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New answer posted

3 weeks ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

R = u x * T

= g h * 0 . 5 g h * 2 g = h

A B = h + h = 2 h

New answer posted

3 weeks ago

0 Follower 3 Views

R
Raj Pandey

Contributor-Level 9

Acceleration of the combined system along the incline is given by

a = g s i n ? 37 ? a = 3 g 5

With respect to block M :

N = m g - m a s i n ? 37 ? = m g - 3 m g 5 3 5 N = 16 25 m g μ N = m a c o s ? 37 ? μ 16 25 m g = 3 m g 5 4 5 μ = 3 4 = 750 1000 I = 750

 

New answer posted

3 weeks ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

I? = P? – P? = 3 * 4? – 3 * 3î = 12? – 9î |I? | = √12² + 9² = 15kgm/s

New answer posted

3 weeks ago

0 Follower 16 Views

V
Vishal Baghel

Contributor-Level 10

N = (mg/√2) + (mω²h/√2)
(mω²h/√2) = (mg/√2) + f?
(mω²h/√2) = (mg/√2) + μ (mg/√2) + (mω²h/√2)
ω = √ (g/h) (1+μ)/ (1-μ) = 10rad/s

New answer posted

3 weeks ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

Let the acceleration of string = a

For 2 kg 20 – T1 = 2a ………… (i)

For monkey T2 – 80 – T1 = 8 (2 – a)  …. (ii)

For 10 kg T2 – 100 = 10a    ……. (iii)

a = 0 . 8 m / s 2

Acceleration of monkey w.r. to ground

= 2 – 0.8 = 1.2 m/s2

  s = u t + 1 2 a t 2        

2 . 4 = 0 + 1 2 * 1 . 2 * t 2            

t = 2 sec

New answer posted

3 weeks ago

0 Follower 1 View

V
Vishal Baghel

Contributor-Level 10

Centripetal acceleration is provided by friction force. Maximum friction force should be equal to or greater that mv² / R.
µmg = mv²/R ; µ = 100/ (100*10) ; µ = 1/10 ; µ = 0.1

New answer posted

3 weeks ago

0 Follower 4 Views

V
Vishal Baghel

Contributor-Level 10

Taking torque at top pt, mg sin30°*L / 2 = F cos30°*L ⇒ F = mg / 2√3

New answer posted

3 weeks ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

Applying conservation of momentum, m (ui + 2uj) + 3m (0) = 4m (vi +vj)
⇒ v = u / 4 and v' = u / 2

New answer posted

3 weeks ago

0 Follower 3 Views

R
Raj Pandey

Contributor-Level 9

F = 1 0 i ^ + 5 j ^

a = 1 0 m i ^ + 5 m j ^

= 1 0 0 . 1 k g i ^ + 5 0 . 1 j ^

a = 1 0 0 i ^ + 5 0 j ^

a x = 1 0 0 , a y = 5 0

S x = u t + 1 2 a t 2

= 0 * 2 + 1 2 * 1 0 0 * 2 * 2

Sx = 200 m

a b = 2 0 0 1 0 0 = 2

New answer posted

3 weeks ago

0 Follower 2 Views

R
Raj Pandey

Contributor-Level 9

F.B.D. of hanging length

F.B.D. of chain lying on the table

f = T = μ N  

λ x g = μ λ ( L x ) g  

x = 0 . 5 ( 6 x )  

x = 3 0 . 5 x    

  1 . 5 x = 3  

x = 2m

 

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