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New answer posted
10 months agoContributor-Level 10
E = P? A? T? [ML²T? ²] = [MLT? ¹]? [L²]? [T]?
a=1, 2b+a=2⇒b=1/2, -a+c=-2⇒c=-1. [PA¹/²T? ¹].
New answer posted
10 months agoContributor-Level 10
Efficiency η = Q_net/Q? = (1915-40+125-Q)/ (1915+125) = 0.5.
(2000-Q)/2040=0.5 ⇒ Q=980J.
New answer posted
10 months agoContributor-Level 10
F = MA = m (-ω²x). F=0 at x=0 (t=T/4, 3T/4.). (a) is correct.
Acceleration is max at extremes (t=0, T/2, T.). (b) is correct.
Speed is max at mean position (t=T/4, 3T/4.). (c) is correct.
PE=KE at x=A/√2. t=T/8. (d) is incorrect.
New answer posted
10 months agoContributor-Level 10
By conservation of angular momentum, I? ω? + I? ω? = (I? +I? )ω
ω = (I? ω? +I? ω? )/ (I? +I? ) = (0.110 + 0.25)/ (0.1+0.2) = 2/0.3 = 20/3 rad/s
K = ½ (I? +I? )ω² = ½ (0.3) (20/3)² = 20/3 J.
New answer posted
10 months agoContributor-Level 10
g_h = g (1-2h/R). g_d = g (1-d/R). Given h=d.
g_h = g (R/ (R+h)² ≈ g (1-2h/R). This differs from the image.
The image has g/ (1+h/R)² = g (1-h/R). This leads to h² + hR - R²=0. h = R (√5-1)/2.
New answer posted
10 months agoContributor-Level 10
ΔE = have = 13.6Z² (1/n? ² - 1/n? ²)eV.
hv = 13.6 (1²) (1/n² - 1/ (n+1)²) = 13.6 (n+1)²-n²)/ (n² (n+1)²)
v = (13.6/h) * (2n+1)/ (n² (n+1)²). For n>>1, v ≈ (13.6/h) (2n/n? ) ∝ 1/n³.
New answer posted
10 months agoContributor-Level 10
R = 100Ω. tanφ = (X_L-X_C)/R. tan45° = X_L/R (since it's an inductor). X_L=R=100Ω.
Lω=100 ⇒ L (2πf)=100 ⇒ L = 100/ (2π1000) ≈ 1.59 * 10? ² H. The closest option is (A).
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