Answer:
Explanation:
Cross sectional area of post = 3.14 x(10.5)² x 10⁻⁴ = 346.18 x 10⁻⁴ m²
Weight = mg = 8300 x 9.8 = 8.134 x 10⁴ N
Stress = weight / area
= 8.134 x 10⁴ / (346.18 x 10⁻⁴)
23.49 x 10⁵ N / m²
Strain = stress / modulus of elasticity of steel
modulus of elasticity = 190 x 10⁹ Pa
strain = Stress \ young modulus
=![\frac{23.49\times10^5}{190\times10^9}](https://tex.z-dn.net/?f=%5Cfrac%7B23.49%5Ctimes10%5E5%7D%7B190%5Ctimes10%5E9%7D)
= 12.36 x 10⁻⁶
strain = change in length / original length
change in length = original length x strain
= 2.7 x 12.36 x 10⁻⁶
= 33.372 x 10⁻³ mm
.033 mm.
Λ = c/f; where λ is the wavelength, c is the velocity (3.0 ×10^8 m/s) and f is the frequency of the wave.
Therefore; λ1 = (2.99792×10^8)/5.5×10^5)
= 545.5 m
λ2 = (2.99792×10^8)/1.6×10^6
= 187.5 m
Therefore; the shortest wavelength is 187.5 m
Answer:
h ’= 12,768 cm
Explanation:
For this exercise let's use the constructor equation
1 / f = 1 / p + 1 / q
where f is the focal length, p the distance to the object and q the distance to the image
the magnification equation is
m = h '/ h = -q / p
let's find the distance to the object
1 / p = 1 / f- 1 / q
1 / p = 1/20 - 1 / (- 37.5)
1 / p = 0.076666
p = 13.04 cm
now let's use the magnification equation
h ’= - q / p h
let's calculate
h ’= - (-37.5) / 13.04 4.44
h ’= 12,768 cm
Answer:
2.0*![10^{17}](https://tex.z-dn.net/?f=10%5E%7B17%7D)
Explanation:
Let us calculate total moles (air + CO)
moles (n) = PV /RT
P = 755 torr/760 = 0.9934 atm
V = 1.00 L
R = gas constant
T = 23°C + 273 = 296 K
So,
total moles (n) = 0.9934 x 1.00/0.0821 x 296 = 0.041 moles
mole fraction of CO = 8.1 x ![10^{-6}](https://tex.z-dn.net/?f=10%5E%7B-6%7D)
moles of CO =8.1 x
*0.041
=3.31 x
moles
number of molecule of CO =3.31 x
*(6.0233x
)
= 2.0*![10^{17}](https://tex.z-dn.net/?f=10%5E%7B17%7D)