Since the track is friction less, block’s kinetic energy at the bottom of the track is equal to its potential energy at the top of the track.
PE = 81 * 9.8 * 3.8 = 3016.44 J
Work = 1/2 * 1888 * d^2
PE = Kinetic energy at the base.
1/2 * 1888 * d^2 = 3016.44
d = 1.78 approx 1.8
F = Ke = 1888 * 1.8 = 3398.4N
Answer:
w₂ = 22.6 rad/s
Explanation:
This exercise the system is formed by platform, man and bricks; For this system, when the bricks are released, the forces are internal, so the kinetic moment is conserved.
Let's write the moment two moments
initial instant. Before releasing bricks
L₀ = I₁ w₁
final moment. After releasing the bricks
= I₂W₂
L₀ = L_{f}
I₁ w₁ = I₂ w₂
w₂ = I₁ / I₂ w₁
let's reduce the data to the SI system
w₁ = 1.2 rev / s (2π rad / 1rev) = 7.54 rad / s
let's calculate
w₂ = 6.0/2.0 7.54
w₂ = 22.6 rad/s
-- Since it's a cube, its length, width, and height are all the same 4 cm .
-- Its volume is (length x width x height) = 64 cm³ .
-- Density = (mass) / (volume)
= (176 g) / (64 cm³)
= 2.75 gm/cm³ .
Explanation:
the vehicles displacement, since displacement deals with position
Answer:
The capillary rise of the glycerin is most nearly 
Explanation:
From the question we are told that
The diameter of the glass tube is 
The density of glycerin is 
The surface tension of the glycerin is 
The capillary rise of the glycerin is mathematically represented as

substituting value


Therefore the height of the glass tube the glycerin was able to cover is