Answer:
0.447 s²
Explanation:
First, convert to SI units.
(354 mg) (45 km) / (0.0356 kN)
(0.354 g) (45000 m) / (35.6 N)
One Newton is kg m/s²:
(0.354 g) (45000 m) / (35.6 kg m/s²)
(0.000354 kg) (45000 m) / (35.6 kg m/s²)
Simplify:
0.447 s²
Answer:
diameter is 14 mm
Explanation:
given data
power = 15 kW
rotation N = 1750 rpm
factor of safety = 3
to find out
minimum diameter
solution
we will apply here power formula to find T that is
power = 2π×N×T / 60 .................1
put here value
15 ×
= 2π×1750×T / 60
so
T = 81.84 Nm
and
torsion = T / Z ..........2
here Z is section modulus i.e = πd³/ 16
so from equation 2
torsion = 81.84 / πd³/ 16
so torsion = 416.75 / / d³ .................3
so from shear stress theory
torsion = σy / factor of safety
so here σy = 530 for 1020 steel
so
torsion = σy / factor of safety
416.75 / d³ = 530 × 
/ 3
so d = 0.0133 m
so diameter is 14 mm
All of the above (hope that helps!)
Answer:
SPCA factor
Single payment compound amount factor.
Total amount pay A = $24,230.95 (Approx)
Interest paid = $4,230.95 (Approx)
Explanation:
Given:
P = $20,000
n = 6 year
r = 3.25%
Find:
Total amount pay A
Computation:
A=p(1+r)ⁿ
A=20,000[1+3.5%]⁶
A=20,000[(1.0325)⁶]
Total amount pay A = $24,230.95 (Approx)
Interest paid = $24,230.95 - 20,000
Interest paid = $4,230.95 (Approx)
Answer: 35.3 °
Explanation:
Body-centered cubic lattice (bcc or cubic-I), just like all lattices, has lattice points at the eight corners of the unit cell with an additional points at the center of the cell. It has unit cell vectors a = b = c and interaxial angles α=β=γ=90°.
The simplest crystal structures are those that have present only a single atom at each lattice point.
body-centered cubic unit cell has atoms at each of the eight corners of a cube (like the cubic unit cell) plus one atom in the center of the cube. Each of the corner atoms is the corner of another cube so the corner atoms are shared between eight unit cells. It is said to have a coordination number of 8. The bcc unit cell consists of a net total of two atoms; one in the center and eight eighths from corners atoms
With the use of BCC unit cell, if a applied stress is in [110] direction, but slip applies in [111] direction, the angle between applied direction and slip direction is given as:
[1 1 0] [1 1 1]
λ = Cos^-1 ( 1×1 + 1×1 + 0×1 ÷ (1^2 + 1^2 +0^2) (1^2 + 1^2+ 1^2))
Cos^-1 2/ sqrt 6
= 35.386°
