# Physics 1 Flashcards Preview

## ALTIUS Cards > Physics 1 > Flashcards

Flashcards in Physics 1 Deck (35)
1
Q
• Name 5 scalars
• “MaTe Speed TiDe”
A
1. Mass
2. Temp
3. Speed
4. Time
5. Density
2
Q
• Name 8 vectors
• “VeDiAcFo MagMoImTo”
A
1. Velocity
2. Displacement
3. Accel
4. Force
5. Mag. Field
6. Momentum
7. Impulse
8. Torque
3
Q

Think of force as…

A
• any influence capable of:
• causing a mass to accelerate
4
Q

Center of Mass equation

• What should you remember to keep in mind with these problems?
A

Cmass=(r1m1+r2m2….)/mtotal

• r=reference point

Hint: choose a reference point from which to measure each displacement vector

5
Q
• Constant net force causes WHAT acceleration?
• and therefore WHAT velocity?
A
• CONSTANT acceleration
• therefore CHANGING velocity!!!
6
Q
• When you see “constant velocity” or “constant speed,” think? (5)**​
A
1. NO acceleration
2. NO net force (Fnet=0)
3. All forces sum to zero
• up forces=down forces, left=right, etc.
4. NO change in direction
5. The object is in EQUILIBRIUM!!!
7
Q
• When you see a LINEAR MOTION GRAPH, you’ll ask yourself these 6 things:
A
1. What does the SLOPE represent?
2. Is this slope (+) or (-)?
3. It the slope….
• Constant (straight line) or
• Non-constant (curved line)?
4. What is the value on the x-axis?
5. Is the y value (+) or (-)
• ​​aka are you above or below the x-axis?
6. At t=0​, do I expect the value on the y-axis to be:
1. LARGE, or
2. SMALL?
8
Q
• When I see the word “projectiles,” I will remember 7 things:
A
1. Horizontal VELOCITY:
• never changes
2. Horizontal ACCELERATION:
• is always =0
3. VERTICAL acceleration:
• is always 10 m/s2 downward
4. Vertical BEHAVIOR
• is always symmetrical
• ex: upward trip=downward trip
5. Time in air (Tair) depends on:
1. VERTICAL COMPONENT of velocity ONLY!!
6. Range depends on both:
1. vertical AND horizontal components
7. Time is always the same for both x and y components of the motion
9
Q

Law of Universal Gravitation formula

A

Fg=Gm1m2/r2

10
Q
• Fg=Gm1m2/r2
• gives WHAT?
A
• the force DUE to gravity
• NOT gravity itself!
11
Q
• Formula for GRAVITY (ITSELF!)
• aka “acceleration due to gravity”
• aka “strength of gravitational field”
A

g=Gm/r2

12
Q
• What are the 2 physics equations that could be used FOR FALLING OBJECTS?
A
• x=½at2
• V=√2gh
13
Q

Time in air equation

• Tair=?
• this equation can ONLY be used to calculate what?
A

Tair=2V/g

• can only be used to calculate “round trip” times
• aka total time in air,V, must be vertical component of INITIAL velocity

14
Q
• At terminal velocity
• What 2 things are happening?
• Give the formula for terminal velocity
A
• object has stopped accelerating
• forces of gravity and air resistance are BALANCED

Fair=mg

15
Q

Vavg=?

A

Vavg=(V1+V2)/2

16
Q
• For PEgrav, which variation will you MOST LIKELY see on the MCAT?
A

PEgrav=mgh

• (At or near earth’s surface; g=10m/s2)

17
Q

Inclined Planes

• When is the equation when solving for:

Force down an IP PARALLEL to the surface?

