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3 years ago

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For states with larger values of n that admit a d subshell, the d subshell is at even higher energy than the p subshell. In fact

, its energy is higher than that of the s subshell for the n+1 state. With this in mind, which of the following are valid ground-state electron structures?1. 1s2 2s2 2p63 s2 3p6 3d22. 1s2 2s2 2p6 3s2 3p6 3d113. 1s2 2s2 2p6 3s2 3p6 4s2 3d5A. 1B. 2C. 3D. 1 and 2E. 1 and 3F. 2 and 3

1 answer:

frosja888 [35]3 years ago

4 0

Answer:

Explanation:

Correct order of energy of energy states within an atom is as follows

1s , 2 s, 2 p 3 s ,3 p, 4s , 3d, 4p, 4p,5s etc.

There is discrepancies in this order due to overlapping of outer orbitals with inner orbitals so sometimes outer orbitals have lower energy.

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A 4.00 m long, massless beam rests horizontally on a support 3.00 m from the left

Rus_ich [418]

If the beam is in static equilibrium, meaning the Net Torque on it about the support is zero, the value of x₁ is 2.46m

Given the data in the question;

Length of the massless beam; $L = 4.00m$

Distance of support from the left end; $x = 3.00m$

First mass; $m1 = 31.3 kg$

Distance of beam from the left end( m₁ is attached to ); $x_1 = ?$

Second mass; $m_2 = 61.7 kg$

Distance of beam from the right of the support( m₂ is attached to ); $x_1 = 0.273m$

Now, since it is mentioned that the beam is in static equilibrium, the Net Torque on it about the support must be zero.

Hence, $m_1g( x-x_1) = m_2gx_2$

we divide both sides by $g$

$m_1( x-x_1) = m_2x_2$

Next, we make $x_1$ , the subject of the formula

$x_1 = x - [ \frac{m_2x_2}{m_1} ]$

We substitute in our given values

$x_1 = 3.00m - [ \frac{61.7kg\ * \ 0.273m}{31.3kg} ]$

$x_1 = 3.00m - 0.538m$

$x_1 = 2.46m$

Therefore, If the beam is in static equilibrium, meaning the Net Torque on it about the support is zero, the value of x₁ is 2.46m

Learn more; brainly.com/question/3882839

6 0

3 years ago

An applied force of 50 N is used to accelerate an object, that weighs 73 N, to the right across a frictional surface. The object

Hunter-Best [27]

Answer:

5.38 m/s^2

Explanation:

NET force causing the object to accelerate = 50 -10 = 40 N

Mass of the object = 73 N / 9.81 m/s^2 = 7.44 kg

F = ma

40 = 7.44 * a a = 5.38 m/s^2

6 0

2 years ago

A boy whirls a stone in a horizontal circle of radius 1.4 m and at height 1.5 m above ground level. The string breaks, and the s

balandron [24]

Answer: $233.23 m/s^2$

Explanation:

Given

radius of circle=1.4 m

Height of stone above ground=1.5 m

Horizontal distance(R)=10 m

It is given at the time of break stone flies horizontally thus stone to cover a height of 1.5 m in time t before reaching ground

$1.5=0+\frac{gt^2}{2}$

t=0.55 s

Initial horizontal velocity at the time of break is given by u

$R=u\times t$

$10=u\times 0.55$

u=18.07 m/s

Therefore magnitude of centripetal acceleration is given by

$a_c=\frac{u^2}{r}=\frac{18.07^2}{1.4}=233.23 m/s^2$

6 0

3 years ago

2. How long must a 400 W electrical engine work in order to produce 300 kJ of work?

yanalaym [24]

Answer:

Explanation:

400 W = 400 J/s

300000 J / 400 J/s = 750 s or 12.5 minutes

7 0

3 years ago

The quartz crystal used in an electric watch vibrates with a frequency of 32,768 Hz. What is the period of the crystal's motion?

Elan Coil [88]

Answer:

Time period, $T=3.05\times 10^{-5}\ s$

Explanation:

Given that,

The quartz crystal used in an electric watch vibrates with a frequency of 32,768 Hz, f = 32768 Hz

We need to find the period of the crystal's motion. The relationship between the frequency and the time period is given by :

$T=\dfrac{1}{f}$

T is the time period of the crystal's motion.

Time period is given by :

$T=\dfrac{1}{32768}$

$T=3.05\times 10^{-5}\ s$

So, the time period of the crystal's motion is $3.05\times 10^{-5}\ s$ . Hence, this is the required solution.

8 0

3 years ago