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New answer posted

2 months ago

0 Follower 5 Views

A
alok kumar singh

Contributor-Level 10

velocity of car at t = 4sec is
v = u + at
v = 0 + 5 (4)
= 20 m/s
At t = 6sec
acceleration is due to gravity ∴ a = g = 10 m/s
v? = 20 m/s (due to car)
v? = u + at
= 0 + g (2) (downward)
= 20 m/s (downward)
v = √ (v? ² + v? ²)
= √ (20² + 20²)
= 20√2 m/s

New answer posted

2 months ago

0 Follower 7 Views

V
Vishal Baghel

Contributor-Level 10

i = (V? /R) (1 - e? /? )
U = ½Li²
dU/dt = Li (di/dt)
(R/2L) * (V? ²/R²) (1 - e? /? )²e? /? = (V? ²/R) (1 - e? /? )
t = (L/R)ln2

New answer posted

2 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

(4/3)πR³ = 27 (4/3)πr³) ⇒ R = 3r (i)
v = Kq/r = V? /r = (q? /q? ) (r? /r? ) (i)
⇒ 220/V? = (q/ (27q) (3r/r) (i)
⇒ 220/V? = 1/9 (i)
⇒ V? = 220 * 9 = 1980 volt (i)

New answer posted

2 months ago

0 Follower 5 Views

V
Vishal Baghel

Contributor-Level 10

p²/2m = Vq => p = √2mVq
θ = sin? ¹ (l/R') = l/R = mV/qB = p/qB

New answer posted

2 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

IgRc = (I - Ig) (S? + S? + S? )
1mA * 10 = 9mA (S? + S? + S? )

New answer posted

2 months ago

0 Follower 1 View

R
Raj Pandey

Contributor-Level 9

Kindly consider the following Image 

 

New answer posted

2 months ago

0 Follower 1 View

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

New answer posted

2 months ago

0 Follower 4 Views

V
Vishal Baghel

Contributor-Level 10

W_agent + W_battery = ΔU
W = ½ (KC - C)V² - [ (KC - C)V]V
= ½ (K - 1)CV²

New answer posted

2 months ago

0 Follower 2 Views

R
Raj Pandey

Contributor-Level 9

r n = a 0 n 2

r 1 = a 0 = 5.3 * 10 - 11 m r 3 = a 0 ( 3 ) 2 = 5.3 * 10 - 11 * 9 = 4.77 ?

New answer posted

2 months ago

0 Follower 4 Views

V
Vishal Baghel

Contributor-Level 10

W = Uf - Ui
= 0 - (-P (λ/2πε? d)
= Pλ/2πε? d

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