A

F=mgsinθ

18
Q

Inclined Planes

When is the equation when solving for:

Normal Force (F<strong>N</strong>) down an IP

A

FN=mgcosθ

19
Q

Inclined Planes

• When is the equation when solving for:

Velocity of a particle at the base of an inclined plane

• What other kind of problem could this equation be used to help solve?
A

Vfinal = √2gh

Can also be used to help solve for

FALLING OBJECT problems

20
Q

Inclined Planes

• When is the equation when solving for:

ACCELERATION DOWN an IP

• What other equation can you derive this from?

HINT:

• Notice you’re solving for acceleration DOWN an IP.
• What other equation solves for something DOWN an IP that you know of?
A

a=gsinθ

Derived from:

• F=mgsinθ (Force DOWN an IP)

F=ma, ∴ “a”=gsinθ

21
Q

Inclined Planes

• Why does Vf = √2gh work for either falling bodies OR a mass on an inclined plane?

HINT:

What is the above equation derived from?

What is happening to an object as it goes from the point where it is dropped until hitting the ground?

A

The formula Vf = √2gh is derived from CONSERVATION OF ENERGY

• by equating mgh to ½mv2
• and solving for “v”

As long as friction, air resistance, etc. are ignored (which they are), energy will be conserved in an identical way…

WHETHER THE OBJECT FALLS DIRECTLY TO THE GROUND OR ROLLS DOWN A PLANE

22
Q

Inclined Planes

As the angle of incline of a plane INCREASES:

1. What happens to the value of a?
2. What happens to the value of sinθ and cosθ?
3. What happens to the normal force?
4. What happens to the force down the plane?
5. What is the maximum value for acceleration down an inclined plane?
6. What is the minimum value for acceleration down an inclined plane?
A
1. ) Because the acceleration down a plane is directly related to the SINE of the angle
* (F=mgsinθ, where gsinθ​=”a,” since F=ma)*
• the greater the angle, the closer the sine of the angle will be to ONE (1.0)
• Because increasing the angle of SIN:
• 0 ⇒ 1.0
• Therefore, the larger the angle, the closer the acceleration will be to 9.8m/s2*
2. ) The normal force is related to the cosine of the angle (Fn=mgcosθOSD) , so as the angle increases, this value gets closer to zero
• Because increasing the angle of COS:
• ​ 1.0 ⇒ 0
• Therefore, as the angle increases the normal force decreases
1. ) The force down an inclined plane is also related to the sine of the angle
* so it too will increase as the angle of incline increases
2. ) The theoretical maximum incline is 90 degrees
* where acceleration would be exactly 9.8 m/s2
3. ) The minimum would be a plane with NO angle of incline
* where acceleration down the plane would be ZERO
23
Q

Tension Forces

What is the tension in a rope being pulled from opposite ends with identical forces of 50N?

A

50N

24
Q

Tension Forces

• A 500kg elevator is being accelerated upward by a cable with a tension of 6,000N

What force does the elevator exert on the cable?

A

TRICK QUESTION!

According to Newton’s 3rd Law, if the elevator CABLE is pulling on the ELEVATOR with 6,000N of force…

…then the ELEVATOR must be pulling on the

CABLE with a force of 6,000N

25
Q

Hooke’s Law

• Give the equation
A

F=kΔx

• Δx is the displacement of the spring from its equilibrium point

26
Q

Hooke’s Law

F=kΔx

A ball rolls along a frictionless table and strikes a spring

• Describe:
• The force experienced by the ball due to the spring
• The acceleration of the ball
• How both change with time
A
• As the ball strikes the spring it experiences an ever-changing force, F
• As the spring compresses, however, that force increases according to Hooke’s Law*
• Because the ball experiences an increasing force, it will also experience an increasing acceleration
• The maximum force and acceleration will occur at the maximum compression of the spring*
• As the ball is pushed backward by the spring these variables will change in a symmetrical way
• such that their value is exactly the same at any given value of x on either the way in, or the way back out*
27
Q

QUICK!

What is the density of water?

• How many cm3 per mL?
• How many L of water in 1 kg?
• How many mL of water in 1 gram?

