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New answer posted
10 months agoContributor-Level 10
The law of radioactive decay is N = N? e? λt, where N is the amount remaining at time t.
Given that at time t, N/N? = 9/16.
So, 9/16 = e? λt
At time t/2, the fraction remaining will be N'/N?
N' = N? e? λ ( t/2 ) = N? (e? λt)¹/²
Substituting the value of e? λt:
N' = N? (9/16)¹/² = N? (3/4)
The fraction remaining is N'/N? = 3/4.
New answer posted
10 months agoContributor-Level 10
For a rotating loop, the induced emf is ε = ε? sin (ωt), where the peak emf ε? = NABω.
Here, Area A = πab. The number of turns N = 1.
So, ε? = B (πab)ω
The average power loss due to Joule heating in a resistor R is given by:
P_avg = (ε_rms)² / R
Where ε_rms = ε? /√2
P_avg = (ε? ²/2) / R = (B (πab)ω)² / (2R) = π²a²b²B²ω² / (2R)
New answer posted
10 months agoContributor-Level 10
By Conservation of Angular Momentum (COAM) about O, just before and after the collision:
Initial angular momentum L? = mv?
Final angular momentum L? = Iω? = (I_rod + I_block)ω? = (M? ²/3 + m? ²)ω?
L? = L?
mv? = (M? ²/3 + m? ²)ω?
ω? = mv / (M? /3 + m? ) = (16) / (21/3 + 1*1) = 6 / (5/3) = 18/5 rad/s
By Conservation of Total Mechanical Energy (COTME) after collision until it comes to rest:
Initial Energy (at the lowest point) = Rotational K.E. = (1/2)Iω? ²
Final Energy (at the highest point) = Potential Energy = mgh_m + Mg h_M
mgh_m = mg (? -? cosθ)
Mg h_M = Mg (? /2 - (? /2)cosθ)
(1/2) * (M? ²/3 + m? ²) * ω? ² = (mg + Mg/2) *
New answer posted
10 months agoContributor-Level 10
If the work function of the metal is φ, then the kinetic energy (K.E.) of the emitted photoelectron is given by Einstein's photoelectric equation:
K.E. = hν - φ = (hc/λ) - φ
Case 1: λ? = 500 nm
K? = (hc/λ? ) - φ
Case 2: λ? = 200 nm
K? = (hc/λ? ) - φ
Given, K? = 3K?
So, (hc/λ? ) - φ = 3 * [ (hc/λ? ) - φ]
(hc/λ? ) - φ = 3 (hc/λ? ) - 3φ
2φ = 3 (hc/λ? ) - (hc/λ? )
2φ = hc * (3/λ? - 1/λ? )
φ = (hc/2) * [3/ (500 nm) - 1/ (200 nm)]
φ = (hc/2) * [ (6 - 5) / 1000 nm]
φ = hc / 2000 nm
Using hc ≈ 1240 eV·nm
φ = 1240 eV·nm / 2000 nm = 0.62 eV
New answer posted
10 months agoContributor-Level 10
At lower end
Tension, T? = 2g = 20 N (due to the 2 kg block)
Velocity, v? = √ (T? /μ) = √ (20/μ)
Wavelength, λ? = 6 cm
At upper end
Tension, T? = (2 kg + 6 kg)g = 8g = 80 N (due to the block and the rope)
Velocity, v? = √ (T? /μ) = √ (80/μ) = √4 * √ (20/μ) = 2v?
Since frequency (f) remains the same:
f = v? /λ? = v? /λ?
⇒ λ? = λ? * (v? /v? )
⇒ λ? = λ? * (2v? /v? ) = 2λ?
⇒ λ? = 2 * 6 cm = 12 cm
New answer posted
10 months agoContributor-Level 10
Through isotropy, we can say it enables equal distribution of velocity components in all directions. This allows us to replace directional velocity terms with a fraction of the total velocity squared. That basically simplifies the pressure formula to involve only rms speed.
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