Abdulkareem F.

Mechanical Engineer Junior at University of California, Irvine

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Physics (Newtonian Mechanics)

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Question:

A 3.00-kg crate slides down a ramp. The ramp is 1.00 m in length and inclined at an angle of 30.08. The crate starts from rest at the top, experiences a constant friction force of magnitude 5.00 N, and continues to move a short distance on the horizontal floor after it leaves the ramp. -Use energy methods to determine the speed of the crate at the bottom of the ramp.

Abdulkareem F.

Answer:

Because vi= 0, the initial kinetic energy of the system when the crate is at the top of the ramp is zero. If the y coordinate is measured from the bottom of the ramp (the final position of the crate, for which we choose the gravitational potential energy of the system to be zero) with the upward direction being positive, then yi = 0.500 m. 1- Write the expression for the total mechanical energy of the system when the crate is at the top: Ei = Ki + Ui = 0 + Ui = m*g*yi 2- Write an expression for the final mechanical energy: Ef = Kf+ Uf = 0.5*m*[(vf)^2] + 0 3- Use conservation of Energy Law: E(mech)= Ef - Ei = 0.5*m*[(vf)^2] - m*g*yi 4- Solve for vf: vf = [(2/m)*(m*g*yf-f*d)]^0.5 5- Substiture numerical values to get the answer: vf = 2.54 m/s

Pre-Calculus

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Question:

Solve the equation : sin^2(β) + sin(β) = 0 for 0≤β≤2π

Abdulkareem F.

Answer:

sin^2(β) + sin(β) = 0 sin(β)* (sin(β)+1) =0 Therefore: sinβ=0 --> β=0 or β=π OR sinβ = -1 --> β= 3π/2

Calculus

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Question:

Find a pair of solutions to the equation below: y'' + 5y' - 6y = 0

Abdulkareem F.

Answer:

The auxiliary equation associated with the equation: y”+5y’-6y=0 is: R^2 + 5R - 6=0 (R-1) (R+6)=0 Therefore, the roots of the previous equation are: R=1 and R=-6 Finally, the solutions of the second order differential equation are: R1= e^t R2= e^-6t

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