Answer:
x = 5.79 m
Explanation:
given,
mass of the car = 39000 Kg
spring constant = 5.7 x 10⁵ N/m
acceleration due to gravity = 9.8 m/s²
height of the track = 25 m
length of spring compressed = ?
using conservation of energy
potential energy is converted into spring energy




x = 5.79 m
the spring is compressed to x = 5.79 m to stop the car.
Answer:
The difference between frictionless ramp and a regular ramp is that on a frictionless ramp the ball cannot roll it can only slide, but on a regular ramp the ball can roll without slipping.
We will use conversation of energy.

Note that initial potential energy is zero because the ball is on the bottom, and the final kinetic energy is zero because the ball reaches its maximum vertical distance and stops.
For the ball B;


The initial velocities of the balls are equal. Their maximum climbing point will be proportional to their final potential energy. Since their initial kinetic energies are equal, their final potential energies must be equal as well.
Hence, both balls climb the same point.
Explanation:
Stress required to cause slip on in the direction [ 1 1 0 ] is 7.154 MPa
<u>Explanation:</u>
Given -
Stress Direction, A = [1 0 0 ]
Slip plane = [ 1 1 1]
Normal to slip plane, B = [ 1 1 1 ]
Critical stress, Sc = 2.92 MPa
Let the direction of slip on = [ 1 1 0 ]
Let Ф be the angle between A and B
cos Ф = A.B/ |A| |B| = [ 1 0 0 ] [1 1 1] / √1 √3
cos Ф = 1/√3
σ = Sc / cosФ cosλ
For slip along [ 1 1 0 ]
cos λ = [ 1 1 0 ] [ 1 0 0 ] / √2 √1
cos λ = 1/√2
Therefore,
σ = 2.92 / 1/√3 1/√2
σ = √6 X 2.92 MPa = 2.45 X 2.92 = 7.154MPa
Therefore, stress required to cause slip on in the direction [ 1 1 0 ] is 7.154MPa
Answer:
W = 3.1 N
Explanation:
moments about any convenient point will sum to zero.
I choose summing about the knife edge mark and will assume the ruler of weight W is of uniform construction.
I will assume the ruler weight makes a positive moment
W[55 - 50) - 0.040(9.8)[ 95 - 55] = 0
5W = 15.68
W = 3.136