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New answer posted

7 months ago

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C
Chandra Pruthi

Beginner-Level 5

The exact formula for the magnetic force per unit length between two parallel current-carrying conductors:

F l = ? 0 I1 I2 2? d For more information, check: Force between parallel conductors

New answer posted

7 months ago

0 Follower 14 Views

A
alok kumar singh

Contributor-Level 10

2SO? (g) + O? (g)? 2SO? (g)
t=0: 250 m bar 750 m bar -
0 625 m bar 2 x 250 = 500
P_total = 625 + 250m bar
= 875mbar

New answer posted

7 months ago

0 Follower 6 Views

V
Vishal Baghel

Contributor-Level 10

Both will show same tension
∴ Reading of is = 5 N.

New answer posted

7 months ago

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V
Vishal Baghel

Contributor-Level 10

P = F.v = 5+3-6 = 2W

New answer posted

7 months ago

0 Follower 7 Views

V
Vishal Baghel

Contributor-Level 10

Maximum range is obtained at 45?
u²/g = 1.6 or u = 4m/s
T = (2u sin 45? )/g = (2*4* (1/√2)/10 = 0.4√2s
Number of jumps in a given time, n = t/T = 2√2 / 0.4√2 = 5

New answer posted

7 months ago

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V
Vishal Baghel

Contributor-Level 10

Gold band is equal to 5% tolerance

New answer posted

7 months ago

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V
Vishal Baghel

Contributor-Level 10

x = 8R/2, y = 5R/2
x/y = 1.60

New answer posted

7 months ago

0 Follower 22 Views

V
Vishal Baghel

Contributor-Level 10

a? = v? ²/4r
a_A? = (v? ²/r²) * r = v? ²/r
a_A = 3v? ²/4r

New answer posted

7 months ago

0 Follower 46 Views

V
Vishal Baghel

Contributor-Level 10

-mv cos 60? + 2mu = 0 => v = 4u
½m [v² + u² + 2uv cos 120? ] + ½mu² = mgx sin 60?
=> v² = (8/7)√3gx => ar = (4√3/7)g
∴ t = √ (2L * 7)/ (4√3g)

New answer posted

7 months ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

z² * (13.6) (1 - ¼) = 3 * (13.6)
z = 2 . (i)
h/√2mk? = (1/2.3) * h/√2mk?
=> k? = (2.3)²k? = 5.25k? (ii)
Now, k? = E? - Φ
k? = E? - Φ = z²E? - Φ
∴ k? /k? = (10.2 - Φ)/ (4 * 10.2 - Φ) = 1/5.25
=> Φ = 3eV

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