Answer:
Explanation:
1 ha = 10⁴ m²
1375 ha = 1375 x 10⁴ m² = 13.75 x 10⁶ m²
In flow in a month = .5 x 10⁶ x 30 m³ = 15 x 10⁶ m³
Net inflow after all loss = 18.5 - 9.5 - 2.5 cm = 6.5 cm = .065 m
Net inflow in volume = 13.75 x 10⁶ x .065 m³= .89375 x 10⁶ m³
Let Q be the withdrawal in m³
Q - 15 x 10⁶ - .89375 x 10⁶ = 13.75 x 10⁶ x .75 = 10.3125 x 10⁶
Q = 26.20 x 10⁶ m³
rate of withdrawal per second
= 26.20 x 10⁶ / 30 x 24 x 60 x 60
= 26.20 x 10⁶ / 2.592 x 10⁶
= 10.11 m³ / s
Answer:
ω' = 0.815 rad/s
Explanation:
Given,
R = 1.20 m
Inertia of merry-go- round= 240 kg.m²
Rotating speed = 9 rpm =
=0.9424 rad/s
mass of the child, m = 26 kg
angular speed of the merry-go-round=?
we know
Angular momentum, L = I ω
Moment of inertia of the child
I' = m r² = 26 x 1.2² = 37.44 kgm²
Conservation of angular momentum
initial angular momentum = Final angular momentum
I ω = (I+I')ω'
240 x 0.9424 = (240+37.44) ω'
226.176= 277.44 ω'
ω' = 0.815 rad/s
new angular speed of the merry-go- round is equal to 0.815 rad/s
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 :
......(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).
= 5.45 x 10⁷
λ = 1.83 x 10⁻⁸ m