A

Density of water:

1000kg/m3 or 1.0g/cm3

• 1cm3 = 1mL
• 1L of water = 1kg
• 1mL of water = 1 gram
28
Q

Define “SPECIFIC GRAVITY”

• HINT: It’s a ratio that compares two things…
• Give the formula for Specific Gravity
A

SPECIFIC GRAVITY=

A ratio that describes how DENSE something is COMPARED TO WATER

SG = Dsubstance / Dwater

29
Q

Specific Gravity

• For objects floating in liquids, the fraction of the object submerged = ?
• in other words, Fractionsubmerged​= ?

**If the liquid in which it is submerged is WATER, the fraction submerged is equal to ?

A

For objects floating in liquids​:

• Fraction of the object submerged is equal to the ratio of the density of the object to the density of the liquid

Fractionsubmerged = Dobject / Dliquid

**If the liquid in which it is submerged is WATER, the fraction submerged is _EQUAL_ to the specific gravity!

30
Q

Specific Gravity

• A ball is floating ¾ submerged in a liquid with a density of 2.0 g/cm3

What is the specific gravity of the liquid and the density of the ball?

A

Because the ball floats with ¾ of its volume submerged, it must be ¾ as dense as the liquid

• Therefore the density of the ball must be 1.5g/cm3

This is 1.5 times as dense as water, so the SG of the ball is 1.5, and the SG of the liquid is:

DLiquid/DH2O = 2g/cm3 / 1g/cm3

=2

31
Q

Define ARCHIMEDES’ PRINCIPLE

• Differentiate b/t whether it’s FULLY submerged or PARTIALLY (aka it’s “floating”)
A

Archimedes’ Principle

Any object displaces an amount of fluid…..

• Exactly EQUAL to its OWN volume
• …if FULLY submerged

…OR…

• To the volume of whatever FRACTION of the object IS submerged (¾, ½, etc.)
• …if FLOATING

The weight of the displaced fluid is exactly equal to the buoyant force pushing UP on the object

32
Q

THE BUOYANT FORCE

• Give the equation
• Describe what each part means

HINT: PUG!

A

The Buoyant Force:

Fbuoyant = ρvg

• v
• volume of fluid that GETS displaced!

NOT the total volume of the fluid itself!

• ρ
• density of the fluid

NOT the object!

33
Q

REMEMBER**!

The buoyant force is always EXACTLY EQUAL to….?

A

… to the weight of the amount of fluid

that is getting DISPLACED BY the object

So, if an object is displacing 4 lb worth of fluid, then the buoyant force would be 4 N

34
Q

The Buoyant Force

1. What causes the buoyant force?
A

The best way to intuit buoyant force is to look at the pressure differential between the top and bottom of an object

• The fluid pressure, ρgh, will be larger at the object’s bottom surface than it is at the top surface
• due to the larger value of h (since it’s at a deeper surface)

Let’s examine a submerged cube with the same surface area both top and bottom

The formula P = F/A tells us:

• If pressure is greater at the bottom, and area stays the same:
• there must be a greater force UP on the bottom surface than there is down on the top surface
• This makes it logical that any submerged object will experience a net upward force–*
• because of the PRESSURE DIFFERENTIAL!!*
35
Q

The Buoyant Force

1. How does the buoyant force change with depth (h) ?
2. How does the buoyant force change with the mass (m) of the object?

HINT: Think of the formula…“FB PUG” (facebook pug)

A

1.)

Buoyant force does NOT change with depth!

• Depth (h) is NOT in the formula ( F = ρvg )
• Therefore, we can confidently conclude that depth does NOT change buoyant force
• Whether the depth is shallow or very large, the pressure difference between ρgh-top and ρgh-bottom will remain the SAME

2.)

Buoyant force does** **NOT** **change with depth!

Similarly, we can say that the mass of the object does NOT affect buoyant force

• Like depth, it is NOT in the formula
• Nor is it accounted for by our understanding of what causes the buoyant force