Physics
Get insights from 5.6k questions on Physics, answered by students, alumni, and experts. You may also ask and answer any question you like about Physics
Follow Ask QuestionQuestions
Discussions
Active Users
Followers
New answer posted
5 months agoContributor-Level 10
Heat generated in the resistance
H = i²RT
H? = 500 = (1.5)²R (20) ⇒ R = 500/45 = 11.11 Ω
H? = (3)²R (20) = 9 * (500/45) * 20 = 2000 J.
New answer posted
5 months agoContributor-Level 10
Voltage across secondary source
P = V? i? = V? i?
V? = P/i? = 60/0.11 ≈ 545 V
Since voltage across secondary source is more than primary source (220V)
⇒ Step- up transformer.
New answer posted
5 months agoContributor-Level 10
Statement-I: F? = mv²/r ≤ f? = µmg ⇒ v ≤ √ (µgR) = √ (0.2*9.8*2) = 1.98 m/s
v_cyclist = 7 km/h = 1.94 m/s. Since 1.94 m/s < 1.98 m/s, statement-I is correct.
Statement-II: v_max = √gR (tanθ+µ)/ (1-µtanθ) = .
v_min = √gR (tanθ-µ)/ (1+µtanθ) = √ (9.8*2 (tan45-0.2)/ (1+0.2tan45) = 3.65 m/s
v_cyclist = 18.5 km/h = 5.14 m/s. This is outside the safe range. Statement-II is incorrect.
New answer posted
5 months agoContributor-Level 10
According to Question, we can write
E = λ/ (2πε? r) ⇒ σ = ε? E = λ/ (2πr)
⇒ σ = 8*10? / (2*3.14*3) = 0.424 nCm? ²
Taking an Exam? Selecting a College?
Get authentic answers from experts, students and alumni that you won't find anywhere else
Sign Up on ShikshaOn Shiksha, get access to
- 66k Colleges
- 1.2k Exams
- 681k Reviews
- 1800k Answers









