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New answer posted

a month ago

0 Follower 87 Views

P
Payal Gupta

Contributor-Level 10

Given Acceleration of m2 is 2m/s2

(upward)

N = 20a              T – m2g = m2a

T – 20g = 20a

T = 200 + 20 * 2

T = 240 N

T = m1 a1

240 = 10a1

a1 = 24 m/s2

Ta+Ta1Ta2=0

a2=24+2=26m/s2

F – T – N = Ma2

F – 240 = 100 * 26 + 20 * 26

F = 3360 N

New answer posted

2 months ago

0 Follower 1 View

V
Vishal Baghel

Contributor-Level 10

For equilibrium net force acting on the system should be zero.

New answer posted

2 months ago

0 Follower 3 Views

R
Raj Pandey

Contributor-Level 9

M V 0 = M V 1 + m V 2 . . . . ( i )

M V 1 = m V 2  . (ii)
M V 0 2 = M V 1 2 + m V 2 2 . (iii)
M V 0 2 = M V 1 2 + m ( M V 1 m ) 2 [ u s i n g e q n ( i i ) ]
( 1 + M m ) = 4
( M m ) m a x = 3

 

New answer posted

2 months ago

0 Follower 4 Views

R
Raj Pandey

Contributor-Level 9

θ = 6 0 ° 2 = 3 0 °

 

New answer posted

2 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

V r e l d m d t m g = F e x t

=>500 d m d t 1 0 0 0 * 1 0 = 1 0 0 0 * 2 0

d m d t = 6 0 k g / s

             

New answer posted

2 months ago

0 Follower 2 Views

R
Raj Pandey

Contributor-Level 9

F. B. D of Beaker 

By NLM2                    

f = m a = m ω 2 R m ω 2 R μ N R μ g / ω 2

New answer posted

2 months ago

0 Follower 3 Views

P
Payal Gupta

Contributor-Level 10

Let additional force is F , so

F+5 (+i^)+6 (i^)+7 (+j^)+8 (j^)=0

F (i^+j^)=0F=i^+j^F=2N, 45°

New answer posted

2 months ago

0 Follower 2 Views

P
Payal Gupta

Contributor-Level 10

Δp = Change in momentum of each ball

Δp=2mv=2*0.05*10=1Ns

F=ΔpΔt=2*0.05*100.005=200N

New answer posted

2 months ago

0 Follower 2 Views

P
Payal Gupta

Contributor-Level 10

Since liquid drop is in equilibrium, so

mg = FB + 2πRT

43πR3ρg=23πR3σg+2πRT

R=3T (2ρσ)g=15 (2ρσ)*102m=15 (2ρσ)cm

New answer posted

2 months ago

0 Follower 4 Views

V
Vishal Baghel

Contributor-Level 10

F t = m Δ v

? F ? = m Δ v t = 1 0 * 1 0 3 * 4 . 5 1 0 0 * 5 9 * 10-4 N.

= 9 * 10-5 N

= 9 dyne.

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