Laws of Motion

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New answer posted

3 weeks ago

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V
Vishal Baghel

Contributor-Level 10

Kindly consider the following figure

New answer posted

4 weeks ago

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A
alok kumar singh

Contributor-Level 10

velocity of car at t = 4sec is
v = u + at
v = 0 + 5 (4)
= 20 m/s
At t = 6sec
acceleration is due to gravity ∴ a = g = 10 m/s
v? = 20 m/s (due to car)
v? = u + at
= 0 + g (2) (downward)
= 20 m/s (downward)
v = √ (v? ² + v? ²)
= √ (20² + 20²)
= 20√2 m/s

New answer posted

a month ago

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A
alok kumar singh

Contributor-Level 10

R = R 0 ( 1 + α Δ T )

6.8 = 2 [ 1 + α * ( 80 - 0 ) ] α = 3.4 - 1 80 = 0.03 = 3 * 10 - 2 ? C - 1

New answer posted

a month ago

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A
alok kumar singh

Contributor-Level 10

By conservation of charge, (50) (10) = 20 (10+C) ⇒ C=15µF

New answer posted

a month ago

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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

a month ago

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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

a month ago

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A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

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