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DIA [1.3K]
3 years ago
7

What happens when a chemical change takes place?

Physics
2 answers:
mart [117]3 years ago
5 0
<span>When chemical change takes place, the atoms in the reactants rearrange and bond together differently to form one or more new products with different characteristics. </span>
Natali [406]3 years ago
3 0
Form one or more new products with different characterstics
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A bullet starting from rest accelerates uniformly at a rate of 1,250 meters per square second. What is the bullet's speed after
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Answer:

125,000

Explanation:

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The relationship between sharks and remoras is characterized<br><br>TRUE <br>   OR<br>FALSE
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False, sharks and remoras have a symbiotic relationship. The remora removes parasites from the sharks scales, and the shark provides protection for the remora
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A ball is thrown at an angle of 38 degrees to the horizontal. What happens to the magnitude of the ball's vertical acceleration
Brut [27]

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<em>I belive it is 38 degrees</em>

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Explanation:

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3 years ago
A proton is moving toward a second, stationary proton. What happens as the protons get closer?
MatroZZZ [7]

Answer:

A. Kinetic energy is converted to electric potential energy, and the proton moves more slowly.

Explanation:

When a moving proton is brought close to a stationary one, the kinetic energy of the moving one is converted to electric potential  and the proton moves more slowly.

Kinetic energy is the energy due to the motion of a body. A moving proton will possess this form of energy.

Two protons according to coulombs law will repel each other with an electrostatic force because they both have similar charges. This will increase their electric potential energy of both of them.

Potential energy is the energy at rest of a body. As it increases, the motion of a body will be slower and it will tend towards being stationary.

5 0
3 years ago
Find the magnitude of the electric field due to a charged ring of radius "a" and total charge "Q", at a point on the ring axis a
34kurt

Answer:

E=\frac{KQ}{2\sqrt 2a^2}

Explanation:

We are given that

Charge on ring= Q

Radius of ring=a

We have to find the magnitude of electric filed on the axis at distance a from the ring's center.

We know that the electric field at distance x from the center of ring of radius R is given by

E=\frac{kQx}{(R^2+x^2)^{\frac{3}{2}}}

Substitute x=a and R=a

Then, we get

E=\frac{KQa}{(a^2+a^2)^{\frac{3}{2}}}

E=\frac{KQa}{(2a^2)^{\frac{3}{2}}}

E=\frac{KQa}{2\sqrt 2a^3}

E=\frac{KQ}{2\sqrt 2a^2}

Where K=9\times 10^9 Nm^2/C^2

Hence, the magnitude of the electric filed due to charged ring on the axis of ring at distance a from the ring's center=\frac{KQ}{2\sqrt 2a^2}

4 0
3 years ago
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