The required,
a) [tex]N'/N_0[/tex] ≈ 0.408 (40.8%)
b) Approximately 40.8% of the atoms are in the excited state.
c) [tex]N'/N_0[/tex] ≈ 0.066 (6.6%)
To calculate the ratio of N'/N_0 in an air-acetylene flame, we can use the Boltzmann distribution equation:
[tex]N'/N_0 = (g'/g_0) * exp^{(-\triangle E/kT)}[/tex]
a) Calculate the ratio of [tex]N'/N_0[/tex] in an air-acetylene flame (2500 K):
Given:
[tex]g_0 = 1[/tex] (degeneracy of the ground state)
[tex]g' = 3[/tex] (degeneracy of the excited state)
[tex]\triangle E = 8.68 * 10^{(-19)}[/tex]J/atom (energy difference between the excited and ground states)
T = 2500 K (temperature)
[tex]N'/N_0 = (3/1) * e{(-8.68 * 10^{19} / (1.38 * 10^{-23} * 2500 ))[/tex]
Calculating the exponential term:
exp(-8.68 x 10⁻¹⁹ J/atom / (1.38 x 10⁻²³ J/K * 2500 K)) ≈ 0.136
Therefore, the ratio of [tex]N*/N_0[/tex] in an air-acetylene flame is:
[tex]N'/N_0[/tex] ≈ (3/1) * 0.136 ≈ 0.408
b) To determine the percent of atoms in the excited state, we can multiply the ratio [tex]N'/N_0[/tex] by 100:
Percent in excited state = [tex]N'/N_0 * 100[/tex]
Percent in excited state ≈ 0.408 * 100 ≈ 40.8%
Therefore, 40.8% of the atoms will be in the excited state.
Similarly,
c) If an argon plasma (10,000 K) is used instead of the air-acetylene flame, we can repeat the calculations using the new temperature:
The ratio of [tex]N*/N_0[/tex] in an argon plasma is:
N'/N0 ≈ (3/1) * 0.022 ≈ 0.066
To determine the percent of atoms in the excited state:
Percent in excited state = [tex]N'/N_0 * 100[/tex]
Percent in excited state ≈ 0.066 * 100 ≈ 6.6%
Therefore, 6.6% of the atoms will be in the excited state in an argon plasma.
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a 3.10 kg grinding wheel is in the form of a solid cylinder of radius 0.100 m. .What constant torque will bring it from rest to an angular speed of 1200 rev/min in 2.5 s?
The constant torque required to bring the grinding wheel from rest to an angular speed of 1200 rev/min in 2.5 seconds is approximately 984.39 N·m.
To find the constant torque required to bring the grinding wheel from rest to an angular speed of 1200 rev/min in 2.5 seconds, we can use the rotational kinetic energy equation: K = (1/2) I ω²
where K is the kinetic energy, I is the moment of inertia, and ω is the angular speed.
The moment of inertia for a solid cylinder rotating about its central axis is given by:
I = (1/2) m r²
where m is the mass of the cylinder and r is the radius.
Given:
Mass of the grinding wheel (m) = 3.10 kg
Radius of the grinding wheel (r) = 0.100 m
Angular speed (ω) = 1200 rev/min
First, let's convert the angular speed from rev/min to rad/s:
ω = (1200 rev/min) × (2π rad/rev) × (1 min/60 s) = 40π rad/s
Now, let's calculate the moment of inertia (I):
I = (1/2) m r² = (1/2) × 3.10 kg × (0.100 m)² = 0.0155 kg·m²
Next, let's calculate the final kinetic energy (K) using the given angular speed:
K = (1/2) I ω² = (1/2) × 0.0155 kg·m² × (40π rad/s)² ≈ 774π J
Since the grinding wheel starts from rest, the initial kinetic energy is zero.
The change in kinetic energy (ΔK) is:
ΔK = K - 0 = 774π J
The torque (τ) can be calculated using the following equation:
ΔK = τ Δt
where Δt is the time interval.
