Answer:
The flexural strength of a specimen is = 78.3 M pa
Explanation:
Given data
Height = depth = 5 mm
Width = 10 mm
Length L = 45 mm
Load = 290 N
The flexural strength of a specimen is given by
78.3 M pa
Therefore the flexural strength of a specimen is = 78.3 M pa
Answer:
The outer diameter of the spacers that yields the most economical and safe design is 25.03 mm
Explanation:
For steel bolt
Stress = 210 MPa or 210 N/mm2
Pressure = Stress* Area
Pbolt = 210 N/mm2 * 16^2 *(pi)/4
Pbolt = 210 N/mm2 * 200.96 mm^2 = 42201.6 N
For Brass spacer
Pressure = 42201.6 N
Area of Brass spacer = Pressure/Stress
Area of Brass spacer = 42201.6 N/145 N/mm^2 = 291.044 mm^2
Area of Brass spacer = (pi) (d^2 - 16^2)/4 = 291.044 mm^2
d^2 - 16^2 = 291.044 mm^2* 4/(pi) = 370.758
d^2 = 370.758 + 16^2
d^2 = 626.758
d = 25.03 mm
The outer diameter of the spacers that yields the most economical and safe design is 25.03 mm
Answer is yes.
Tire rotation is undertaken to ensure that the tires wear evenly. This can extend tire life and save you money.
For example, failure to rotate tires on a front-wheel-drive vehicle will eventually result in the front tires having significantly less tread than the rear tires.
The answer is T because a task is getting something done
Answer:
The value of R is 10101
Explanation:
As per the given data
D = 1000100100
G = 100101
Redundant bit = 6-bits - 1-bit = 5-bits
No add fice zero to D
D = 100010010000000
Now calculate R as follow
R = D / G
R = 100010010000000 / 100101
R = 10101
Workings are attached with this question
