Answer:
Ionization potential of C⁺⁵ is 489.6 eV.
Wavelength of the transition from n=3 to n=2 is 1.83 x 10⁻⁸ m.
Explanation:
The ionization potential of hydrogen like atoms is given by the relation :
.....(1)
Here <em>E</em> is ionization potential, <em>Z</em> is atomic number and <em>n</em> is the principal quantum number which represents the state of the atom.
In this problem, the ionization potential of Carbon atom is to determine.
So, substitute 6 for <em>Z</em> and 1 for <em>n</em> in the equation (1).
<em> E = </em>489.6 eV
The wavelength (λ) of the photon due to the transition of electrons in Hydrogen like atom is given by the relation :
![\frac{1}{\lambda} =RZ^{2}[\frac{1}{n_{1} ^{2}}-\frac{1}{n_{2} ^{2} }]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5Clambda%7D%20%3DRZ%5E%7B2%7D%5B%5Cfrac%7B1%7D%7Bn_%7B1%7D%20%5E%7B2%7D%7D-%5Cfrac%7B1%7D%7Bn_%7B2%7D%20%5E%7B2%7D%20%7D%5D)
......(2)
R is Rydberg constant, n₁ and n₂ are the transition states of the atom.
Substitute 6 for Z, 2 for n₁, 3 for n₂ and 1.09 x 10⁷ m⁻¹ for R in equation (2).
![\frac{1}{\lambda} =1.09\times10^{7} \times6^{2}[\frac{1}{2 ^{2}}-\frac{1}{3 ^{2} }]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5Clambda%7D%20%3D1.09%5Ctimes10%5E%7B7%7D%20%5Ctimes6%5E%7B2%7D%5B%5Cfrac%7B1%7D%7B2%20%5E%7B2%7D%7D-%5Cfrac%7B1%7D%7B3%20%5E%7B2%7D%20%7D%5D)
= 5.45 x 10⁷
λ = 1.83 x 10⁻⁸ m
Answer:
3.16 m·s⁻¹ at an angle of 71.6°
Explanation:
Assume that the diagram is like Fig. 1 below.
The boat is heading straight across the river and the current is directed straight downstream.
We have two vectors at right angles to each other.
1. Calculate the magnitude of the resultant
We can use the Pythagorean theorem (Fig. 2).
R² = (3 m·s⁻¹)² + (1 m·s⁻¹)² = 9 m²·s⁻² + 1 m²·s⁻² = 10 m²·s⁻²
R = √(10 m²·s⁻²) ≈ 3.16 m·s⁻¹
2. Calculate the direction of the resultant
The direction of the resultant is the counterclockwise angle (θ) that it makes with due East
.
tanθ = opposite/adjacent = 3/1 = 3
θ = arctan 3 = 71.6°
To an observer at point O, the velocity of the boat is 3.16 m·s⁻¹ at an angle of 71.6°.
The Atwood's machine is in motion starting from rest, then Vf = Vo + a(t).
<span>Final Velocity is given as 6.7 m/s and the time is 1.9 s thus 6.7= 0+ a(1.9) </span>
<span>then a = 6.7/1.9 = 3.526 m/s². </span>
<span>The Atwood's Machine also has the formula d= distance = 1/2a(t²) </span>
<span>distance given is 6.365 m , then 6.365 = 1/2 a (1.9)², </span>
<span>a = 3.526 m/s² the same acceleration. </span>
<span>a= g(m1-m2) / m1+m2) </span>
<span>m1a + m2a = m1g - m2g </span>
<span>m1a - m1g = -m2g - m2a </span>
<span>3.526 m1 - 9.81 m1 = -9.81m2 - 3.526 m2 </span>
<span>-6.28 m1 = -13.34 m2 </span>
<span>0.47 m1= m2 </span>
<span>if 24J = 1/2mv² </span>
<span>then 24J = 1/2 m1 ( 6.7)² </span>
<span>48/ 44.89 = m1 </span>
<span>1.069 kg = m1 , then </span>
<span>0.47(1.069) = m2 </span>
<span>0.503 kg = m2</span>
If the roller coaster is moving, it will want to keep moving, along the direction of motion, unless something causes it to speed up or slow down. This resistance of the moving roller coaster to changing its velocity is another example of its inertia. Again, the greater the mass of the body, the more inertia it has.
Answer:
Part a)
Part b)
Part c)
Part d)
Explanation:
As we know that disc rotates by 32.5 rad in 5 s from rest position
Since we know that angular acceleration of the disc is constant here so we can use the condition of kinematics
Part a)
Part b)
average angular speed is know as the ratio of total angular displacement and total time
so we can say
Part c)
instantaneous angular speed is given as
Part d)
Total angle turn by the disc in total t = 10 s
now the angle turned in next 5 s will be