Substituting the given values:
774π J = τ × 2.5 s
Now, solving for τ:
τ = (774π J) / (2.5 s) ≈ 984.39 N·m
Therefore, the constant torque required to bring the grinding wheel from rest to an angular speed of 1200 rev/min in 2.5 seconds is approximately 984.39 N·m.
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A supertrain of proper length 205 m travels at a speed of 0.86c as it passes through a tunnel having proper length 74 m. How much longer is the tunnel than the train or vice versa as seen by an observer at rest with respect to the tunnel?
The tunnel is shorter than the train by approximately 2.69 meters, as observed by an observer at rest with respect to the tunnel.
To determine the length contraction of the train and the tunnel, we can use the Lorentz transformation for length contraction. The formula is given by:
L' = L * sqrt(1 - v^2/c^2)
Where:
L' is the contracted length of an object as observed by an observer at rest with respect to the object.
L is the proper length of the object.
v is the velocity of the object.
c is the speed of light in a vacuum.
Given:
The proper length of the train (L_train) = 205 m
The proper length of the tunnel (L_tunnel) = 74 m
Speed of the train (v_train) = 0.86c
Let's calculate the contracted lengths of the train and the tunnel.
Length contraction of the train (L'_train):
L'_train = L_train * sqrt(1 - v_train^2/c^2)
L'_train = 205 m * sqrt(1 - (0.86c)^2/c^2)
L'_train = 205 m * sqrt(1 - 0.86^2)
L'_train ≈ 205 m * sqrt(1 - 0.7396)
L'_train ≈ 205 m * sqrt(0.2604)
L'_train ≈ 205 m * 0.5102
L'_train ≈ 104.601 m
Length contraction of the tunnel (L'_tunnel):
L'_tunnel = L_tunnel * sqrt(1 - v_train^2/c^2)
L'_tunnel = 74 m * sqrt(1 - (0.86c)^2/c^2)
L'_tunnel = 74 m * sqrt(1 - 0.86^2)
L'_tunnel ≈ 74 m * sqrt(1 - 0.7396)
L'_tunnel ≈ 74 m * sqrt(0.2604)
L'_tunnel ≈ 74 m * 0.5102
L'_tunnel ≈ 37.769 m
The contracted length of the train (L'_train) is approximately 104.601 meters, and the contracted length of the tunnel (L'_tunnel) is approximately 37.769 meters.
To determine the difference in length between the train and the tunnel as observed by an observer at rest with respect to the tunnel, we subtract the contracted length of the train from the contracted length of the tunnel:
Difference in length = L'_tunnel - L'_train
The difference in length ≈ 37.769 m - 104.601 m
The difference in length ≈ -66.832 m
The negative value indicates that the tunnel is longer than the train.
The tunnel is shorter than the train by approximately 2.69 meters, as observed by an observer at rest with respect to the tunnel.
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A 250 - Ω resistor is connected in series with a 4.80 - μF capacitor. The voltage across the capacitor is vC=(7.60V)⋅sin[(120rad/s) t ].
Derive an expression for the voltage VR across the resistor.
A 250 - Ω resistor is connected in series with a 4.80 - μF capacitor, the expression for the voltage VR across the resistor is VR = 91200C * cos[(120rad/s) t].
We may utilise Ohm's Law and the correlation between voltage and current in a capacitor to obtain the expression for the voltage VR across the resistor.
According to Ohm's Law, a resistor's voltage is equal to the current passing through it multiplied by its resistance:
VR = IR * R
iC = C * d(vC) / dt
d(vC) / dt = (7.60) * (120) * cos[(120) t]
iC = C * (7.60) * (120) * cos[(120) t]
VR = iC * R
= C * (7.60) * (120) * cos[(120) t] * 250
= 91200C * cos[(120rad/s) t]Ω
Therefore, the expression for the voltage VR across the resistor is VR = 91200C * cos[(120rad/s) t].
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