In a one-dimensional collision, a particle of mass 2m collides with a particle of mass m at rest. If the particles stick together after the collision, what fraction of the initial kinetic energy is lost in the collision?
First we apply momentum conservation since the collosion is inelastic to find the final velocity of both the masses together. This is followed by finding difference in initial kinetic energy and final kinetic energy using the velocity found from first step.
Water flows over a sharp flat plate 3m long, 3 m wide with an approach velocity of 10m/s. Estimate the error in drag force in the flow over the entire plate is assumed turbulent. Assumed the mixed regions can be expressed by the following coefficient of drag relationship. Cd=(0.074/(Re)^0.2)-(1742/ReL). For water, density is 1000kg/m3 and kinematic viscosity as 1*10^-6 m/s^2
Assume turbulent, ReL=Vinf*L/kinematic viscosity=3*10^7 which is greater than 5*10^5 and thus turbulent Cd= 0.074/(Re)^0.2=0.0024 Fd=0.5*row*A*Vinf^2*Cd=o.5*1000*(3*3)*10^2*0.0024=1064N b) Now we treat the plate as two parts, laminar section, followed by a turbulent section Cd=0.074/(Re)^0.2 -1742/ReL= (0.074/(3*10^7)^0.2)-(1742/3*10^7)=0.0023 Fd=0.5*1000*9*10^2*0.0023=1038N Hence, error=1064-1038/1038=2.5% high
The diagonal [𝐵𝐷] of parallelogram 𝐴𝐵𝐶𝐷 is divided by points 𝑀, 𝑁, in 3 segments. Prove that 𝐴𝑀𝐶𝑁 is a parallelogram and find the ratio between 𝜎[𝐴𝑀𝐶𝑁] and 𝜎[𝐴𝐵𝐶𝐷].
Given :Let O be the intersection point of the diagonals of parallelogram ABCD. ‖D𝑀‖ = ‖𝑁𝑀‖ = ‖𝑁𝐵‖ ‖𝐷𝐶‖?⟹ ∆𝑀𝑂𝐶 = ∆𝑁𝐵𝐴 ⟹ ‖𝑀𝐶‖ = ‖𝐴𝑁‖ It is proved in the same way that ∆𝐷𝐴𝑀 = ∆𝐵𝐶𝑁 ⟹ ‖𝑀𝐶‖ = ‖𝑁𝐶‖. Thus 𝐴𝑁𝐶𝑀 is a parallelogram. Area of triangle AOB= 0.5*||OA|| ||OB|| sin @ Area of triangle AOD=0.5*||OA|| ||OD|| sin(pi-@) Thus, Area of parallelogram ABCD=(Area(AOB)+Area(AOD)= ||OA|| ||DB|| sin @ Area of parallelogram AMCN=||OA|| ||MN|| sin@ =||OA|| ||BD/3|| sin@ Thus; Area of AMCN/Area of ABCD= 1/3