Answer:
8100 lbin²
Explanation:
Moment or inertia is expressed using the formula
I = mr²
M is the mass of the body
r is the radius of gyration
Given
W = 100lb
r = 9in
Required
Moment of inertia
I = Wr²
I = 100(9)²
I = 100×81
I = 8100lbin²
Hence the moment of inertia about the center of gravity for the rotator is
8100 lbin²
Answer:
1.747 lb ft⋅s^2
Explanation:
I = m * r^2
r = 9in = 9/12 = 0.75ft
m = W/g = 100lb/32.2ft/s^2 = 3.106
I = 3.106 * (0.75)^2
I = 1.747 lb ft⋅s^2
Two stepped bar is supported at both ends.At the join point of two segments,the force F is applied(downwards).Calculate reactive forces R1 and R3 at the supports.What is value of absolute maximal stress?
Choose one answer nearest your result.
given= d1=10mm d2=20mm L1=20mm L2=10mm E=200GPa F=20kN
Answer:
F=200kN
Explanation:
A universal shift register can shift in both the left-to-right and right-to-left directions, and it has parallel-load capability. Draw a circuit for such a shift register.
Answer:
Explanation:
A unidirectional shift register allows for the capability of shifting in one direction as the name unidirectional implies.
A bidirectional shift register has the capabilities of shifting in both the left to right and right to left directions.
For a Universal shift register, there are possibilities of bidirectional shifts and parallel-load capabilities that have the following properties.
There is the existence of a clear control input whose main FUNCTION is to set all the register to 0.'
A shift control to both the right and left direction to enable both the shift right operation and shift left operation.
Finally, a parallel-load control whose function is to activate a parallel transfer from input to output.
The diagram for the circuit for such a shift register can be seen in the diagram attached below.
to determine cam ring speed you must use a ___________________,____________________or a________________
Answer:
=> The total numbers of cylinders on the engine.
=> Total number of lobes in the cam ring.
=> The direction at which the cam ring rotates.
Explanation:
A cam is a kind of ring and a device that is being used in engines. One or the main purpose of using cams in engines us because it helps in the change or transformation of the rotational movement of the engine to a translational one. Instead of using a cam ring, a cam roller can be used in place.
There are three things that can be used to to determine the speed of cam ring speed and they are given below as;
=> The total numbers of cylinders on the engine.
=> The Total number of lobes in the cam ring.
=> The direction at which the cam ring rotates.
The three suspender bars AB, CD, and EF are made of A-36 steel and have equal cross-sectional areas of 500 mm2. Determine the average normal stress in each bar if the rigid beam is subjected to a force of P
Answer:
hello a diagram attached to your question is missing attached below is the missing diagram
The three suspender bars AB, CD, and EF are made of A-36 steel and have equal cross-sectional areas of 500 mm2. Determine the average normal stress in each bar if the rigid beam is subjected to a force of P = 70kN , let d = 2.4 m , L = 4m
Answer :
Stress = force / area
for Bar 1 = (40.03 * 1000) / 500 = 80.06 MPa
Bar 2 = ( 23.33 * 1000 ) / 500 = 46.66 MPa
for Bar 3 = ( 5.33 * 1000 ) / 500 = 10.66 MPa
Explanation:
Given data:
Type of steel = A-36
cross-sectional area = 500 mm^2
Calculate the average normal stress in each bar
we have to make some assumptions
assume forces in AB, CD, EF to be p1,p2,p3 respectively
∑ Fy = 0 ; p1 + p2 + p3 = 70kN ---------- ( 1 )
∑ Mc = 0 ; P1 * d - p * d/2 - p3 * d = 0
where d = 2.4 hence ; p1 - p3 = 35 -------- ( 2 )
Take ; Tan∅
Tan∅ = MN / 2d = OP/d
i.e. s1 - 2s2 - s3 = 0
[tex]\frac{P1L}{AE} - \frac{2P2}{AE} + \frac{P3L}{AE} = 0[/tex]
L , E and A are the same hence
P1 - 2p2 + p3 = 0 ----- ( 3 )
Next resolve the following equations
p1 = 40.03 kN, p2 = 23.33 kN, p3 = 5.33 kN
Stress = force / area
for Bar 1 = (40.03 * 1000) / 500 = 80.06 MPa
Bar 2 = ( 23.33 * 1000 ) / 500 = 46.66 MPa
for Bar 3 = ( 5.33 * 1000 ) / 500 = 10.66 MPa