Physics

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New answer posted

8 months ago

0 Follower 4 Views

R
Raj Pandey

Contributor-Level 9

i b = 5 0 . 7 8 . 6 = 0 . 5 m A

I c = β I b = 1 0 0 * 0 . 5   m A

By using 

V C E = V C C I C R L = 1 8 5 0 * 1 0 3 * 1 0 0 = 1 3   V

New answer posted

8 months ago

0 Follower 1 View

R
Raj Pandey

Contributor-Level 9

The charge on hole is positive.

New answer posted

8 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

F ? = e E 0 j ?

a ? y = e E 0 m j ?

= v 0 i ? + e E 0 t m j ?

v ? = v 0 2 + e E 0 t m 2 = v 0 1 + e E 0 t m 2

λ = λ 0 1 + e E 0 t m 2

New answer posted

8 months ago

0 Follower 2 Views

R
Raj Pandey

Contributor-Level 9

Two successive β decays increase the charge no. by 2.

New question posted

8 months ago

0 Follower 2 Views

New answer posted

8 months ago

0 Follower 6 Views

A
alok kumar singh

Contributor-Level 10

C 1 = 1 μ F V 1 = 100 V ? U 1 = 1 2 * 1 * 10 - 6 * 100 2 = 5 * 10 - 3 J  

C 2 = 1 μ F , common potential V = C 1 V 1 C 1 C 2 = 100 2 = 50 V

Final electrostatic energy stored in both the capacitors.

= 1 2 C 1 + C 2 V 2 = 1 2 * 2 * 50 2 * 10 - 6

= 2.5 * 10 - 3 J

E n e r g y l o s t = 2.5 * 10 - 3

% l o s s o f e n e r g y = 2.5 * 10 - 3 5 * 10 - 3 * 100

= 50 %

New answer posted

8 months ago

0 Follower 17 Views

A
alok kumar singh

Contributor-Level 10

When wire is stretched uniformly

R R

R R ' = l n l 2 = 1 n 2  

R ' = n 2 R

When this elongated wire is cut into 5 parts, resistance of each part  

Effective resistance between A and C = n 2 R 5  

New answer posted

8 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

Given   F 1 r n

m ω 2 R = K R n

? m . 4 π 2 R T 2 = K R n

? T 2 R n + 1

T R n + 1 2

New answer posted

8 months ago

0 Follower 5 Views

A
alok kumar singh

Contributor-Level 10

Angular momentum imparted to sprinkler per

s e c o n d = v r d m d t = 0.6 60 * 0.5 * 1 = 0.005 N . m

? l d ω d t = 0.005

? d ω d t = 0.005 500 * 10 - 7

= 100 r a d s 2

New answer posted

8 months ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

Applying conservation of momentum for collision of blocks

2 m v + 0 = 3 m v '

? v ' = 2 v 3 1

Now, from conservation of energy

1 2 * 3 m * v ' 2 = 1 2 k x 0 2

x 0 = 3 m v ' 2 k = v ' 3 m k

= 2 v 3 3 m k = v 4 m 3 k

 

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