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New answer posted
4 months agoContributor-Level 10
V_AB = 2Ω (1A) + 2Ω (3A) = 8V.

New answer posted
4 months agoContributor-Level 10
V_PQ * (49/100). 1.02 = V_PQ * 0.49. V_PQ ≈ 2.08V.
Potential gradient = 2.08V/100cm = 0.0208 V/cm.
New answer posted
4 months agoContributor-Level 10
Propagation direction v? = E? * B? .
v? = k? * (2i-2j) = 2 (k*i) - 2 (k*j) = 2j + 2i.
Unit vector is (i+j)/√2.
New answer posted
4 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
4 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
4 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
4 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
4 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.
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