We use the Rydberg Equation for this which is expressed as:
<span>1/ lambda = R [ 1/(n2)^2 - 1/(n1)^2]
</span>
where lambda is the wavelength, where n represents the final and initial states. Brackett series means that the initial orbit that electron was there is 4 and R is equal to 1.0979x10^7m<span>. Thus,
</span>
1/ lambda = R [ 1/(n2)^2 - 1/(n1)^2]
1/1.0979x10^7m = 1.0979x10^7m [ 1/(n2)^2 - 1/(4)^2]
Solving for n2, we obtain n=1.
Using your periodic table if you look at it 3-11 are tansition metals so the horizontal Group Number will help if the group number has to digits just remove the one so if it were to be 13, the valence would be 3, if it were 14 the valence would be ,4 if it were 15, the valence would be 5, if it were 16 the valence would be 6, if it were 17 the valence would be 7 if it were group 18 the valence would be 8 so if anymore help needed to explain hit me up
Answer:
As the skydiver falls to the Earth, she experiences positive acceleration only due to gravity.
Explanation:
As the skydiver falls to the Earth, she experiences friction in the form of air resistance which tries to slow her down and is proportional to the her velocity. So it cannot have a positive acceleration as it acts in opposite direction to slow her down.
Inertia during skydiving is experienced when we open an parachute, the parachute slows down the speed of are descent hence changing our inertia of motion with a velocity.
Only the Earth's, gravitational field has an positive acceleration as it pulls us towards the Earth, hence increasing our velocity.
Answer:
a). M = 20.392 kg
b). am = 0.56 
(block), aM = 0.28 (bucket)
Explanation:
a). We got N = mg cos θ,
f = 
= 
If the block is ready to slide,
T = mg sin θ + f
T = mg sin θ + .....(i)
2T = Mg ..........(ii)
Putting (ii) in (i), we get
M = 20.392 kg
b). 
.............(iii)
Here, l = total string length
Differentiating equation (iii) double time w.r.t t, l, h and h' are constants, so
.....................(iv)
We got, N = mg cos θ
∴ 
................(v)
Mg - 2T = M
(from equation (iv))
.....................(vi)
Putting (vi) in equation (v),
![$\frac{g\left[\frac{M}{2}-m \sin \theta-\mu_K m \cos \theta\right]}{(\frac{M}{4}+m)}=a_m$](https://tex.z-dn.net/?f=%24%5Cfrac%7Bg%5Cleft%5B%5Cfrac%7BM%7D%7B2%7D-m%20%5Csin%20%5Ctheta-%5Cmu_K%20m%20%5Ccos%20%5Ctheta%5Cright%5D%7D%7B%28%5Cfrac%7BM%7D%7B4%7D%2Bm%29%7D%3Da_m%24)
![$\frac{9.8\left[\frac{20.392}{2}-10(\sin 30+0.5 \cos 30)\right]}{(\frac{20.392}{4}+10)}=a_m$](https://tex.z-dn.net/?f=%24%5Cfrac%7B9.8%5Cleft%5B%5Cfrac%7B20.392%7D%7B2%7D-10%28%5Csin%2030%2B0.5%20%5Ccos%2030%29%5Cright%5D%7D%7B%28%5Cfrac%7B20.392%7D%7B4%7D%2B10%29%7D%3Da_m%24)
Using equation (iv), we get,
