Answer:
Explanation:
Bohr's energy expression is as follows
E_n = 13.6 z² /n² where z is atomic no and n is principal quantum no of the atom .
z for helium is 2 and for ionised atom is 5 . Let energy of n₁ level of He is equal to energy level n₂ of ionised atom
so
13.6 x 2² / n₁² = 13.6 x 5² / n₂²
n₁ / n₂ = 2/5 , ie 2nd energy level of He matches 5 th energy level of ionised atom .
For quantum numbers less than or equal to 9 , If we take n₁ = 8 for He
Putting it in the equation above
2² / 8² = 5² / n₂²
n₂ = 5 x 8 / 2
= 20 .
energy
= - 13.6 x2² / 8²
= - 0.85 eV .
The frequency of the red light is 428 terahertz. To get the value of the red light's frequency, use the formula F = velocity/wavelength. The velocity of light is 3.00 x 10^8 m/s. For easier computation, convert 700.5 nanometers to meter. 1 nanometer is equal to 1 x 10^-9 meters. 700.5 nanometers is equal to 7.005 x 10^-7 meters. Divide the velocity 3.00 x 10^8m/s by wavelength 7.005 x 10^-7 meters. The result will be 4.28 x 10^14 Hertz or 428 terahertz.
<span>As it is descended from a vertical height h,
The lost Potential Energy = Mgh
The gained Kenetic Energy = (1/2)Mv^2; The rotational KE = (1/2)Jw^2
The angular speed w = speed/ Radius = v/R
So Rotational KE = (1/2)Jw^2 = (1/2)J(v/R)^2; J is moment of inertia
Now Mgh = (1/2)Mv^2 + (1/2)J(v/R)^2 => 2gh/v^2 = 1 + (J/MR^2)
As v = (5gh/4)^1/2, (J/MR^2) = 2gh/v^2 - 1 => (J/MR^2) = (8gh/5gh) - 1
so (J/MR^2) = 3/5 and therefore J = (3/5)MR^2.</span>
Answer:
a) a= 8.33 m/s², T = 12.495 N
, b) a = 2.45 m / s²
Explanation:
a) this is an exercise of Newton's second law. As the upper load is secured by a cable, it cannot be moved, so the lower load is determined by the maximum acceleration.
We apply Newton's second law to the lower charge
fr₁ + fr₂ = ma
The equation for the force of friction is
fr = μ N
Y Axis
N - W₁ –W₂ = 0
N = W₁ + W₂
N = (m₁ + m₂) g
Since the beams are the same, it has the same mass
N = 2 m g
We replace
μ₁ 2mg + μ₂ mg = m a
a = (2μ₁ + μ₂) g
a = (2 0.30 + 0.25) 9.8
a= 8.33 m/s²
Let's look for cable tension with beam 2
T = m₂ a
T = 1500 8.33
T = 12.495 N
b) For maximum deceleration the cable loses tension (T = 0 N), so as this beam has less friction is the one that will move first, we are assuming that the rope is horizontal
fr = m₂ a₂
N- w₂ = 0
N = W₂ = mg
μ₂ mg = m a₂
a = μ₂ g
a = 0.25 9.8
a = 2.45 m / s²
Answer:
R₂ / R₁ = D / L
Explanation:
The resistance of a metal is
R = ρ L / A
Where ρ is the resistivity of aluminum, L is the length of the resistance and A its cross section
We apply this formal to both configurations
Small face measurements (W W)
The length is
L = W
Area
A = W W = W²
R₁ = ρ W / W² = ρ / W
Large face measurements (D L)
Length L = D= 2W
Area A = W L
R₂ = ρ D / WL = ρ 2W / W L = 2 ρ/L
The relationship is
R₂ / R₁ = 2W²/L
