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Nick T > Physics 1 > Flashcards

Physics 1 Flashcards

1
Q

1st equation of motion

A
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2
Q

2nd equation of motion

A
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3
Q

3rd equation of motion

A
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4
Q

4th equation of motion

A
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5
Q

5th equation of motion

A
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6
Q

How do you know if there is a change in acceleration?

A

Change in speed, direction, or both

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7
Q

What is the velocity and acceleration at the peak of a projectile

A

0 m/s; 9.8 m/s^2

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8
Q

What is velocity in relation to a position vs. time graph?

A

Slope of tangent line at given point

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9
Q

What is velocity in relation to a position vs. time graph?

A

Slope

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10
Q

Since there’s no acceleration in the horizontal for a projectile what is the relation of initial and final velocity?

A

They are equal

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11
Q

What is represented by equation v/|v|?

A

Unit vector

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12
Q

True or false, the direction of net force is always the same as direction of acceleration?

A

True

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13
Q

We can calculate acceleration using the derivative of velocity. Fundamentally, acceleration is determined by ______, not _______.

A

force, velocity

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14
Q

|a| = √(ax)^2 + (ay)^2

A

Equation for magnitude of a vector

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15
Q

For general projectile motion, when the projectile is at the highest point of its trajectory:

a. Its acceleration is zero.
b. Its velocity is perpendicular to the acceleration
c. Its velocity and acceleration are both zero
d. The horizontal component of its velocity is zero
e. The horizontal and vertical components of its velocity are zero

A

b. Its velocity is perpendicular to the acceleration

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16
Q

A frog leaps vertically into the air and encounters no appreciable air resistance. Which statement about the frog’s motion is correct?

A) On the way up its acceleration is 9.8 m/s^2 directed upwards, and on the way down its acceleration is 9.8m/s^2 directed upwards.
B) On the way up and the way down its acceleration is 9.8 m/s^2 directed downwards, and at the highest point its acceleration is zero.
C) On the way up, on the way down, and at the highest point its acceleration is 9.8 m/s^2 directed downwards.
D) At the highest point of the motion it reverses the direction of its acceleration.
E) We do not have enough information to know.

A

On the way up, on the way down, and at the highest point its acceleration is 9.8 m/s^2 directed downwards.

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17
Q

A contestant starts a race with an acceleration of 3.1 m/s^2

If a contestant accelerates at that rate for16 m, and then maintains that velocity he has at that point for the rest of the 100m race, what will be his time for the entire race?

A

11.65 seconds

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18
Q

The car shown in the figure is moving towards the left, which is the negative x
direction. It travels 89.2 m during a 10.73-s time interval. Assuming that the car does not change direction, must there be an instant at which the speed of the car is equal to its average speed?

A

Yes

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19
Q
A
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20
Q

Always opposes motion between objects in contact

A

Friction

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21
Q

Initial force applied to an object

A

Normal force

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22
Q

Force due to gravity

A

Weight

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23
Q

Push or pull

A

Contact force

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24
Q

Gravitational pull

A

Field force

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25
Q

How would the total force across x and y directions be calculated in this diagram?

A
26
Q

What would the vector sum of this diagram be?

A
27
Q

An object at rest remains at rest or if in motion, remains in motion at constant velocity unless acted on by a net external force.

  • Cause-and-effect: Force is the cause for any change in velocity
A

Newton’s first law

28
Q

The acceleration of an object is directly proportional to and in the same direction as the net force acting on the object and is inversely proportional to its mass.

a = Fnet/m

A

Newton’s second law

29
Q

When an object exerts a force on a second object, the first object experiences a force that is equal in magnitude and opposite in direction to the force that it exerts.

A

Newton’s third law

30
Q

If two systems are in contact and stationary relative to one another, then the friction between them is called ______ friction.

A

Static

31
Q

What is the type of friction between M and m?

A

Static friction

32
Q

What is the direction of the friction between M and m?

A

Right, as “m” moves relative to M.

33
Q

Consider a bowling ball of mass “M” attached to two ropes. The bowling ball is in equilibrium, with one rope tied to the ceiling and the second rope pulling horizontally.

How is the tension T2 related to the weight of the bowling ball.

A

It is greater than the ball’s weight

34
Q

Draw the body diagram given the problem below.

A
35
Q

Draw the free body diagram given the following problem.

