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New answer posted
4 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
4 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
4 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
4 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.
New answer posted
4 months agoContributor-Level 10
This is because pressure is a result of how momentum transfers during collisions. We also know that momentum depends on velocity. The derivation shows that pressure is proportional to the average of squared velocities. That we know from the root-mean-square (rms) speed.
New answer posted
4 months agoContributor-Level 10
. τ? about B is zero at equilibrium
T?100 - mg*50 – 2mg*25 = 0
T?100 = 100mg
T? = 1mg
New answer posted
4 months agoContributor-Level 10
V? = 36 km/hr = 10 m/s
V? = -72 km/hr = -20 m/s
V? = -1.8 km/hr = -0.5 m/s
V? = V? + V?
= V? + V? - V?
= -0.5 + 10 – (-20)
= -0.5 + 30 = 29.5 m/s
New answer posted
4 months agoContributor-Level 10
P = E/t
= 2/235 * (6.023*10²? *200*1.6*10? ¹? )/ (30*24*60*60) = 60W
New answer posted
4 months agoContributor-Level 10
Pitch = (Vcosθ)T = (Vcosθ)2πm/eB
= (4*10? cos60°) (2π (1.67*10? ²? )/ (0.3*10? ¹? *1.69*10¹? )
= 4 cm
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