All categories

1 year ago

How many sub levels does the third energy level have?

Physics

1 answer:

EastWind [94]1 year ago

3 0

The energy levels are also called principal levels. The number of sublevels in the third energy level is three.

<h3>What is energy level?</h3>

The Energy levels are at a fixed distance from the nucleus of an atom, where electrons are arranged in shells.

In the third energy shell, electrons are present. The third principal energy level has three sublevels, s, p and d.

The number of sublevels in the third energy level is three.

Learn more about energy levels.

brainly.com/question/17396431

#SPJ

You might be interested in

A car is traveling at 39.7 mi/h on a horizontal highway. The acceleration of gravity is 9.8 m/s 2 . If the coefficient of fricti

777dan777 [17]

Answer:

The minimum distance in which the car will stop is

x=167.38m

Explanation:

$39.7\frac{mi}{h}*\frac{1km}{0.621371mi}*\frac{1000m}{1km}*\frac{1h}{3600s}=17.747\frac{m}{s}$

∑F=m*a

∑F=u*m*g

The force of friction is the same value but in different direction of the force moving the car so it can stop so

$F=m*a\\a=\frac{F}{m}\\a=\frac{u*m*g}{m}\\a=u*g\\a=0.096*-9.8\frac{m}{s^{2} }$

$a=-0.9408 \frac{m}{s^{2}}$

$v_{f}^{2}=v_{o}^{2}+2*a*(x_{f}-x_{o})\\v_{f}=0 \\x_{o}=0\\0=v_{o}^{2}+2*a*x_{f}\\x_{f}=\frac{v_{o}^{2}}{2*a} \\x_{f}=\frac{(-17.747\frac{m}{s})^{2}}{2*(-0.9408)} \\x_{f}=167.38m$

4 0

2 years ago

A guitar string is 0.620m long, and oscillates at 234Hz. What is the velocity of the waves in the string? m/s

9966 [12]

Answer:

v = 72.54 m/s

Explanation:

We have,

Length of a guitar string is 0.62 m

Frequency of a guitar string is 234 Hz

For guitar string,

$L=2\lambda\\\\\lambda=\dfrac{L}{2}\\\\\lambda=\dfrac{0.62}{2}\\\\\lambda=0.31\ m$

The velocity of the wave in the string is given by :

$v=f\lambda\\\\v=234\times 0.31\\\\v=72.54\ m/s$

So, the velocity of the waves in the string is 72.54 m/s.

3 0

3 years ago

Help pls i need this right now

pantera1 [17]

Answer:

The x-component of $F_{3}$ is 56.148 newtons.

Explanation:

From 1st and 2nd Newton's Law we know that a system is at rest when net acceleration is zero. Then, the vectorial sum of the three forces must be equal to zero. That is:

$\vec F_{1} + \vec F_{2} + \vec F_{3} = \vec O$ (1)

Where:

$\vec F_{1}$ , $\vec F_{2}$ , $\vec F_{3}$ - External forces exerted on the ring, measured in newtons.

$\vec O$ - Vector zero, measured in newtons.

If we know that $\vec F_{1} = (70.711,70.711)\,[N]$ , $\vec F_{2} = (-126.859, 46.173)\,[N]$ , $F_{3} = (F_{3,x},F_{3,y})$ and $\vec O = (0,0)\,[N]$ , then we construct the following system of linear equations: