Class 12th

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New answer posted

6 months ago

0 Follower 4 Views

V
Vishal Baghel

Contributor-Level 10

 m1=1kg, r1=i^+2j^+k^

m1=3kg, r2=3i^2j^+k^

rcom=m1r1+m2r2m1+m2=14 (i^+2j^+k^+9i^6j^+3k^)

2i^j^+k^

|rcom|=4+1+1=6.

New answer posted

6 months ago

0 Follower 4 Views

V
Vishal Baghel

Contributor-Level 10

 E=12mu2

EHighestpoint=12m (u2)2=18mu2=E4

New answer posted

6 months ago

0 Follower 4 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

6 months ago

0 Follower 9 Views

V
Vishal Baghel

Contributor-Level 10

N = m v 2 R

 

New answer posted

6 months 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

6 months ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

 t=x+4

then dtdx=12x (dxdt)t=4= (2x)t=4= [2 (t4)]att=4

= 0

New answer posted

6 months 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

6 months ago

0 Follower 5 Views

V
Vishal Baghel

Contributor-Level 10

 h=u22gu=2gh

h3=2ght5t2

5t220ht+h3=0

t=20h±20h4*5*h310

t1t2=20h403h20h+403h=1231+23=323+2

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