Class 12th

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New answer posted

a year ago

0 Follower 7 Views

V
Vishal Baghel

Contributor-Level 10

 m1=1kg, r→1=i^+2j^+k^

m1=3kg, r→2=−3i^−2j^+k^

r→com=m1r→1+m2r→2m1+m2=14 (i^+2j^+k^+−9i^−6j^+3k^)

= 2i^−j^+k^

|r→com|=4+1+1=6.

New answer posted

a year ago

0 Follower 4 Views

V
Vishal Baghel

Contributor-Level 10

 E=12mu2

EHighest point=12m (u2)2=18mu2=E4

New answer posted

a year ago

0 Follower 6 Views

A
alok kumar singh

Contributor-Level 10

Sol.

↑ 1 ?   (1)   ε = 3   (2)   ε - I r = 2.5 V ⇒ I r = 0.5   Now, IR   = 2.5 ⇒ R r = 5 . ⇒ P R R r = I 2 R I 2 r = R r = 5 ⇒ P r = 0.5 5 = 0.1

 

New answer posted

a year ago

0 Follower 14 Views

V
Vishal Baghel

Contributor-Level 10

N = m v 2 R

 

New answer posted

a year ago

0 Follower 4 Views

P
Payal Gupta

Contributor-Level 10

 

C 1 : | z − ( 4 + 3 i ) | = 2     a n d     C 2 : | x | + | z − 4 | = 6 , Z ∈ C  

C1 : circle centre (4, 3) radius 2

C2 : Ellipse foci (0, 0) & (4, 0)

length of major axis = 6,

length of semi-major axis 2 5  

Now, (4, 2) lies inside the both C1 and C2 and (4, 3) lie outside C2

∴  Number of point of intersection = 2

New answer posted

a year ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

 t=x+4

then dtdx=12x∴ (dxdt)t=4= (2x)t=4= [2 (t−4)]at  t=4

= 0

New answer posted

a year ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

Sol. ? v

⇒mv=16mv1

⇒V1=V16

⇒Δk loss =12mv2-12 (16m)V162

⇒12mv21516

% loss =1516*100=93.75%

New answer posted

a year ago

0 Follower 17 Views

V
Vishal Baghel

Contributor-Level 10

 h=u22g⇒u=2gh

h3=2gh t−5t2

5t2−20h t+h3=0

t=20h±20h−4*5*h310

∴t1t2=20h−403h20h+403h=1−231+23=3−23+2

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