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2 years ago

An electron in a(n) __________ subshell experiences the greatest effective nuclear charge in a many-electron atom.

Physics

1 answer:

Anon25 [30]2 years ago

6 0

Answer:

D. 3s

Explanation:

As we know that the electrons in an atom are arranged in subshells of the shells that are present around the nucleus of the atom in their increasing configuration number, viz. K shell is the inner most shell of any atom ans L,M,N and so on are the subsequent outer shells of the atom containing electron.

Among the subshells, s is having the electrons of the lowest energy of a particular shell and hence are the nearest to the nucleus of the atom.

From Coulomb's law we know that the force due to electrostatic charges in proximity is inversely proportional to the square of the distance between them and hence the nearest electron feels the most effect of the protons of the nucleus.

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One end of a piano wire is wrapped around a cylindrical tuning peg and the other end is fixed in place. The tuning peg is turned

liq [111]

Answer:

$T = 3183 N$

Explanation:

When the screw is turned by two turns then change in the length of the wire is given as

$\Delta L = 2(2\pi r)$

$\Delta L = 4\pi r$

$\Delta L = 4(\pi)(1.0 mm)$

$\Delta L = 12.56 mm$

now we know by the formula of Young's modulus

$Y = \frac{T/A}{\Delta L/L}$

so we have

$T = \frac{AY \Delta L}{L}$

$T = \frac{\pi (0.55 \times 10^{-3})^2(2.0 \times 10^{11})(12.56 \times 10^{-3})}{0.75}$

$T = 3183 N$

3 0

2 years ago

A 0.250 kgkg toy is undergoing SHM on the end of a horizontal spring with force constant 300 N/mN/m. When the toy is 0.0120 mm f

konstantin123 [22]

Answer:

(a) The total energy of the object at any point in its motion is 0.0416 J

(b) The amplitude of the motion is 0.0167 m

Explanation:

Given;

mass of the toy, m = 0.25 kg

force constant of the spring, k = 300 N/m

displacement of the toy, x = 0.012 m

speed of the toy, v = 0.4 m/s

(a) The total energy of the object at any point in its motion

E = ¹/₂mv² + ¹/₂kx²

E = ¹/₂ (0.25)(0.4)² + ¹/₂ (300)(0.012)²

E = 0.0416 J

(b) the amplitude of the motion

E = ¹/₂KA²

$A = \sqrt{\frac{2E}{K} } \\\\A = \sqrt{\frac{2*0.0416}{300} } \\\\A = 0.0167 \ m$

$E = \frac{1}{2} mv_{max}^2\\\\v_{max} = \sqrt{\frac{2E}{m} } \\\\v_{max} = \sqrt{\frac{2*0.0416}{0.25} } \\\\v_{max} = 0.577 \ m/s$

8 0

2 years ago

You use a slingshot to launch a potato horizontally from the edge of a cliff with speed v0. The acceleration due to gravity is g

Ray Of Light [21]

Answer:

$\displaystyle t=\frac{2v_o}{g}$

Explanation:

<u>Horizontal Launch</u>

When an object is launched horizontally at a speed vo, it describes a curved called parabola as the speed in the x-direction does not change and the speed in the y-direction increases with time because the gravity makes it return to the ground.

The vertical distance the object (potato) travels downwards is:

$\displaystyle y=\frac{gt^2}{2}$

The horizontal distance is

$x=v_ot$

We need to find the time when both distances are equal, thus

$\displaystyle \frac{gt^2}{2}=v_ot$

Simplifying by t

$\displaystyle \frac{gt}{2}=v_o$

Solving for t

$\displaystyle \boxed{t=\frac{2v_o}{g}}$

8 0

2 years ago

A pair of glasses is dropped from the top of a 32.0m stadium. A pen is dropped 2.Os later. How high above the ground is the pen

Svetllana [295]

Answer:

$h_p = 30.46\ m$

Explanation:

<u>Free Fall Motion</u>

A free-falling object refers to an object that is falling under the sole influence of gravity. If the object is dropped from a certain height h, it moves downwards until it reaches ground level.

The speed vf of the object when a time t has passed is given by:

$v_f=g\cdot t$