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Physics (12) (ELCA)
- Physical quantities that are described only be a magnitude are called scalar.
- Temperature, pressure, and mass are examples of scalar quantities.
Representing Forces GraphicallyEdit
- Because force is a vector, with magnitude and direction, it can be represented by or drawn as an arrow (free body diagram).
- Let arrow length represent the magnitude.
- Let the arrow heading represent the direction.
- Scale. Ex. Scale is 1 cm = 50 m
Forces that act along the lineEdit
- If two forces act in the same direction we add them as ordinary numbers.
- If there as in the opposite direction we subtract them as ordinary numbers.
How do we add forces that act at an angle?Edit
- A circle has 360º
- A circle is divided into 4, 90º quadrants.
Forces acting on an angleEdit
If forces on an object don't act along the same line:
- you can not add or subtract them as ordinary numbers.
- you have to solve them graphically or mathematically.
Forces acting on an angleEdit
- Select a scale: 1 ft, 1 cm = 1 m
- Draw a suitable horizontal line on graph paper using your scale.
- Draw a suitable vertical line on graph paper using your scale.
- Draw the vector by attaching the tail of one vector to the head of the other.
- Draw a line to connect the point of origin with the head of the 2nd vector.
- Measure the length of the resultant.
- Measure the angle of the resultant based on the horizontal.
- The resultant of two forces always acts in a direction between the direction of the two forces.
- When you want to add forces that do not act along a line, if the two vectors are at a 90º to each other, you can solve the problem mathematically.
- Pythagorean Theorem
A2 + B2 = C2
- The Radius of a unit circle is one.
- sinø = opposite / hypotenuse
- cosø = adjacent / hypotenuse
- tanø = opposite / adjacent
Non-Right Angle TrianglesEdit
- ∆x1 = d1(cos(ø1))
- ∆y1 = d1(sin(ø1))
- ∆x2 = d2(cos(ø2))
- ∆y2 = d2(cos(ø2))
- ∑x = ∆x1 + ∆x2
- ∑y = ∆y1 + ∆y2
- d = √(∑x 2 + ∑y2 )
- ø = tan-1(∑y/∑x)
- vx = vi(cos(ø))
- ∆x = vi(cos(ø))∆t
- vyf = vi(sin(ø)) – gt
- vyf2 = vi 2(sin(ø))2 – 2gy
- ∆y = vi(sin(ø)) - g(∆t)/2
- (- b+/-√(b2 - 4ac))/2a
- ∆t = ∆x/vi cos(ø)
- vi2 = ((g)(∆x)/(2(cos(ø))(sin(ø)))
Projectile and TrajectoryEdit
- Projectile - any object moving through the air only affected by gravity.
- Trajectory- the path a projectile follows. The shape of the path is an arch-shaped curve called a parabola.
- Projectiles launched at steep angles result in a tall narrow parabola
- Projectiles launched at an angle close to the horizontal result in a wide, low parabola
- Range – the distance a projectile travels horizontally.
Range depends upon:
- initial angle
- height of ground
Range traveled by a projectile depends on launch speed and launch angle. The greater the launch speed, the greater the range is proportional to the square of the velocity. Doubling the launch speed quadruples the range of the projectile.
The range of the projectile is calculated from the horizontal velocity and the time of flight. The vertical velocity is responsible for giving the projectile its airtime. The time of flight is twice the time it takes the projectile to reach the top of its trajectory where Vy = 0. A projectile travels farthest when launched 45º. At this angle, its velocity is evenly divided between the horizontal and the vertical.
Range of a projectile:
x = vixt=((vi)(cos(ø))(((2vi)(sin(ø))/g)
Range = 2/g • (vi2 • sin(ø) • cos(ø))
To calculate “hang time,” rearrange the vertical distance formula to solve for time:
∆y = -1/2g∆t2Hangtime of Projectile:
t = √(2y/g)
The more the launch angle varies from 45º, the smaller the range. A projectile launch at 30º will have the same range as a projectile launched at 60º.
Velocity measurements depend on the frame of reference of the observer. Example: you’re driving north on a road at 80 km/hr and are passed by someone driving 90 km/hr. To you it appears that the other car is moving 10 km/hr compared to your own speed. An observer on the side of the road would measure the velocity of the faster car at 90 km/hr north.
Observers using frames of reference may measure different displacements or velocities for an object in motion.
Example: a stunt dummy is dropped from an airplane flying horizontally over Earth with a constant velocity.
A passenger on the airplane would observe that the dummy fell straight down to earth.
An observer on the ground would view the dummy’s fall as a projectile (curving down to the ground.
When solving relative velocity problems, write down all the information that is given and what you want to know in the form of velocities with subscripts added.
- vse= +80 km/hr
- se = slower car with respect to the earth
- vfe = +90 km/hr
- fe = faster car with respect to the earth
- vfs = ? km/hr
To solve for vfs, write an equation for vfs in terms of the other velocities. On the right side of the equation, the subscript starts with f and eventually ends with s. Each velocity subscript starts with the letter that ended the preceding velocity subscript.
vfs = vfe + vesIf north is taken to be the positive direction, we know that ves = - vse because an observer in the slow car perceives Earth as moving south at a velocity of 80 km/hr while a startionary observer on the ground (Earth) views the car as moving north at a velocity of 80 km/hr
vfs = +90 km/hr – 80 km/hr = 10 km/hr
Notes from Outside ELCAEdit
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