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New answer posted

a year ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

From momentum conservation
mui + 0 = mvj + 3mv'
v' = u/3 I - v/3 j
From kinetic energy conservation 1/2 mu² = 1/2 mv² + 1/2 (3m) (u/3)²+ (v/3)²)
Solving, v = u/√2

New answer posted

a year ago

0 Follower 13 Views

V
Vishal Baghel

Contributor-Level 10

Flux as a function of time Φ = B.A = ABcos (ωt) emf induced,
e = -dΦ/dt = ABωsin (ωt)
Maximum value of emf = ABω = πR²Bω
= 3.14*0.1*0.1*3*10? * (0/0.2) = 15

New answer posted

a year ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution 

 

New answer posted

a year ago

0 Follower 22 Views

A
alok kumar singh

Contributor-Level 10

ρdrω²r = ρgdh
ω²∫? rdr = g∫? dh
ω²R²/2 = gh
h = ω²R²/2g = 25ω²/2g

 

New answer posted

a year ago

0 Follower 7 Views

A
alok kumar singh

Contributor-Level 10

mω²acosθ = mgsinθ
ω = √ (gtanθ/a)
y = 4cx²

tanθ = dy/dx = 8xC
(tanθ)? , b = 8aC
ω = √ (g*8aC/a) = 2√ (2gC)

 


New answer posted

a year ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

ρ? = 98 * 10?
ρ? = 2.65 * 10?
ρc = 1.724 * 10?
ρT = 5.65 * 10?

New answer posted

a year ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

ΔP = dsinθ = dθ
dy/D = (10? ³ * 1.270mm)/1m = 1.27µm

New answer posted

a year ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

M = ∫ ρdV
M = ∫? (k/r) 4πr²dr
M = 4πkR?²/2 = 2πkR?²
F? = GMm/R?² = 2ω?²R

 

New answer posted

a year ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

FR > mgcosθR
F > mgcosθ
F > mg √ (R²- (R-a)²)/R ⇒ Mg√ (1- (R-a)²/R²)

 

New answer posted

a year ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

Zero error = 0 + 7 * 0.1 = 0.070
Vernier reading = (3.1 + 4 * 0.01) – 0.07 = 3.07

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