Physics Laws of Motion

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New answer posted

3 months ago

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S
Syed Aquib Ur Rahman

Contributor-Level 10

As per Galileo's Law of Inertia, objects in motion have a natural tendency to stay in motion. This property is called inertia. But, they stop moving as there are external forces. Now, we should know that friction is a type of force. It acts in parallel and opposes motion when two surfaces are in contact. Then we have air resistance, which is a type of friction that acts on objects as they move through the air.

In an ideal scenario, as Galileo and Newton would have proved through their observations and mathematical enquiries, there will be no friction or air resistance. Then an object in motion would continue to move indefinitely in

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New answer posted

3 months ago

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A
alok kumar singh

Contributor-Level 10

for smooth surface

a = g s i n 3 0 ° = g 2

S 1 = u t + 1 2 a t 2         

S 1 = 1 2 g 2 t 2 = g 4 t 2 . . . . . . . ( i )               

for rough Surface

S = 1 2 g 2 ( 1 μ 3 ) α 2 t 2 . . . . . . . . ( i i )               

By (i) and (ii)

μ = 1 3 ( α 2 1 α 2 ) x = 3            

 

New answer posted

3 months ago

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V
Vishal Baghel

Contributor-Level 10

x = | P + Q | = P 2 + Q 2          

=> y = | P Q | = P 2 + Q 2

| x + y | = x 2 + y 2 + 2 x y

3 ( P 2 + Q 2 ) = ( P 2 + Q 2 ) 2 + ( P 2 + Q 2 ) 2 + 2 ( P 2 + Q 2 ) c o s θ 1

θ 1 = 6 0 °

=>Using same formula : θ2 = 90°

 

New answer posted

3 months ago

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A
alok kumar singh

Contributor-Level 10

Suppose acceleration of wedge is a and acceleration of block w. r.t. wedge is a1 then N cos60° = Ma = 16 a -> N = 32 a

For block w.r.t. wedge

N + 8a sin 30° = 8g cos 30°

N = 8g cos 30° - 8a sin 30°

->32a = 8g cos 30° - 8a sin 30°

->a = 3 9 g  

Now for 8 kg,

8 g s i n 3 0 ° + 8 a c o s 3 0 ° = m a 1  

a 1 = g * 1 2 + 3 9 g * 3 2  


= 2 3 g  

 

New answer posted

3 months ago

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V
Vishal Baghel

Contributor-Level 10

For equilibrium net force acting on the system should be zero.

New answer posted

3 months ago

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R
Raj Pandey

Contributor-Level 9

M V 0 = M V 1 + m V 2 . . . . ( i )

M V 1 = m V 2  . (ii)
M V 0 2 = M V 1 2 + m V 2 2 . (iii)
M V 0 2 = M V 1 2 + m ( M V 1 m ) 2 [ u s i n g e q n ( i i ) ]
( 1 + M m ) = 4
( M m ) m a x = 3

 

New answer posted

3 months ago

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R
Raj Pandey

Contributor-Level 9

θ = 6 0 ° 2 = 3 0 °

 

New answer posted

3 months ago

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A
alok kumar singh

Contributor-Level 10

For 8 kg block,

8g – 2T = 8a       . (i)

For 2kg block,

T – 2g = 2 * 2a   . (ii)

Solving equations (i) and (ii),

a = g 4 = 2 . 5 m / s 2              

t = 2 s a = 2 * 0 . 2 2 . 5 = 0 . 4 s              

New answer posted

3 months ago

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V
Vishal Baghel

Contributor-Level 10

F x = [ 1 0 * 3 2 + 2 0 * 1 2 + 2 0 2 1 5 2 1 5 3 2 ] = 9 . 2 5 i ^

F y = [ 1 5 * 1 2 + 2 0 * 3 2 + 1 0 * 1 2 1 5 2 2 0 2 ] = 5 j ^

New answer posted

3 months ago

0 Follower 2 Views

R
Raj Pandey

Contributor-Level 9

F. B. D of Beaker

By NLM2    

f = m a = m ω 2 R m ω 2 R μ N R μ g / ω 2

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