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Tutor profile: Sid I.

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Sid I.
Eager to help in Math (any level up to Calculus), Physics and Chemistry
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Questions

Subject: Pre-Calculus

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Question:

Two functions of x are defined as follows f(x) = 2x + 1; g(x) = sin(x/2) Evaluate: a) f(1) + g(pi) b) f(-2)*g(pi) c) f(g(2pi)) d) g(f(x)) e) At the value x = -1, which one of these two functions is greater

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Sid I.
Answer:

a) f(x) + g(x) = [2(1) + 1] + sin(pi/2) = 3 + 1 = 4 b) f(x)(g(x)) = [2*(-2)+1] * sin(pi/2) = -3*1 = -3 c) f(g(x)) = 2(sin(2pi/2)) + 1 = 0 + 1 d) sin[(2x+1)/2] e) 2(-1) + 1 = -1; sin(-1/2) = -0.479 g(-1) is larger than f(-1)

Subject: Calculus

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Question:

Using a u-substitution, evaluate the following integral ∫x*sin(2x)

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Sid I.
Answer:

Let u = x; dv = sin(2x) ∫u*dv = u*v - ∫v*du du = 1; v = -cos(2x)/2 ∫u*dv = -xcos(2x)/2 - ∫=2cos(2x) = -xcos(2x)/2 + (sin(2x)/2)/2 = -xcos(2x)/2 + sin(2x)/4 + C

Subject: Physics

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Question:

A 1kg box moving at 4 m/s hits a 2kg box standing still at the edge of an incline that's 1.5m in height. The 2kg box goes down the incline onto a flat surface, where it slides until it halts to a stop. The collision is perfectly elastic. a) Calculate the velocity of the 5kg box, after the collision, (you may neglect friction) b) Calculate the velocity of the 2kg immediately after reaching the bottom of the incline (you may neglect friction on the incline) c) After reaching the bottom of the incline the box slides a total of 5m, calculate the kinetic friction coefficient of the bottom surface.

Inactive
Sid I.
Answer:

a) m1v1 + m2v2 = m1v1' + m2v2' m1 = 1 kg; m2 = 2 kg; v1 = 4 m/s; v2 = 0; (1 kg*4m/s) + 0 = (5 kg*v1') + (2 kg*v2') = 4 kg*m/s 5*(v1') + 2(v2') = 20 v1 + v1' = v2 + v2' 4 m/s = 0 + v2' - v1' v1' + 2v2' = 4 (-v1' + v2')*-2 = 4(-2) 3v1' = -4 --> v1' = -1.33 m/s b) (1 kg*4m/s) + 0 = (1 kg*-1.33 m/s) + (2 kg*v2') = 4 kg*m/s v2' = [4 - (-1.33 m/s)]/2 = 2.66 m/s vf^2 = vi^2 + 2ay vi = v2' = 2.66 m/s; a = 9.8 m/s^2; y = 1.5 m vf^2 = 2.66^2 + 2*9.8*1.5 = 32.07 --> vf = 32.07^.5 = 5.66 m/s c) vf^2 = vi^2 + 2ax a = (vf^2 - vi^2)/2x vf = 0; vi = 5.66 m/s; x = 5m a = -(5.66^2)/(2*3m) = -3.21 m/s^2 Ff = μsFn; Fn = mg = 2kg*9.8 m/s^2 = 19.6 N μs = Ff/Fn = m|a|/Fn = 3.21 m/s^2*2 kg/19.6 N = 0.328 Don't forget your free body diagarms!

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