Physics

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New answer posted

6 months ago

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Payal Gupta

Contributor-Level 10

A=A0eλt1  [Radio active decay law]

A5=A0eλ (t2t1)

ln5=λ (t2t1)

Averagelife=1λ= (t2t1)ln5

New question posted

6 months ago

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New answer posted

6 months ago

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P
Payal Gupta

Contributor-Level 10

 y=αxβx2

dydx=α2βx=0

x=α2β

ymax=α*α2ββ* (α2β)2=α24β

Range = 2x=αβ=2u2sinθ.cosθg

On comparing with

y = x tan θ- gx22u2cos2θ

New answer posted

6 months ago

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Payal Gupta

Contributor-Level 10

mg – T = ma. (1)

T * R = l α. (2)

a = α R. (3)

With the help of equations (1), (2) and (3), we get

a=mgm+lR2

v=2ah=ωR

ω2=2mghl+mR2

New answer posted

6 months ago

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A
alok kumar singh

Contributor-Level 10

 Electric field due to infinite sheet is uniform.

New answer posted

6 months ago

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alok kumar singh

Contributor-Level 10

Kindly go through the solution 

New answer posted

6 months ago

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Payal Gupta

Contributor-Level 10

Using Ampere's law for long hollow cylinder carrying current on its surface

Bin = 0

Bout=μ0i2πr

New answer posted

6 months ago

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alok kumar singh

Contributor-Level 10

Mono atomic ? Cv=3R2Cp=5R2

Di- atomic ? CV=5R2Cp=7R2

(Rigid)

Di-atomic ? Cv=7R2Cp=9R2

(Non-Rigid)

Tri-atomic ? CV=3RCP=4R

New answer posted

6 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

Uinitial =k (4q) (q) (d/2)+k (q) (-q) (d/2)

6kq2dUfinal =4 (4q) (q)3d2+k (q) (-q) (d/2)23kq2dΔU=23-6kq2d-163kq2d

New answer posted

6 months ago

0 Follower 7 Views

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution 

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