Disk A has a mass of 8 kg and an initial angular velocity of 360 rpm clockwise; disk B has a mass of 3.5 kg and is initially at
rest. The disks are brought together by applying a horizontal force of magnitude 20 N to the axle of disk A. Assume that μk = 0.15 between the disks. Neglect bearing friction.
Required:
Determine:
a. the angular acceleration of each disk.
b. the final angular velocity of each disk.
Required:
Determine:
a. the angular acceleration of each disk.
b. the final angular velocity of each disk.
2 answers:
Answer:
a. the angular acceleration of each disk;
For disk A, the angular acceleration is 9.36 rad/s²
For disk B, the angular acceleration is 28.57 rad/s²
b. the final angular velocity of each disk;
For disk A, the final angular velocity is 250.57 rpm
For disk A, the final angular velocity is 334.00 rpm
Explanation:
While slipping occurs, and upward friction force F is applied to disk A, and downward friction force F to disk B.
Assumption;
Radius of disk A is 80 mm
Radius of disk B is 60 mm.
Pls find the explanation of the question in the attached files as typing it here may be difficult due to the equations and diagrams involved.
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Answer:
A) 1040 steel is not a possible candidate for this application
B) 35.94%
Explanation:
Initial length = 100 mm = 0.1 m
Initial diameter ( d ) = 7.5 mm = 0.0075 m
Tensile load ( p ) = 18,000 N
Condition : The 1040 steel must not experience plastic deformation or a diameter reduction of more than 1.5 * 10^-5 m
<u>A) would the 1040 steel be a possible candidate for this application</u>
<em>Yield strength of 1040 steel < stress ( in order to be a possible candidate )</em>
stress = p / A0 = ( 18000 ) / ( ) * 0.0075^2
= 18,000 / (4.418 * 10^-5 ) = 407.424 MPa
Yield strength of 1040 steel = 450 MPa
stress = 407.424 MPa
∴ Yield strength ( 450 MPa ) > stress ( 407.424 MPa )
Therefore 1040 steel is not a possible candidate for this application
<u>B) Determine How much cold work would be required to reduce the diameter of the steel to 6.0 mm</u>
Area1 = ( ) ( 0.006 )^2 = 2.83 * 10^-5 m^2
therefore % of cold work done = ( A0 - A1 ) / A0 * 100 = 35.94%
Answer:
44.95 tonnes
Explanation:
According to principle of buoyancy the object will just sink when it's weight is more than the weight of the liquid it displaces
It is given that empty weight of box = 40 tons
Let the mass of the stones to be placed be = M tonnes
Thus the combined mass of box and stones = (40+M) tonnes..........(i)
Since the box will displace water equal to it's volume V we have
Now the weight of water displaced = is density of water = 1000kg/
Thus weight of liquid displaced = ..................(ii)
Equating i and ii we get
40 + M = 84.95
thus Mass of stones = 44.95 tonnes
Answer:
#See solution for details.
Explanation:
#The commutative property of multiplication tells us that it doesn't matter in what order you multiply numbers. The formula for this property is a * b = b * a:
Hence, the product of the four numbers remains 320,000 irrespective of their order.