Class 12th

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New answer posted

10 months ago

0 Follower 13 Views

P
Payal Gupta

Contributor-Level 10

Put one part of shaded region to another side, shape becomes.

? =q24ε0

New answer posted

10 months ago

0 Follower 6 Views

P
Payal Gupta

Contributor-Level 10

When capacitor is removed

tan 45° = XLXRωL=R

When inductor is removed.

tan45°=XCXR1ωC=R

i0=V0Z=V0R2+ (ωL1ωC)2=V0R=220110=2A

New answer posted

10 months ago

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P
Payal Gupta

Contributor-Level 10

Fermi level of p-type semiconductors will go downward

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10 months ago

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P
Payal Gupta

Contributor-Level 10

Stopping potential defined in terms of wavelength as :

eV=hcλ? 0.710eV=hc491? 1.43eV=hcλ?

Using above two equation we can calculate :

λ=382nm

New answer posted

10 months ago

0 Follower 10 Views

P
Payal Gupta

Contributor-Level 10

Using truth table of logic gates

Y= (AB¯)+ (A¯B)¯= ( (AB¯))¯ ( (A¯B))¯

Y= (A¯+B¯) (A¯+B¯)= (A¯+B) (A+B¯)=AB+A¯B¯

New question posted

10 months ago

0 Follower 7 Views

New answer posted

10 months ago

0 Follower 19 Views

A
alok kumar singh

Contributor-Level 10

A 2 B 3 ? 2 A + 3 + 3 B 2

1-α        2α        3α

i = 1 + 4 α = 1 + 4 * 0 . 6 = 1 + 2 . 4 = 3 . 4

Δ T b = i k b m = 3 . 4 * 0 . 5 2 * 1 = 1 . 7 6 8 1 . 7 7 K

T b = 3 7 4 . 7 7 K 3 7 5 K .         

New answer posted

10 months ago

0 Follower 11 Views

A
alok kumar singh

Contributor-Level 10

l o g K = l o g A E a 2 . 3 0 3 R T

S l o p e = E a 2 . 3 0 3 R = 1 0 , 0 0 0 K  

E a 2 . 3 0 3 R = 1 0 4            

l o g ( 1 0 5 ) = l o g A 1 0 4 * 1 5 0 0   (at 500 K temperature)

T = 1 0 4 1 9 K = 1 0 0 1 9 * 1 0 2 K = 5 . 2 6 3 1 * 1 0 2 K = 5 2 6 . 3 1 K 5 2 6 K

New answer posted

10 months ago

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P
Pallavi Arora

Beginner-Level 5

When two or more individual charges are present in a system, the total charge will be an algebraic sum of all individual charges and not the vector sum. Therefore, an electric charge is considered as a scalar quantity.

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10 months ago

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C
Chandra Pruthi

Beginner-Level 5

This is an interesting question! Even though a proton has a positive charge, the net positive charge in a conducting material is always due to the removal of free electrons. 

This happens due to the availability of only free electrons in all conductors. Since protons are present in the nucleus, they can not roam freely in the conductor, as electrons travel freely in the conductor due to their presence in the outer shell.

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