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New answer posted
5 months agoContributor-Level 9
Given: Power in R_B = 200 W, R_B = 50 Ω, V_source = 200 V
I² R_B = 200
I² (50) = 200
I² = 4 ⇒ I = 2A
Also, I = V_source / (R + R_B)
2 = 200 / (R + 50)
2 (R + 50) = 200
R + 50 = 100
R = 50Ω
New answer posted
5 months agoContributor-Level 9
From conservation of momentum:
2*4 = 2v? + mv?
Given v? = 1 m/s (interpreted from intermediate steps)
8 = 2 (1) + mv?
mv? = 6 . (i)
From coefficient of restitution (e=1 for elastic collision):
e = (v? - v? )/ (u? - u? )
1 = (v? - v? )/ (4 - 0)
-1 = (v? - v? )/ (0 - 4) (as written in the image)
⇒ 4 = v? - 1
⇒ v? = 5 . (ii)
Put (2) in (1), m (5) = 6
m = 1.2kg
New answer posted
5 months agoContributor-Level 10
E = ½mω²a² . (i)
When KE = 3E/4, PE = E - KE = E/4
E/4 = ½mω²y² . (ii)
Divide eq? (i) by eq? (ii) we get
4 = a²/y² => y = a/2
New answer posted
5 months agoContributor-Level 10
h = u²/2g, u = √2gh
Now, S = h/3
S = ut + ½at²
h/3 = √2ght - ½gt²
t² - 2√ (2h/g)t + 2h/3g = 0
Using quadratic formula for t:
t = ( 2√ (2h/g) ± √ (8h/g) - 4 (2h/3g) / 2
t = √ (2h/g) ± √ (2h/g - 2h/3g) = √ (2h/g) ± √ (4h/3g)
t? /t? = (√ (2h/g) - √ (4h/3g) / (√ (2h/g) + √ (4h/3g)
t? /t? = (√2 - √ (4/3) / (√2 + √ (4/3) = (√6 - 2)/ (√6 + 2)
(Note: There is a calculation error in the provided solution. Re-evaluating the physics.)
h/3 = (√2gh)t - ½gt²
(g/2)t² - (√2gh)t + h/3 = 0
t = (√2gh ± √ (2gh - 4 (g/2) (h/3)/g = (√2gh ± √ (4gh/3)/g
t? /t? = (√2gh - 2√gh/√3)/ (√2gh + 2√gh/√3)
New answer posted
5 months agoContributor-Level 9
(1/2)mu² = (1/2)mv² + mgl (1-cosθ)
⇒ v² = u² - 2gl (1-cosθ)
⇒ v² = 3² - 2*10*0.5* (1 - 1/2)
⇒ v² = 9 - 5 = 4
v = 2m/s
New answer posted
5 months agoContributor-Level 9
I = I? e? /?
= (20/10000) e^- (1*10? / 10*10? ³)
= 2 * 10? ³ e? ¹
The provided solution calculates as:
= 2 * 10? ³ e? ¹
= 2e? ¹ mA
= 2 * 0.37mA
= 0.74 mA = 74/100 mA
(Note: There seems to be a calculation discrepancy in the source image, the steps shown lead to I = 2e? ¹ mA)
New answer posted
5 months agoContributor-Level 10
β = λD/d or β ∝ λ
Also, we know that λ_blue < _orange
So, β_orange > β_blue
So, dist b/w consecutive fringes will decrease.
New answer posted
5 months agoContributor-Level 9
F_net = -kq²/ (l+x)² + kq²/ (l-x)²
= kq² [ (l+x)² - (l-x)² / (l²-x²)² ]
= kq² [ 4lx / (l²-x²)² ]
For x << l,
ma ≈ -4kq²x / l³
a = - (4kq²/ml³)x
ω² = 4kq²/ml³
ω = √ (4kq²/ml³)
= √ (4 * 9*10? * 10 / (1*10? * 1)
= 6 * 10? rad/s
= 6000 * 10? rad/s
New answer posted
5 months agoContributor-Level 10
As charge remains same. So,
Initial total charge Q = (2C)V + CV = 3CV
When dielectric is inserted in C, its new capacitance is KC.
The capacitors 2C and KC are in parallel.
Equivalent capacitance C_eq = 2C + KC = (K+2)C
New potential Vc = Q/C_eq = 3CV/ (K+2)C) = 3V/ (K+2)
New answer posted
5 months ago
Contributor-Level 9
ω = √k_eq/μ [μ = (m? )/ (m? +m? ) (Reduced mass)]
k_eq = (k * 4k)/ (k+4k) = 4k/5
ω = √ (4k/5) / (m? / (m? +m? )
= √ (4*20/5) / (0.2*0.8)/ (0.2+0.8)
= √ (16/0.16)
= 10 rad/s
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