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New answer posted
4 months agoContributor-Level 10
In y axis u? = v? a? = -E? q / m
s? = 0, u? t + (1/2)a? t² = 0 ⇒ t = 2u? /a? = 2v? m/E? q
x coordinate at that time = v? * t = (2v? m/0) * v? = (2v? ²m)/E? q
New answer posted
4 months agoContributor-Level 10
Let the origin be at the CM of the particles, let the initial positions of the particles be x = a/2 and x = -a/2 and let the instantaneous positions of the particles be x = r and x = -r
Let the instantaneous velocity of each particle be v
Let the time after which the distance between the particles has reduced to a/2 be T
Then, for the particle that was initially at x = -a/2,
(Gm²/ (2r)²) = -m (vdv/dr) ⇒ (Gm/r²) = - (4vdv/dr) ⇒ -4∫vdv = Gm∫ (dr/r²)
[v is negative because the velocity is towards the –X direction]
dr/dt = -√ (Gm/2) (1/r - 2/a)¹? ²
⇒ ∫ (a/2)^ (a/4) (r/√ (a-2r)dr = -√ (Gm/2a) ∫? dt
⇒ ∫ (a/2)^ (a/4
New answer posted
4 months agoContributor-Level 10
For permanent magnetic Y is good choice, As it has very high coercivity (comparitively)
New answer posted
4 months agoContributor-Level 10
Taking torque at top pt, mg sin30°*L / 2 = F cos30°*L ⇒ F = mg / 2√3
New answer posted
4 months agoNew answer posted
4 months agoContributor-Level 10
The time spent in magnetic field (t) is independent of velocity of charge.
t = (1/4) (2πm/qB)
New answer posted
4 months agoContributor-Level 10
At minima, I = (√I? - √25I? )² = 16I?
At maxima, I = (√I? + √25I? )² = 36I?
∴ Ratio = 16/36 = 4/9
New answer posted
4 months agoContributor-Level 10
Time taken by trains to collide = 300/150 = 2hrs
Speed of bird = 200 km/hr
Total distance travelled by bird = 200 * 2 = 400 km
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