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Alekssandra [29.7K]
3 years ago
15

During which type of collision is none of the energy converted to heat

Physics
2 answers:
Gre4nikov [31]3 years ago
4 0
This would be typical of an elastic collision.
8090 [49]3 years ago
4 0

Answer:

thank you to the other person, the answer was elastic

Explanation:

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What do electrons in the same shell have in common? AThey have the same amount of energy incorrect answer BThey are all positive
Keith_Richards [23]

Answer:

They have the same amount of energy

Explanation:

Electrons are said to be the subatomic particles that move around the nucleus of an atom. These electrons are negatively charged particles that are seen to be quite smaller than the nucleus of an atom.  

The electron shells of these atoms are usually being filled from the inside out with the low-energy shells closer to the nucleus being filled before they can go into the much higher-energy shells that are a bit out

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A constant net force acts on an object. which of the following best describes the object's motion?
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All ions are atoms with a charge
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Which method of testing substances is a sure way to identify a chemical reaction?
Sergeu [11.5K]

Answer:

B

Explanation:

That's the answer. Hope it helped!

7 0
2 years ago
A 15x10^-6c charge is placed at the origin and a 9x10^-6C charge is placed on the x-axis at x=1.00m. where, on the x-axis is the
Harlamova29_29 [7]

The Electric field is zero at a distance 2.492 cm from the origin.

Let A be point where the charge 15\times10^-6 C is placed which is the origin.

Let B be the point where the charge 9\times 10^-6 C is placed. Given that B is at a distance 1 cm from the origin.

Both the charges are positive. But charge at origin is greater than that of B. So we can conclude that the point on the x-axis where the electric field = 0 is after B on x - axis.

i.e., at distance 'x' from B.

Using Coulomb's law, \frac{kQ_A^2}{d_A^2} = \frac{kQ_B^2}{d_B^2},

Q_A = 15\times 10^-6 C

Q_B=9\times10^-6C

d_A = 1+x cm

d_B=x cm

k is the Coulomb's law constant.

On substituting the values into the above equation, we get,

\frac{(15\times10^-6)^2}{(1+x)^2} =\frac{(9\times10^-6)^2}{x^2}

Taking square roots on both sides and simplifying and solving for x, we get,

1.67x = 1+x

Therefore, x = 1.492 cm

Hence the electric field is zero at a distance 1+1.492 = 2.492 cm from the origin.

Learn more about Electric fields and Coulomb's Law at brainly.com/question/506926

#SPJ4

3 0
1 year ago
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