Answer:
3.65 m/s
Explanation:
= Mass of first car = 103 kg
= Mass of second car = 81 kg
= Initial Velocity of first car = 5 m/s
= Initial Velocity of second car = -3.3 m/s
= Velocity of center of mass
For elastic collision
Velocity of center of mass = 1.35 m/s
The final velocity of car 1 is 3.65 m/s
Answer:
Explanation:
Given data
Length L=2.5 m
Radius R=d/2=30/2 = 15 mm
Torque based on allowable stress
Allowable shear stress τ=50 Mpa
Allowable torque T=(π/2)τc³
Torque based on allowable angle of twist
Allowable Angle of twist
Ф=7.5°
Ф=7.5×(π/180)=130.90×10⁻³ rad
Allowable torque
T=(GJФ)/L
T=(G(π/2)c⁴)Ф)/L
T=(πGc⁴Ф)/2
Maximum Power Transmitted
Maximum power transmitted is given by
Never is the correct answer
Answer:
Option D is the correct answer.
Explanation:
We equation for elongation
Here we need to find weight required,
We need to stretch a steel road by 2 mm, that is ΔL = 2mm = 0.002 m
E = 200GPa = 2 x 10¹¹ N/m²
L=2m
Substituting
Option D is the correct answer.
You have two possible ways to connect the springs: in parallel or in series.
The equivalent stiffness of three springs in parallel is given by:
k_eq = k₁ + k₂ + k₃
In order to keep this number the smallest possible, you need to take the three springs with smaller k:
k_eq_min = 3.5 + 6 + 8.5 = 18 N/m
The equivalent stiffness of three springs in series is given by:
1 / k_eq = 1 / <span>k₁ + 1 / k₂ + 1 / k₃
In order to get the smallest k_eq possible, 1 / k_eq must be the biggest possible, therefore you need to take again the three springs with smaller k:
k_eq = 1 / (</span>1 / <span>k₁ + 1 / k₂ + 1 / k₃)
= 1 / (1 / 3.5 + 1 / 6 + 1 / 8.5)
= 1.754 N/m
Therefore, in order to get the smallest equivalent stiffness, you need to connect the first three springs in series (one after the other).</span>