Answer:
50.0543248872 ft
Explanation:
F = Load = 20 ton = 
d = Diameter = 1.25 in
= Initial length = 50 ft
= Final length
A = Area = 
Y = Young's modulus = 
Young's modulus is given by

The length during the lift is 50.0543248872 ft
The magnitude of the electrostatic force acting on a charge q is the product between the charge and the intensity of the electric field E. The magnitude of the electron charge is

(we are not interested in the sign), so the electrostatic force magnitude is
Answer:
<em>You would use the kinematic formula:</em>

Explanation:
The upwards vertical motion is ruled by the equation:

Where:





Naming Δy = y - y₀, the equation becomes:

Then, you just need to substitute with Δy = 0.1m, t = 2s, and g = 9.8m/s², ans solve for the intital vertical velocity.
Answer:
2.39 revolutions
Explanation:
As she jumps off the platform horizontally at a speed of 10m/s, the gravity is the only thing that affects her motion vertically. Let g = 10m/s2, the time it takes for her to fall 10m to water is




Knowing the time it takes to fall to the pool, we calculate the angular distance that she would make at a constant acceleration of 15 rad/s2:


As each revolution is 2π, the total number of revolution that she could make is: 15 / 2π = 2.39 rev
The mass of the puck is
m = 0.15 kg.
The diameter of the puck is 0.076 m, therefore its radius is
r = 0.076/2 = 0.038 m
The sliding speed is
v = 0.5 m/s
The angular velocity is
ω = 8.4 rad/s
The rotational moment of inertia of the puck is
I = (mr²)/2
= 0.5*(0.15 kg)*(0.038 m)²
= 1.083 x 10⁻⁴ kg-m²
The kinetic energy of the puck is the sum of the translational and rotational kinetic energy.
The translational KE is
KE₁ = (1/2)*m*v²
= 0.5*(0.15 kg)*(0.5 m/s)²
= 0.0187 j
The rotational KE is
KE₂ = (1/2)*I*ω²
= 0.5*(1.083 x 10⁻⁴ kg-m²)*(8.4 rad/s)²
= 0.0038 J
The total KE is
KE = 0.0187 + 0.0038 = 0.0226 J
Answer: 0.0226 J