Answer
given,
length of slender rod =80 cm = 0.8 m
mass of rod = 0.39 Kg
mass of small sphere = 0.0200 kg
mass of another sphere weld = 0.0500 Kg
calculating the moment of inertia of the system
using conservation of energy
we know,
v = r ω
v = 1.084 m/s
Explanation :
When an electron jumps from one energy level to another, the energy of atom gets changed.
If a photon gets absorbed, the electron will move to higher energy levels and then fall back to the lower energy levels. Then each time a photon will be absorbed whose energy is given by difference between the initial and final energy levels i.e
In Balmer series, the transition is from higher energy levels to n = 2.
So, the necessary condition for Balmer series is that the electron should be at first excited state or n = 2 level as shown in figure.
Answer:
a) T_A = , b) T_B = g [m₂ (
) + m₁ (
) ]
c) x = , d) m₂ = m₁ (
)
Explanation:
After carefully reading your long sentence, I understand your exercise. In the attachment is a diagram of the assembly described. This is a balancing act
a) The tension of string A is requested
The expression for the rotational equilibrium taking the ends of the bar as the turning point, the counterclockwise rotations are positive
∑ τ = 0
T_A d - W₂ x -W₁ L/2 = 0
T_A =
b) the tension in string B
we write the expression of the translational equilibrium
∑ F = 0
T_A - W₂ - W₁ - T_B = 0
T_B = T_A -W₂ - W₁
T_ B = - g m₂ - g m₁
T_B = g [m₂ ( ) + m₁ (
) ]
c) The minimum value of x for the system to remain stable, we use the expression for the endowment equilibrium, for this case the axis of rotation is the support point of the chord A, for which we will write the equation for this system
T_A 0 + W₂ (d-x) - W₁ (L / 2-d) - T_B d = 0
at the point that begins to rotate T_B = 0
g m₂ (d -x) - g m₁ (0.5 L -d) + 0 = 0
m₂ (d-x) = m₁ (0.5 L- d)
m₂ x = m₂ d - m₁ (0.5 L- d)
x =
d) The mass of the block for which it is always in equilibrium
this is the mass for which x = 0
0 = d - \frac{m_1}{m_2} \ \frac{L}{2d}
m₂ = m₁
m₂ = m₁ ( )
