Work, Energy and Power

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New answer posted

a month ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

New answer posted

a month ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

Kinldy go through the solution

 

New answer posted

a month ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

New answer posted

a month ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

P = constant S = 8 P 9 m t 3 / 2

S t 3 / 2

New answer posted

a month ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

P ? i = P ? f 0.1 * 20 = 2 * v x

v x = 1 m / s

v y = 2 g h = 2 * 10 * 1 , K E = 1 2 m V x 2 + V y 2

K E = 1 2 * 2 * ( 1 + 20 ) = 21 J

New answer posted

a month ago

0 Follower 20 Views

A
alok kumar singh

Contributor-Level 10

First, find the initial velocity (v) of the ball as it leaves the machine, using the kinematic equation for maximum height:
v² = u² + 2as ⇒ 0 = v² - 2gh_max
v² = 2gh_max = 2 * 10 * 20 = 400
v = 20 m/s.
Now, apply the work-energy theorem to the ball while it is being pushed by the machine. The work done by the constant force F equals the change in kinetic energy of the ball.
Work done W = F * d
Change in K.E. = (1/2)mv² - 0
F * 0.2 = (1/2) * 0.15 * (20)²
F * 0.2 = (1/2) * 0.15 * 400 = 0.15 * 200 = 30
F = 30 / 0.2 = 150 N.

New answer posted

a month ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

ΔE = have = 13.6Z² (1/n? ² - 1/n? ²)eV.
hv = 13.6 (1²) (1/n² - 1/ (n+1)²) = 13.6 (n+1)²-n²)/ (n² (n+1)²)
v = (13.6/h) * (2n+1)/ (n² (n+1)²). For n>>1, v ≈ (13.6/h) (2n/n? ) ∝ 1/n³.

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