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3 years ago

What is the maximum number of atoms that can be bonded in an ionic bond

Physics

1 answer:

Dima020 [189]3 years ago

6 0

A maximum of four bonds can be used between two atoms.

<u>Explanation:</u>

Ionic bond is one of the strongest bond used in chemical reaction where the valence electrons are bonded and shared between the atoms. Sometimes four covalent bonds between two atoms might make them unstable as the bonds are formed in their outermost shell to make them to form or complete a total of 8 in their shell. For example carbon has 3 as maximum.

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3 years ago

A particle with mass 1.81*10^-3kg and a charge of 1.22*10^-8C has, at a given instant, a velocity v= (3.0*10^4 m/s)j.

charle [14.2K]

Answer:

a = -0.33 m/s² k^

Direction: negative

Explanation:

From Newton's law of motion, we know that;

F = ma

Now, from magnetic fields, we know that;. F = qVB

Thus;

ma = qVB

Where;

m is mass

a is acceleration

q is charge

V is velocity

B is magnetic field

We are given;

m = 1.81 × 10^(−3) kg

q = 1.22 × 10 ^(−8) C

V = (3.00 × 10⁴ m/s) ȷ^.

B = (1.63T) ı^ + (0.980T) ȷ^

Thus, since we are looking for acceleration, from, ma = qVB; let's make a the subject;

a = qVB/m

a = [(1.22 × 10 ^(−8)) × (3.00 × 10⁴)ȷ^ × ((1.63T) ı^ + (0.980T) ȷ^)]/(1.81 × 10^(−3))

From vector multiplication, ȷ^ × ȷ^ = 0 and ȷ^ × i^ = -k^

Thus;

a = -0.33 m/s² k^

7 0

3 years ago

according to newtons first law of motion, what will an object in motion do when no external force acts on it?

shutvik [7]

Answer:

The object will not move.

Explanation:

If nothing pushes against it it will not move. If its not on a slant it will not move.

4 0

3 years ago

Read 2 more answers

The train passes point A with a speed of 30 m/s and begins to decrease its speed at a constant rate of at = - 0.25 m/s^2. Determ

prisoha [69]

Explanation:

At point B, the velocity speed of the train is as follows.

$\nu^{2}_{B} = \nu^{2}_{A} + 2a_{t} (s_{B} - s_{A})$

= $(30)^{2} + 2(-0.25(412 - 0))$

= 26.34 m/s

Now, we will calculate the first derivative of the equation of train.

y = $200 e^{\frac{x}{1000}}$

$\frac{dy}{dx} = 0.2 e^{\frac{x}{1000}}$

Now, second derivative of the train is calculated as follows.

$\frac{dy}{dx} = 0.2 e^{\frac{x}{1000}}$

$\frac{d^{2}y}{dx^{2}} = 0.2 (10^{-3}) e^{\frac{x}{1000}}$

Radius of curvature of the train is as follows.

$\rho = \frac{[1 + (\frac{dy}{dx})^{2}]^{\frac{3}{2}}}{\frac{d^{2}y}{dx^{2}}}$

= $\frac{[1 + 0.2e^{\frac{400}{1000}}^{2}]^{\frac{3}{2}}}{0.2(10^{-3})e^{\frac{400}{1000}}}$

= 3808.96 m

Now, we will calculate the normal component of the train as follows.

$a_{n} = \frac{\nu^{2}_{B}}{\rho}$

= $\frac{(26.34)^{2}}{3808.96}$

= 0.1822 $m/s^{2}$

The magnitude of acceleration of train is calculated as follows.

a = $\sqrt{(a_{t})^{2} + (a_{n})^{2}}$

= $\sqrt{(-0.25)^{2} + (0.1822)^{2}}$

= $0.309 m/s^{2}$

Thus, we can conclude that magnitude of the acceleration of the train when it reaches point B, where sAB = 412 m is $0.309 m/s^{2}$ .

6 0

3 years ago