Physics Laws of Motion

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New answer posted

4 months ago

0 Follower 15 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

4 months ago

0 Follower 5 Views

A
alok kumar singh

Contributor-Level 10

Upward journey

0 - v 0 2 = - 2 g 2 + m g 3 2 s

S = v 0 2 g ( 1 + μ 3 )

Downward journey

v 0 2 4 - 0 = 2 g 2 - 4 g 3 2 v 0 2 g ( 1 + μ 3 )

1 4 = 1 - 3 μ 1 + 3 μ

5 3 = 2 2 3 μ μ = 3 5 = 1732 5

μ = 0.346

I = 346.00

New answer posted

4 months ago

0 Follower 11 Views

V
Vishal Baghel

Contributor-Level 10

V 2 = U 2 + 2 g S V 2 = 0 + 2 g ( h - y ) V 2 = 2 g h - 2 g y V = 2 g h - 2 g y

New answer posted

4 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

By conservation of charge, (50) (10) = 20 (10+C) ⇒ C=15µF

New answer posted

4 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

σ4πr² + σ4πR² = Q ⇒ σ = Q/ (4π (R²+r²)
V_c = kq? /r + kq? /R = k (σ4πr²)/r + k (σ4πR²)/R = kσ4π (r+R)
= K (Q/ (R²+r²) (R+r)

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