Physics Laws of Motion

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New answer posted

2 months ago

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A
alok kumar singh

Contributor-Level 10

σ4πr² + σ4πR² = Q ⇒ σ = Q/ (4π (R²+r²)
V_c = kq? /r + kq? /R = k (σ4πr²)/r + k (σ4πR²)/R = kσ4π (r+R)
= K (Q/ (R²+r²) (R+r)

New answer posted

2 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

By Conservation of Angular Momentum (COAM) about O, just before and after the collision:
Initial angular momentum L? = mv?
Final angular momentum L? = Iω? = (I_rod + I_block)ω? = (M? ²/3 + m? ²)ω?
L? = L?
mv? = (M? ²/3 + m? ²)ω?
ω? = mv / (M? /3 + m? ) = (16) / (21/3 + 1*1) = 6 / (5/3) = 18/5 rad/s

By Conservation of Total Mechanical Energy (COTME) after collision until it comes to rest:
Initial Energy (at the lowest point) = Rotational K.E. = (1/2)Iω? ²
Final Energy (at the highest point) = Potential Energy = mgh_m + Mg h_M
mgh_m = mg (? -? cosθ)
Mg h_M = Mg (? /2 - (? /2)cosθ)
(1/2) * (M? ²/3 + m? ²) * ω? ² = (mg + Mg/2) *

...more

New answer posted

2 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

From momentum conservation
mui + 0 = mvj + 3mv'
v' = u/3 I - v/3 j
From kinetic energy conservation 1/2 mu² = 1/2 mv² + 1/2 (3m) (u/3)²+ (v/3)²)
Solving, v = u/√2

New answer posted

2 months 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

2 months 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

2 months 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

2 months ago

0 Follower 6 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

2 months 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

3 months 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

3 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

T 2 = 100 T 2 = F

F = 100 N

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