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New answer posted

a month ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

mω²acosθ = mgsinθ
ω = √ (gtanθ/a)
y = 4cx²

tanθ = dy/dx = 8xC
(tanθ)? , b = 8aC
ω = √ (g*8aC/a) = 2√ (2gC)

 


New answer posted

a month ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

FR > mgcosθR
F > mgcosθ
F > mg √ (R²- (R-a)²)/R ⇒ Mg√ (1- (R-a)²/R²)

 

New answer posted

a month ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

mv? *cosθ*2=2m (v? /2)
⇒ cosθ=1/2 ⇒ θ=60°
∴ 2θ=120°

New answer posted

a month ago

0 Follower 5 Views

V
Vishal Baghel

Contributor-Level 10

µ=tanθ ⇒ 3/4=tanθ ⇒ θ=37°
h = R-Rcosθ = 1-1 (4/5)=0.2m

New answer posted

a month ago

0 Follower 6 Views

V
Vishal Baghel

Contributor-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

a month ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

50.00

For no toppling

F a 2 + b m g a 2

μ a 2 + μ b a 2

0.2 a + 0.4 b 0.5 a

0.4 b 0.3 a

b 3 a 4

b 0.75 a (in limiting case)

But is not possible as maximum value of  can be equal to  only.

100 b a m a x = 50.00

New answer posted

a month ago

0 Follower 1 View

A
alok kumar singh

Contributor-Level 10

T 2 = 100 T 2 = F

F = 100 N

New answer posted

a month ago

0 Follower 1 View

R
Raj Pandey

Contributor-Level 9

Kindly consider the following Image

 

New question posted

a month ago

0 Follower 4 Views

New answer posted

a month ago

0 Follower 1 View

R
Raj Pandey

Contributor-Level 9

Kindly consider the following Image

 

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