A
36
Q

What is the tension, in newtons, of rope 1?

A

58.7 N

37
Q

What is the tension, in newtons, of rope 2?

A

69.48 N

38
Q

1 radian

A

approx. 57.3 degrees

39
Q

Formula for average angular velocity

A
40
Q

Magnitude of centripetal acceleration

A
41
Q

An object is in a uniform circular motion. Which of the following statements
must be true?

A. The net force acting on the object is zero.
B. The acceleration of the object is constant.
C. The velocity of the object is constant.
D. The speed of the object is constant.
E. The angular velocity of the object is constant.

A

D. The speed of the object is constant.

42
Q

What is the angular velocity of Earth’s orbit around the Sun?

A
43
Q

Formula for tangential velocity

A

Angular velocity * radius

44
Q

What is the tangential velocity of Earth’s orbit around the Sun?

A
45
Q

What is the magnitude of centripetal acceleration of Earth’s orbit around the Sun?

A
46
Q

When a force acts on an object that undergoes a displacement from one position to another

A

Work

47
Q

1 Joule

A

1 N * m

48
Q

How is work of a varying force calculated?

A

Integration from A to B

49
Q

A stone in a sling is swung in a circle of radius r = 0.65 m with a constant angular velocity of ω = 8.5 rad/s.

Calculate the magnitude of the linear velocity of the stone in the sling v in m/s.

A

5.525

50
Q

Magnitude of linear velocity formula

A

radius * angular velocity

51
Q

Magnitude of centripetal acceleration formula

A

radius * angular velocity squared

52
Q

A stone in a sling is swung in a circle of radius r = 0.65 m with a constant angular velocity of ω = 8.5 rad/s.

A

46.96

53
Q

A car is driving along a level and unbanked circular track of diameter d=0.64km at a constant speed of v=25.7m/s.

What is the magnitude, in meters per squared second, of the acceleration of the car?

A

2.064 m/s^2

54
Q

Work done in 1D motion

A

Force * distance * cosine(theta)

55
Q

Find the amount of work done under the given conditions.
F = 100 N
theta = 30 degrees
d = 4 m

A

346 Joules

56
Q

Gravity does _______ work on an object that moves upward. In other words, you must do _______ work against gravity to
lift an object

A

negative, positive

57
Q

Work due to gravity formula

A

WG = (mg) * h * cos(theta)
m = mass
g = gravity
h = height
theta = angle

58
Q

A shopper pushes a grocery cart for a distance 19 m at constant speed on level ground, against a 27.5 N frictional force. He pushes in a direction 27.5° below the horizontal.

a) What is the work done on the cart by friction, in joules?

A

a) W = Fd * cos(theta)
(27.5 N)(19 m)(cos(180 degrees)) = -522.5 J

e) From the work-energy theorem, the net work done will be zero because the cart is moving with constant speed. Hence, the total work done on the cart is zero.

59
Q

A shopper pushes a grocery cart for a distance 19 m at constant speed on level ground, against a 27.5 N frictional force. He pushes in a direction 27.5° below the horizontal.

b) What is the work done on the cart by the gravitational force, in joules?

A

b) W = Fd * cos(theta)
(9.81 N)(0 m)(cos(90 degres) = 0J

60
Q

A shopper pushes a grocery cart for a distance 19 m at constant speed on level ground, against a 27.5 N frictional force. He pushes in a direction 27.5° below the horizontal.

c) What is the work done on the cart by the shopper, in joules?

A

c) Net Force Equation
Fa + Fb + Fc = 0;
Fc = -Fa - Fb
Fc = 522.5 J

61
Q

A shopper pushes a grocery cart for a distance 19 m at constant speed on level ground, against a 27.5 N frictional force. He pushes in a direction 27.5° below the horizontal.

d) Find the magnitude of the force, in newtons, that the shopper exerts on the cart.

A

d) W = Fd * cos(theta); F = W/(d*cos(theta))
F = 522.5/((19)(cos(27.5degrees))
F = 31.0 N

62
Q

A shopper pushes a grocery cart for a distance 19 m at constant speed on level ground, against a 27.5 N frictional force. He pushes in a direction 27.5° below the horizontal.

e) What is the total work done on the cart, in joules

A

From the work-energy theorem, the net work done will be zero because the cart is moving with constant speed. Hence, the total work done on the cart is zero.

Nick T (4 decks)

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