Answer:
The question has some details missing : The 35-kg block A is released from rest. Determine the velocity of the 13-kgkg block BB in 4 ss . Express your answer to three significant figures and include the appropriate units. Enter positive value if the velocity is upward and negative value if the velocity is downward.
Explanation:
The detailed steps and appropriate calculation is as shown in the attached file.
The load is 17156 N.
<u>Explanation:</u>
First compute the flexural strength from:
σ = FL / π
= 3000 (40
10^-3) / π (5
10^-3)^3
σ = 305 10^6 N / m^2.
We can now determine the load using:
F = 2σd^3 / 3L
= 2(305 10^6) (15
10^-3)^3 / 3(40
10^-3)
F = 17156 N.
Explanation:
Air fuel ratio:
Air fuel ratio is the ratio of mass of air to the mass of fuel.So we can say that
As we know that fuel burn in the presence of air that is why we have to maintain a proper amount of air fuel ratio.
When we need more power then we have supply more fuel and to burn this fuel ,require a specified amount of air.So for different loading condition of engine different air fuel ratio is required.
When air is less and fuel is more then it is called rich air fuel ratio .when air is more and fuel is less then it is called poor air fuel ratio.
Answer:
c = 18.0569 mm
Explanation:
Strategy
We will find required diameter based on angle of twist and based on shearing stress. The larger value will govern.
Given Data
Applied Torque
T = 750 N.m
Length of shaft
L = 1.2 m
Modulus of Rigidity
G = 77.2 GPa
Allowable Stress
г = 90 MPa
Maximum Angle of twist
∅=4°
∅=4*/180
∅=69.813 *10^-3 rad
Required Diameter based on angle of twist
∅=TL/GJ
∅=TL/G*/2*c^4
∅=2TL/G**c^4
c=∅
c=18.0869 *10^-3 rad
Required Diameter based on shearing stress
г = T/J*c
г = [T/(J*/2*c^4)]*c
г =[2T/(J**c^4)]*c
c=17.441*10^-3 rad
Minimum Radius Required
We will use larger of the two values
c= 18.0569 x 10^-3 m
c = 18.0569 mm