The total kinetic energy before the collision is not equal to the total kinetic energy after the collision. A portion of the kinetic energy is converted to other forms of energy such as sound energy and thermal energy. A collision in which total system kinetic energy is not conserved is known as an inelastic collision.
Specific Gravity of the fluid = 1.25
Height h = 28 in
Atmospheric Pressure = 12.7 psia
Density of water = 62.4 lbm/ft^3 at 32F
Density of the Fluid = Specific Gravity of the fluid x Density of water = 1.25 x 62.4
Density of the Fluid p = 78 lbm/ft^3
Difference in pressure as we got the differential height, dP = p x g x h dP = (78 lbm/ft^3) x (32.174 ft/s^2) x (28/12 ft) [ 1 lbf / 32.174 ft/s^2] [1 ft^2 /
144in^2]
Difference in pressure = 1.26 psia
(a) Pressure in the arm that is at Higher
P = Atmospheric Pressure - Pressure difference = 12.7 - 1.26 = 11.44 psia
(b) Pressure in the tank that is at Lower
P = Atmospheric Pressure + Pressure difference = 12.7 + 1.26 = 13.96psia
Answer:
d= 7.32 mm
Explanation:
Given that
E= 110 GPa
σ = 240 MPa
P= 6640 N
L= 370 mm
ΔL = 0.53
Area A= πr²
We know that elongation due to load given as
![\Delta L=\dfrac{PL}{AE}](https://tex.z-dn.net/?f=%5CDelta%20L%3D%5Cdfrac%7BPL%7D%7BAE%7D)
![A=\dfrac{PL}{\Delta LE}](https://tex.z-dn.net/?f=A%3D%5Cdfrac%7BPL%7D%7B%5CDelta%20LE%7D)
![A=\dfrac{6640\times 370}{0.53\times 110\times 10^3}](https://tex.z-dn.net/?f=A%3D%5Cdfrac%7B6640%5Ctimes%20370%7D%7B0.53%5Ctimes%20110%5Ctimes%2010%5E3%7D)
A= 42.14 mm²
πr² = 42.14 mm²
r=3.66 mm
diameter ,d= 2r
d= 7.32 mm
C. Acceleration is the rate of change of velocity. So at the top of the path, while the velocity is zero, the CONSTANT GRAVITATIONAL ACCELERATION is about 10 m/s^2 (9.8)
It's important for Aerodynamics and Balance of the plane.