B) Diphosphorus pentoxide

Explanation:

tigry1 [53]2 years ago

5 0

Answer:

diphosphorus pentoxide

Explanation:

just checked it!! <3 gl

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Bob and John are pulling in different directions. If Bob is pulling to the right with a force of 10N, and John is pulling to the

kykrilka [37]

Answer:

A) -2N

B) Left

C) -0.5

Explanation:

A) -12 + 10

B) More force is acted on in that direction

C) Net force/Mass (-2/4)

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2 years ago

Which statement best explains acceleration?

Alexxx [7]

Acceleration is the ratio of a change in velocity to the time over which the change happend

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3 years ago

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What are the three properties of a magnet?

Mamont248 [21]

1. Magnets attract iron and other ferromagnetic materials such as neodymium and cobalt.

2. Magnets attract or repel other magnets.

3. In addition one part of a magnet will always point north when allowed to swing freely.

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2 years ago

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Two horizontal rods are each held up by vertical strings tied to their ends. Rod 1 has length L and mass M; rod 2 has length 2L

antiseptic1488 [7]

Answer:

Rod 1 has greater initial angular acceleration; The initial angular acceleration for rod 1 is greater than for rod 2.

Explanation:

For the rod 1 the angular acceleration is

$\tau_1 = I_1\alpha _1 \\\\\alpha_1 = \dfrac{\tau_1}{I_1}$

Similarly, for rod 2

$\alpha_2 = \dfrac{\tau_2}{I_2}.$

Now, the moment of inertia for rod 1 is

$I_1 = \dfrac{1}{3}ML^2$ ,

and the torque acting on it is (about the center of mass)

$\tau_1 = Mg\dfrac{L}{2};$

therefore, the angular acceleration of rod 1 is

$\alpha_1 = \dfrac{Mg\dfrac{L}{2}}{\dfrac{1}{3}ML^2},$

$\boxed{\alpha_1 = \dfrac{3g}{2L} }$

Now, for rod 2 the moment of inertia is

$I_2 = \dfrac{1}{3}(2M)(2L)^2$

$I_2 = \dfrac{8}{3} ML^2,$

and the torque acting is (about the center of mass)

$\tau _2 = (2M)g \dfrac{(2L)}{2}$

$\tau _2 = 2MgL;$

therefore, the angular acceleration $\alpha_2$ is

$\alpha_2 = \dfrac{2MgL;}{\dfrac{8}{3} ML^2,}.$

$\boxed{\alpha_2 = \dfrac{3g}{4L}}$

We see here that

$\dfrac{3g}{2L} > \dfrac{3g}{4L}$

therefore

$\boxed{\alpha_1 > \alpha_2.}$

In other words , the initial angular acceleration for rod 1 is greater than for rod 2.

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2 years ago

Who will it benefit if they do butterfly groin stretch

igor_vitrenko [27]

Answer:

ummmmm mmmmm butterflies :)

Explanation:

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2 years ago

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