Answer:
Atomic Size and Mass:
convert given density to kg/m^3 = 8900kg/m^3 2) convert to moles/m^3 (kg/m^3 * mol/kg) = 150847 mol/m^3 (not rounding in my actual calculations) 3) convert to atoms/m^3 (6.022^23 atoms/mol) = 9.084e28 atoms/m^3 4) take the cube root to get the number of atoms per meter, = 4495309334 atoms/m 5) take the reciprocal to get the diameter of an atom, = 2.2245e-10 m/atom 6) find the mass of one atom (kg/mol * mol/atoms) = 9.7974e-26 kg/atom Young's Modulus: Y=(F/A)/(dL/L) 1) F=mg = (45kg)(9.8N/kg) = 441 N 2) A = (0.0018m)^2 = 3.5344e-6 m^2 3) dL = 0.0016m 4) L = 2.44m 5) Y = 1.834e11 N/m^2 Interatomic Spring Stiffness: Ks,i = dY 1) From above, diameter of one atom = 2.2245e-10 m 2) From above, Y = 1.834e11 N/m^2 3) Ks,i = 40.799 N/m (not rounding in my actual calculations) Speed of Sound: v = ωd 1) ω = √(Ks,i / m,a) 2) From above, Ks,i = 40.799 N/m 3) From above, m,a = 9.7974e-26 kg 4) ω=2.0406e13 N/m*kg 5) From above, d=2.2245e-10 m 6) v=ωd = 4539 m/s (not rounding in actual calculations) Time Elapsed: 1) length sound traveled = L+dL = 2.44166 m 2) From above, speed of sound = 4539 m/s 3) T = (L+dL)/v = 0.000537505 s
The answer is 4.0 kg since the flywheel comes to rest the
kinetic energy of the wheel in motion is spent doing the work. Using the
formula KE = (1/2) I w².
Given the following:
I = the moment of inertia about the
axis passing through the center of the wheel; w = angular velocity ; for the
solid disk as I = mr² / 2 so KE = (1/4) mr²w². Now initially, the wheel is spinning
at 500 rpm so w = 500 * (2*pi / 60) rad / sec = 52.36 rad / sec.
The radius = 1.2 m and KE = 3900 J
3900 J = (1/4) m (1.2)² (52.36)²
m = 3900 J / (0.25) (1.2)² (52.36)²
m = 3.95151 ≈ 4.00 kg
Explanation:
Given that,
Radius in which the satellite orbits, r = 6588 km
Solution,
The centripetal force acting on the satellite is balanced by the gravitational force acting between earth and the satellite. Its expression can be written by :
![\dfrac{GmM}{r^2}=\dfrac{mv^2}{r}](https://tex.z-dn.net/?f=%5Cdfrac%7BGmM%7D%7Br%5E2%7D%3D%5Cdfrac%7Bmv%5E2%7D%7Br%7D)
, M is the mass of earth
![v=\sqrt{\dfrac{6.67259\times 10^{-11}\times 5.98\times 10^{24}}{6588\times 10^3}}](https://tex.z-dn.net/?f=v%3D%5Csqrt%7B%5Cdfrac%7B6.67259%5Ctimes%2010%5E%7B-11%7D%5Ctimes%205.98%5Ctimes%2010%5E%7B24%7D%7D%7B6588%5Ctimes%2010%5E3%7D%7D)
v = 7782.53 m/s
Let t is the time required to complete one orbit. It can be calculated as :
![t=\dfrac{d}{v}](https://tex.z-dn.net/?f=t%3D%5Cdfrac%7Bd%7D%7Bv%7D)
![t=\dfrac{2\pi r}{v}](https://tex.z-dn.net/?f=t%3D%5Cdfrac%7B2%5Cpi%20r%7D%7Bv%7D)
![t=\dfrac{2\pi \times 6588\times 10^3}{7782.53}](https://tex.z-dn.net/?f=t%3D%5Cdfrac%7B2%5Cpi%20%5Ctimes%206588%5Ctimes%2010%5E3%7D%7B7782.53%7D)
t = 5318.78 seconds
or
t = 1.47 hour
Therefore, this is the required solution.
Cloud formation is a process which various types of clouds are formed, generally involving adiabatic cooling of ascending moist air.
acceleration = Velocity changes ÷ time of the velocity changes
4 m/s^2 =
4 × 10^(-3) × 3600 km / h =
4 × 3.6 =
14.4 km / h
Thus :
14.4 = V(2) - V(1) / t(2) - t(1)
14.4 = V(2) - 20 / 10
Multiply both sides by 10
10 × 14.4 = 10 × ( V(2) - 20 ) / 10
144 = V(2) - 20
Add both sides 20
144 + 20 = V(2) - 20 + 20
V(2) = 164 Km/h
Thus the final velocity after 10 seconds is 164 Km/h .