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New answer posted
4 months agoContributor-Level 10
FR > mgcosθR
F > mgcosθ
F > mg √ (R²- (R-a)²)/R ⇒ Mg√ (1- (R-a)²/R²)
New answer posted
4 months agoContributor-Level 10
dm (t)/dt = bv²
F_thrust = v ( dm/dt )
Force on satellite = -v (dm (t)/dt)
M (t)a = -v (bv²)
a = - (bv³)/M (t)
New answer posted
4 months agoContributor-Level 10
For no toppling
(in limiting case)
But is not possible as maximum value of
can be equal to
only.
New answer posted
4 months agoContributor-Level 10
Let vel of A and B just after collision be VA & VB respectively.
m * 9 + 0 = m * VA + 2mVB
9 = VA + 2VB
again e = (VB - VA)/ (9-0) = 1
9 = VB - VA
From (i) & (ii)
VB = 6 m/s & VA = -3 m/s
Now, for B and C collision (completely inelastic):
2m * 6 + 0 = (2m + 2m)Vc
12m = 4mVc
Vc = 3 m/s
New answer posted
4 months agoContributor-Level 10
a = F/m = (8î + 2? )
r = ut + ½ at²
r = 0 + ½ (8î + 2? ) 10²
r = 400î + 100?
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