Answer:
the required diameter of the rod is 9.77 mm
Explanation:
Given:
Length = 1.5 m
Tension(P) = 3 kN = 3 × 10³ N
Maximum allowable stress(S) = 40 MPa = 40 × 10⁶ Pa
E = 70 GPa = 70 × 10⁹ Pa
δ = 1 mm = 1 × 10⁻³ m
The required diameter(d) = ?
a) for stress
The stress equation is given by:
A is the area = πd²/4 = (3.14 × d²)/4
Substituting the values, we get
d = (9.77 × 10⁻³) m
d = 9.77 mm
b) for deformation
δ = (P×L) / (A×E)
A = (P×L) / (E×δ) = (3000 × 1.5) / (1 × 10⁻³ × 70 × 10⁹) = 0.000063
d² = (4 × A) / π = (0.000063 × 4) / 3.14
d² = 0.0000819
d = 9.05 × 10⁻³ m = 9.05 mm
We use the larger value of diameter = 9.77 mm
Answer:
The Estimated uncertainty in a nominal displacement of 2 cm at the design stage is plus or minus 0.0124cm
Explanation:
uncertainty in a nominal displacement
= (u^2 + v^2)^(1/2)
assume from specifications that k = 5v/5cm
= 1v/cm
u^2 = (0.0025*2)^(2) + (0.005*10*2)^2 + (0.0025*2)^2
= 0.01225v
v = 2v * 0.001
= 0.002v
uncertainty in a nominal displacement
= (u^2 + v^2)^(1/2)
= ((0.01225)^2 + (0.002)^2)^(1/2)
= 0.0124 cm
Therefore, The Estimated uncertainty in a nominal displacement of 2 cm at the design stage is plus or minus 0.0124cm
Answer:
The current through each lamp is 0.273 Amperes
Power dissipated in each lamp is 0.082W
Explanation:
Battery v = 1.5 V
Each lamp has resistance, r = 1.1 Ohms
The 5 lamps in series will therefore have total resistance, R = 5 * 1.1 = 5.5 Ohms
The current through each lamp, I = v/R = 1.5 / 5.5 = 0.273 Amperes
Power dissipated in each lamp = I² * r = 0.273² * 1.1 = 0.082W
OA bloom is smaller than a bar
Answer:
A. True
Explanation:
MATLAB may be defined as a programming platform that is designed specifically for the engineers as well as the scientists to carry out different analysis and researches.
MATLAB makes use of a desktop environment which is tuned for certain iterative analysis and the design processes with a programming language which expresses matrix as well as array mathematics directly.
Thus the answer is true.
