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2 years ago

An atom of arsenic has how many electron-containing orbitals

Physics

1 answer:

irina [24]2 years ago

8 0

Answer:

The Arsenic has three electron-containing orbitals. The orbitals s, p and d.

Explanation:

Arsenic is an element with an atomic number equal of 33, it means that it has 33 electrons in its orbitals in the following way:

$1s^{2}$

$2s^{2}$

$2p^{6}$

$3s^{2}$

$3p^{6}$

$3d^{10}$

$4s^{2}$

$4p^{3}$

Therefore, the Arsenic has three electron-containing orbitals (s, p d).

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Point charges q1=+2.00μC and q2=−2.00μC are placed at adjacent corners of a square for which the length of each side is 5.00 cm.

8_murik_8 [283]

The electric potential is a scalar unit, so we don't have to struggle with the vectors. The formula that gives electric potential is

$V = \frac{1}{4\pi\epsilon_0}\frac{q}{r}$

1) At point a, the electric potential is the sum of the potentials due to q1 and q2. So,

$V_a = \frac{1}{4\pi\epsilon_0}\frac{q_1}{r_1} + \frac{1}{4\pi\epsilon_0}\frac{q_2}{r_2}$

The distance from the center of the square to one of the corners is $\sqrt2 L/2 = 0.035m$

$V_a = \frac{1}{4\pi\epsilon_0}\frac{2\times10^{-6}}{0.035} + \frac{1}{4\pi\epsilon_0}\frac{-2\times10^{-6}}{0.035} = 0$

The answer is zero, because the point charges are at equal distances and their magnitudes are also equal but their directions are opposite.

2) $V_b = \frac{1}{4\pi\epsilon_0}\frac{q_1}{r_1} + \frac{1}{4\pi\epsilon_0}\frac{q_2}{r_2}$

$r_1 = 0.05\sqrt2m\\r_2 = 0.05m$

$V_b = \frac{1}{4\pi\epsilon_0}\frac{2\times10^{-6}}{0.05\sqrt2} + \frac{1}{4\pi\epsilon_0}\frac{-2\times10^{-6}}{0.05}\\V_b = \frac{1}{4\pi\epsilon_0}\frac{2\times10^{-6}}{0.05} (\frac{1}{\sqrt2}-1)\\V_b = \frac{1}{4\pi\epsilon_0} (4\times 10^{-5})(-0.29)\\V_b = (-\frac{2.9\times10^{-6}}{\pi\epsilon_0})[tex]3) The work done on q3 by q1 and q2 is equal to the difference between energies. This is the work-energy theorem. So,[tex]W = U_b - U_a$

$U = \frac{1}{4\pi\epsilon_0}\frac{q_1q_3}{r} = Vq_3$

$W = q_3(V_b - V_a) = q_3(V_b - 0)\\W = (-2\times10^{-6})(-\frac{2.9\times10^{-6}}{\pi\epsilon_0})\\W = \frac{5.8\times10^{-12}}{\pi\epsilon_0}$

4 0

2 years ago

The ball is displaced to the left and then oscillates backwards and forwards between the two plates. The ball touches a plate on

ELEN [110]

Answer:

Average current produced by the repeated transfer of charge is 5.6 × 10⁻⁷ ampere

Explanation:

The formula to be used here is

Q = It

where Q is the quantity of electricity and it is measured coulombs (C); 2.8 × 10⁻⁸ C or 0.000000028 C

I is current and it is measured in ampere (amps or A); unknown

t is time and it is measured in seconds (s); 0.05 s

Since, average current is what is unknown

I =Q/t

I = 0.000000028/0.05

I = 5.6 × 10⁻⁷ A

Average current produced by the repeated transfer of charge is 5.6 × 10⁻⁷ ampere

4 0

2 years ago

Three boxes in contact rest side-by-side on a smooth, horizontal floor. Their masses are 5.0-kg, 3.0-kg, and 2.0-kg, with the 3.

Ivahew [28]

Answer:

(a)Look at the attached graphic

(b)

(b)-1 Equation 1 : m1= 5kg

50-F1= 5 *a

(b)-2 Equation 2 : m2= 3kg

F1-F2= 3 *a

(b)-3 Equation 3 : m3= 2kg

F2 = 2*a

(d) F2 =10 N

Explanation:

We apply Newton's second law:

∑F = m*a (Formula 1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

(a) Draw the free-body diagrams for each of the boxes

Look at the attached graphic

(b) Write Newton’s equation for each mass along the horizontal direction.

Data: m1= 5.0-kg ,m2= 3.0-kg , ,m3= 2.0-kg

Look m1 free-body diagram:

∑Fx = m1*a

50-F1= 5 *a Equation 1

Look m2 free-body diagram:

∑Fx = m2*a

F1-F2= 3 *a Equation 2

Look m3 free-body diagram:

∑Fx = m3*a

F2 = 2*a Equation 3

(c) What magnitude force does the 3.0-kg box exert on the 5.0- kg box?

Look Free body diagram of the mass set

∑Fx = m*a m= m1+m2+m3= 5+3+2 = 10 kg

50 = 10*a

a= 50/10 = 5 m/s²

We replace a = 5 m/s² in the equation 1:

50-F1= 5 *5

50-25= F1

F1 = 25 N

(d) What magnitude force does the 3.0-kg box exert on the 2.0kg box?

We replace a= 5 m/s² in the equation 3

F2 = 2*5 = 10 N

4 0

3 years ago

Net force needed to accelerate a 1000-kg car at 0.5g

just olya [345]

It would be 2000N ( newtons )

6 0

3 years ago

Two workers are sliding 350 kgkg crate across the floor. One worker pushes forward on the crate with a force of 390 NN while the

svetoff [14.1K]

Answer:

$\mu_k=0.18$

Explanation:

First, we write the equations of motion for each axis. Since the crate is sliding with constant speed, its acceleration is zero. Then, we have:

$x: T+F-f_k=0\\\\y:N-mg=0$

Where T is the tension in the rope, F is the force exerted by the first worker, f_k is the frictional force, N is the normal force and mg is the weight of the crate.

Since $f_k=\mu_k N$ and $N=mg$ , we can rewrite the first equation as:

$T+F-\mu_k mg=0$

Now, we solve for $\mu_k$ and calculate it:

$\mu_k=\frac{T+F}{mg}\\ \\\mu_k =\frac{220N+390N}{(350kg)(9.8m/s^{2})} =0.18$

This means that the crate's coefficient of kinetic friction on the floor is 0.18.

6 0

3 years ago