Laws of Motion

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New answer posted

a month ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

Statement-I: F? = mv²/r ≤ f? = µmg ⇒ v ≤ √ (µgR) = √ (0.2*9.8*2) = 1.98 m/s
v_cyclist = 7 km/h = 1.94 m/s. Since 1.94 m/s < 1.98 m/s, statement-I is correct.
Statement-II: v_max = √gR (tanθ+µ)/ (1-µtanθ) = .
v_min = √gR (tanθ-µ)/ (1+µtanθ) = √ (9.8*2 (tan45-0.2)/ (1+0.2tan45) = 3.65 m/s
v_cyclist = 18.5 km/h = 5.14 m/s. This is outside the safe range. Statement-II is incorrect.

New answer posted

a month ago

0 Follower 4 Views

V
Vishal Baghel

Contributor-Level 10

Let the spring is in extension state and a B > a A , a A B = a A i ^ ( a B i ^ ) = ( a A + a B ) i ^

Hence we can say that block moves away from block B in the frame of B

F – kx = MaB . (1)

kx = Ma. (2)

a B = F M a

New answer posted

a month ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

T = m ? ^ ω 2

T = m : ( 2 ω ) 2

T = 4 T

New answer posted

a month ago

0 Follower 6 Views

A
alok kumar singh

Contributor-Level 10

Capacitive Reactance X C = 1 ω C = 1 2 π f C = 1 2 * 3 . 1 4 * 5 0 * 1 0 * 1 0 6

= 1 0 0 0 3 . 1 4

Vrms=210 V

i rms = V rms X C = 2 1 0 X C

 Peak current =2ims=2*2101000*3.14=0.932

?0.93 A

New answer posted

a month ago

0 Follower 5 Views

A
alok kumar singh

Contributor-Level 10

Y 1 = A · A ¯

= A ¯

Y 2 = B + B ¯

= B ¯

Y = Y 1 + Y 2 ¯

= A ¯ + B ¯ ¯

= A ¯ ¯ · B ¯ ¯

=A·B is similar to output of AND Gate 

New answer posted

a month ago

0 Follower 2 Views

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

a month ago

0 Follower 5 Views

A
alok kumar singh

Contributor-Level 10

F = ( M 1 + M 2 ) a

a=102+3=2 ms2

F=M2 (2)=3*2 N=6 N

New answer posted

a month ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

Sol. Before collision

It undergoes completely inelastic collision

Using conservation of linear momentum

Initial momentum = Final momentum

m v 1 = m v 2 + m v 2

m v 1 = 2 m v 2

v 1 v 2 = 2 1

New answer posted

2 months ago

0 Follower 1 View

A
alok kumar singh

Contributor-Level 10

70 g – N = 70 a = 70 * 0.2 = 14

->N = 700 – 14 = 686 N

New answer posted

2 months ago

0 Follower 4 Views

V
Vishal Baghel

Contributor-Level 10

For ascent

t a = 2 l a a s c e n t = 2 l g ( s i n θ + μ c o s θ )

For descent

t d = 2 l a d e s c e n t = 2 l g ( s i n θ μ c o s θ )

According question, we can write

μ = 3 5 t a n θ = 3 5 t a n 3 0 ° = 3 5

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