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New answer posted

a year ago

0 Follower 10 Views

V
Vishal Baghel

Contributor-Level 10

sx? dg (x? /2) + sx? dg (x? /2)
Uf = (S (x? +x? )/2)gd (x? +x? )/4) x 2
= S (x? + x? )²gd/4
U? -Uf = (Sgd/4) {2x? ² + 2x? ² − (x? + x? )²}
= (Sgd/4) (x? -x? )²

New answer posted

a year ago

0 Follower 10 Views

A
alok kumar singh

Contributor-Level 10

Total mass of reactant should be greater than that of product.
This condition is only fulfilled in case- 3

New answer posted

a year ago

0 Follower 5 Views

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

New answer posted

a year ago

0 Follower 19 Views

V
Vishal Baghel

Contributor-Level 10

i? = 8/8 = 1 A    

New answer posted

a year ago

0 Follower 13 Views

V
Vishal Baghel

Contributor-Level 10

E = E? (1 − ax²)

F = qE?
acceleration = F/m = (qE? /m) (1 - ax²) = v (dv/dx)
(qE? /m) ∫? (1 - ax²)dx = ∫? vdv; (qE? /M) (x - ax³/3) = 0
x (1 - ax²/3) = 0; x = 0 & x = √ (3/a)

New answer posted

a year ago

0 Follower 9 Views

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

New answer posted

a year ago

0 Follower 21 Views

A
alok kumar singh

Contributor-Level 10

mdv? /dt = kv? (1) and mdv? /dt = kv?
(2)/ (1) ⇒ dv? /dv? = v? /v?
v? dv? = v? dv?
v? ² = v? ² + C
v? ² – v? ² = C = Constant
Now, v? * a? = (v? î + v? ) * (k/m) (v? î + v? )
= (k/m) [v? ²k? – v? ²k? ] = (k/m) (v? ² – v? ²)k? = Constant.

New answer posted

a year ago

0 Follower 7 Views

V
Vishal Baghel

Contributor-Level 10

l? = 0.6M
l? = 0.8M
√l? ² + l? ² = 1
MI about 0 = M/12 (l? ² + l? ²)
I? = M/12

MI about O' = M/12 (l? ² + l? ²) + M (l? ²+l? ²)/4
I? = M/12 + M/4 = (4m)/12; I? /I? = 1/4

New answer posted

a year ago

0 Follower 5 Views

A
alok kumar singh

Contributor-Level 10

KIndly go through the solution

 

New answer posted

a year ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

x = (IFV²)/ (WL? )
I = [ML²]
F = [MLT? ²]
V² = [L² T? ²]
W = [ML² T? ²]
Q? = [L? ]
X = [ML? ¹ T? ²]

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