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__**Week three (06th of February, 2012)**__ If **B** is added to **A** the result is 6.0 iˆ + 1.0 jˆ. If **B** is subtracted from **A**, the result is - 4.0 iˆ + 7.0 jˆ.

1- Find **A** and **B**

A+B = 6.0 iˆ + 1.0 jˆ (Equation 1)

A-B = - 4.0 iˆ + 7.0 jˆ (Equation 2)

By using algebra between 1 and 2, A = 1 iˆ + 4 jˆ (Equation 3)

By inserting Equation 3 in Equation 1 B will be equal to 5 iˆ - 3 jˆ

2- Find the magnitudes of **A** and **B**


 * A =** ((1)^2+(4)^2)^(1/2)
 * B =**((5)^2+(-3)^2)^(1/2)

3- What is the angle between **A** and **B**

By using tan^-1(B/A), the angle between A and B will show up.

**Professor's Note : Good, for part iii, check my page !**

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__**Week four (13 th February, 2012)**__ 1. A 0.5 kg rock is projected from the edge of the top of a building with an initial velocity of 10.5 m/s at an angle 54º above the horizontal. Due to gravity, the rock strikes the ground at a horizontal distance of 15.3 m from the base of the building. Assume: The ground is level and that the side of the building is vertical. The acceleration of gravity is 9.8 m/s². What is the height of the building?

Initial Velocity= 10.5 m/s, Angle= 54 , Accelaration in horizontal= 0 , in vertical= 9.81 m/s^2 , (X-Xo) = 15.3

1- I can use the formula (X-Xo) = (Vx)(cos Theta).t - (1/2).ax.t^2 ,, ax = 0 ,so (X-Xo) = (Vx)(cos Theta).t Then I solve it to get the time (t).

2- By using the formula(Y-Yo) = (Vy)(sin Theta).t - (1/2).g.t^2 I can get the hieght of the building.

**Professor's Note : Good, check my page for more hints for tomorrow's Quiz**

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__**Week 05 ( 20th February, 2012)**__

A pelican flying along a horizontal path drops a fish from a height of 5.4 m. The fish travels 8.0 m horizontally before it hits the water below. The acceleration of gravity is 9.81 m/s2. a) What was the pelican’s initial speed? b) If the pelican was traveling at the same speed but was only 2.7 m above the water, how far would the fish travel horizontally before hitting the water below?

1- (y-yo), Voy=0 and a are known we can use the equation (y-yo) = (Voy)t+1/2at^2 then we solve it for t and find the (time) Then by using the equation (x-xo) = (Vox)t+1/2at^2 and solve it for Vox find (Vox) Now we can find the initial velocity v = sqrt((Vox)^2+(Voy)^2)

2- (y-yo), Voy and a are known we can find t by solving the equation (y-yo) = (Voy)t+1/2at^2

we have (Vox) and t Finally we can find (x-xo) by using the equation (x-xo) = (Vox)t+1/2at^2

__ **Week 07 ( 05th of March, 2012)** __ 01 . An object is hung from a spring balance attached to the celing of an elevator cab. The balance reads 65 N, when the cab is standing still. What is the reading when the cab is moving upward, (a) with a constant speed of 8 m/s and (b) with a speed 8 m/s while decelerating at a rate of 2.4 m/s²

(a) Acceleration equal zero becuase the speed is constant , so when T = mg + ma ( ma = 0 ) T=mg=65 N

(b) Because it is decelarating a= -2.4 m/s^2 T-mg = ma mg = 65 N, and mass is w / g  65=m/9.8 m= 6.632 Kg  so T = ma + mg  = -15.916 + 65 T = 49.084 N

Professor's Note: Good, you have the right steps

__** Week 08 (March, 12th, 2012) **__

01. A loaded sled weighing 85 N rests on a plane inclined at an angle of 20⁰ with the horizontal. The coefficents of static and kinetic friction between the inclined plane and the sled are 0.25 and 0.15 respectively. a). What is the least magnitude of the force F, parallel to the plane, that will prevent the sled from slipping down the plane b). What is the minimum force required to start the sled moving upward? c). What value of F (force) is required to move the sled up the plane at constant velocity?

a) fs= Ms.Fn, and Fn= mg.cos( θ), and M static is given = 0,25, and the ∑ forces = 0

So F + fs - mg.sin(θ) = 0, can be solved to find the least magnitude of the force F

b) fs= Ms.Fn, and Fn= mg.cos( θ), and M static is given = 0,25

So F - fs - mg.sin(θ) = 0, can be solved to find the minimum force required

c) fk= Mk.Fn, Fn= mg.cos(θ), and M kinetic is given = 0.15 and a=0 because of the constant speed

So F-fk-mg.sin(θ)=0 can be solved to find F

__** Week 10 (April, 16th, 2012) **__ 01**. A steel ball of mass 0.5 kg is fastened to a cord that is 70 cm long and fixed at the far end. The ball is then released when the cord is horizontal. At the bottom of it's path the ball strikes, a 2.5 kg steel block initially at rest on a frictionless surface. The collision is elastic. Find, a) the speed of the ball b) the speed of the block, both just after the collision

for a) we will use the conservation of energy, KEi +PEi = KEf +PEf KEi=0 and PEf = 0 , the mass will cancels out from both sides , and we will solve for Vf = sqrt(2*g*h)

for b) we will use conservation of momentum, Pi = Pf m1Vi1 + m2Vi2 = m1Vf1 + m2Vf2, where Vi2 = 0 , and Vf1=0 , so we will get m1Vi1 = m2Vf2

we will solve for Vf2 then will find the speed of the block.


 * You have done a very good job ! -- professor**