REVIEW FOR MID TERM 1

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REVIEW FOR MID TERM 1

description

REVIEW FOR MID TERM 1. English  S.I. 1 inch = 2.540 cm (centimeter = 1/100 meter) 1mile = 1.609 km (kilometer= 1000 meters) 1 pound = 454 g (gram) 1.06 quart = 1 L (litre). Components Of Vectors A = A x + A y. Using Components To Add Vectors. C x = A x + B x - PowerPoint PPT Presentation

Transcript of REVIEW FOR MID TERM 1

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REVIEW FOR MID TERM 1

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SI CGS BE

Length Meter (m) Centimeter (cm)

Foot (ft)

Mass Kilogram (kg)

Gram (g) Slug (sl)

Time Second (s) Second (s) Second (s)

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English S.I.

• 1 inch = 2.540 cm (centimeter = 1/100 meter)

• 1mile = 1.609 km (kilometer= 1000 meters)

• 1 pound = 454 g (gram)

• 1.06 quart = 1 L (litre)

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Components Of Vectors

A = Ax + Ay

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Using Components To Add Vectors

Cx = Ax + Bx

Cy = Ay + By

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Vav,x = (x2-x1)/(t2-t1) = Δx/Δt SI unit: m/s

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Average Accelerationaav,x = (v2x – v1x) / (t2 - t1) = Δvx / Δt

SI unit: m/s2

Instantaneous Accelerationax = lim (Δvx/Δt) Δt 0SI unit: m/s2

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Cars Accelerating or Decelerating

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Constant-Acceleration Equations of Motion

Variables Related Equation

Velocity, time, acceleration

vx = v0x + axt

Position, time, acceleration

x = x0 + v0xt + (½ )axt2

Velocity, position, acceleration

vx2 = v0x

2 + 2ax(x – x0) = v0x2 + 2axx

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Constant-Acceleration Equations of Motion in Two-Dimensions

vx = v0x + axt vy = v0y + ayt

x = x0 + v0xt + (½ )axt2 y = y0 + v0yt + (½ )ayt2

vx2 = v0x

2 + 2ax(x – x0) vy2 = v0y

2 + 2ay(y – y0)

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Equations of Motion for projectileax = 0m/s2 ay = -9.81m/s2

vx = v0x + axt vy = v0y + ayt

x = x0 + v0xt + (½ )axt2 y = y0 + v0yt + (½ )ayt2

vy2 = v0y

2 + 2ay(y – y0)

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Determination of key itemsfor projectiles

• x = (vocos o)t

= tan-1(vy/vx)

• y = (vosin o)t - ½gt2

• vx = vocoso

• vy = vosino- gt

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R = F1 + F2 + F3 + ……..= Σ F, (resultant, and vector sum, of forces)

Rx = Σ Fx , Ry = Σ Fy

(components of vector sum of forces)

Once we have the components Rx and Ry, we can find the magnitude and direction of the vector R.

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Newton’s First Law – Figure 4.7

•“Objects at rest tend to stay at rest and objects in motion tend to stay in motion in a straight line unless it is forced to change that state by forces acting on it”

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Newton’s Second Law of Motion (Vector Form)

The vector sum (resultant) of all the forces acting on an object equals the object’s mass times its acceleration :

ΣF = ma

The acceleration a has the same direction as the resultant force ΣF.

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Newton’s Second Law of Motion (Vector Form)

The vector sum (resultant) of all the forces acting on an object equals the object’s mass times its acceleration :

ΣF = ma

The acceleration a has the same direction as the resultant force ΣF.

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Forces are the origin of motionForces Acceleration a = F/m

Velocityv= v0 + at

Positionx = x0 + v0t + ½ at2

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x

y

x positivey positive

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x negativey positive

x

y

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x

y

x negativey negative

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x

y

x positivey negative

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x

y

??????

x positivey positive

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Drawing a FBD of forces on an object (on, not by)

1. Choose the object to analyze. Draw it as a dot.2. What forces physically touch this object?

This object, not some other3. What “action at a distance” forces act on the object?

Gravity is the only one for this PHYS2053

4. Draw these forces as arrows with tails at the dot (object).5. Forces only! No accelerations, velocities, …

Get components of Newton’s 2nd Law

Choose a convenient xy coordinate systemFind the x and y components of each force in the FBDAdd the x and y components separately

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In a rescue, the 70.0  police officer is suspended by two cables, as shown in the figure below Find the tension in the cables.

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w

T2T1

T2 cos480

T2 sin480

T1 sin350

T1 cos350

35o48o

x

y