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1 year ago

What is the symbol of the ion having 12 protons and 10 electrons?

1 answer:

katrin2010 [14]1 year ago

3 0

The symbol of the ion having 12 protons and 10 electrons would be Mg²⁺ , therefore the correct answer is option A.

<h3>What is the atomic number?</h3>

An atom's atomic number is determined by the total number of protons it contains. Both the number of neutrons and the number of electrons contained inside an atom have no connection to the atomic number.

We must determine the symbol for the ion with 12 protons and 10 electrons according to the problem.

As is well knowledge, the magnesium atom, symbolized by Mg2+, contains 12 electrons and 10 protons.

Thus, the correct answer is option a.

To learn more about the atomic number from here, refer to the link;

brainly.com/question/14190064

#SPJ4

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Firlakuza [10]

Answer:

$3.5\:\mathrm{m/s^2}$

Explanation:

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To find the acceleration in the horizontal direction, you need the horizontal component of the force being applied.

Using trigonometry to find the horizontal component of the force:

$\cos 45^{\circ}=\frac{x}{5},\\\frac{\sqrt{2}}{{2}}=\frac{x}{5},\\x=\frac{5\sqrt{2}}{2}$

Use this horizontal component of the force to solve for for the acceleration of the object:

$\frac{5\sqrt{2}}{2}=1.0\cdot a,\\a=\frac{5\sqrt{2}}{2}\approx \boxed{3.5\:\mathrm{m/s^2}}$

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2 years ago

A crane lifts a car with 9023 J of kinetic energy. The crane lifts the car with a speed

lapo4ka [179]

Answer:

m = 2301.8 [kg]

Explanation:

This problem can be solved in an easy way using the definition of kinetic energy, in this way you can clear the mass, depending on the known variables.

$E_{k}=0.5*m*v^2\\replacing\\\\9023 = 0.5 *m*(2.8)^2\\m = 2301.8[kg]$

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3 years ago

An electron moves at 2.80 ×106 m/s through a region in which there is a magnetic field of unspecified direction and magnitude 7.

d1i1m1o1n [39]

Answer:

$a = 3.73*10^{16}m/s^2$

Explanation:

The magnetic force is given by the equation,

$\vec{F}_{magnetic} = q (\vec{v}X\vec{B})$

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Where $\theta$ is the angle between velocity vector and the magnetic field,

That angle is 90°.

We know as well that

F=ma, replacing the mass and the acceleration in our previous equation we have

$ma = qvBsin\theta$

$a= \frac{qvBsin\theta}{m}$

Our values are,

$q= 1.6*10^{-19}c$

$v=2.8*10^6m/s$

$B=7.6*10^{-2}T$

$m=9.11*10^{-31}kg$

Substituting,

$a = \frac{(1.6*10^{-19})(2.8*10^6)(7.6*10^{-2})}{9.11*10^{-31}}$

$a = 3.73*10^{16}m/s^2$

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3 years ago

An elevator of mass m is initially at rest on the first floor of a building. It moves upward, and passes the second and third fl

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Answer:

6. W = 3 m g h

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