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New answer posted
7 months agoBeginner-Level 5
The exact formula for the magnetic force per unit length between two parallel current-carrying conductors:
For more information, check: Force between parallel conductors
New answer posted
7 months agoContributor-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 agoContributor-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 agoContributor-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 agoContributor-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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