뒤로Exam 1 - Study Guide D
스터디 가이드 - 스마트 노트
자료에 맞춘 맞춤형 노트, 핵심 정의, 예시, 맥락을 확장해 제공합니다.
Q1. How many picoseconds are in 2.56 years? Use 1 year = 365.25 days.
Background
Topic: Unit Conversion (Time)
This question tests your ability to convert between different units of time, specifically from years to picoseconds, using scientific notation and metric prefixes.
Key Terms and Formulas
1 year = 365.25 days
1 day = 24 hours
1 hour = 60 minutes
1 minute = 60 seconds
1 picosecond (ps) = seconds
Step-by-Step Guidance
Start by converting 2.56 years to days using the given conversion factor.
Convert the number of days to hours, then to minutes, and then to seconds.
Once you have the total number of seconds, convert seconds to picoseconds using the metric prefix.
Express your answer in scientific notation, matching the format of the answer choices.
Try solving on your own before revealing the answer!
Final Answer: (1) 8.08 × 1019 ps
2.56 years × 365.25 days/year × 24 hours/day × 60 min/hour × 60 s/min × ps/s = ps.
We used the correct conversion factors and scientific notation to arrive at the answer.
Q2. How many significant figures are in the number 0.00378260000?
Background
Topic: Significant Figures
This question tests your understanding of how to count significant figures in a decimal number, including leading and trailing zeros.
Key Terms and Rules
Leading zeros (zeros before the first nonzero digit) are not significant.
All nonzero digits are significant.
Zeros between nonzero digits are significant.
Trailing zeros after a decimal point are significant.
Step-by-Step Guidance
Identify and ignore any leading zeros in the number.
Count all nonzero digits and any zeros that are between nonzero digits.
Count all trailing zeros after the decimal point, as they are significant.
Add up the total number of significant digits.
Try solving on your own before revealing the answer!
Final Answer: (1) 9
The number 0.00378260000 has 9 significant figures: 3, 7, 8, 2, 6, 0, 0, 0, 0 (all digits after the first nonzero digit, including trailing zeros).
Q3. You jog 7 laps around a 400-meter track in 16 minutes, stopping at your starting point. What are your average speed and the magnitude of your average velocity?
Background
Topic: Kinematics – Average Speed vs. Average Velocity
This question tests your understanding of the difference between average speed (total distance over total time) and average velocity (displacement over total time).
Key Terms and Formulas
Average speed:
Average velocity:
Displacement is the straight-line distance from start to finish (with direction).
Step-by-Step Guidance
Calculate the total distance run: number of laps × length of track.
Convert the total time from minutes to seconds.
Compute the average speed using the total distance and total time.
Determine the displacement: since you end where you started, what is your net displacement?
Use the displacement and total time to find the magnitude of the average velocity.
Try solving on your own before revealing the answer!
Final Answer: (1) 2.92 m/s, 0 m/s
Total distance = 7 × 400 m = 2800 m; total time = 16 × 60 = 960 s; average speed = 2800/960 ≈ 2.92 m/s. Displacement is 0, so average velocity is 0 m/s.
Q4. You are running at 2.2 m/s and see an alligator 32 m ahead. You stop 5 m away from it. What was your deceleration?
Background
Topic: Kinematics – Constant Acceleration
This question tests your ability to use kinematic equations to solve for acceleration (deceleration) when initial and final velocities and distances are given.
Key Terms and Formulas
Initial velocity: m/s
Final velocity: m/s (since you stop)
Distance covered: m
Kinematic equation:
Step-by-Step Guidance
Calculate the total distance over which you decelerate.
Write down the known values: initial velocity, final velocity, and distance.
Use the kinematic equation and solve for .
Plug in the known values and rearrange to isolate .
Try solving on your own before revealing the answer!
Final Answer: (2) 0.041 m/s2
Using , solve for : m/s2 (deceleration).
Q5. You throw a baseball straight up, catching it every 1.5 s. It reaches a max height of 2.76 m. How fast are you throwing the ball?
Background
Topic: Kinematics – Vertical Motion under Gravity
This question tests your ability to relate time of flight, maximum height, and initial velocity for an object thrown vertically under gravity.
Key Terms and Formulas
Time to reach max height:
Total time in air:
Max height:
Acceleration due to gravity: m/s2
Step-by-Step Guidance
Recognize that the total time in the air is 1.5 s, so time to max height is half of that.
Use to relate initial velocity to time up.
Use to relate initial velocity to max height.
Set up the two equations and solve for using the given values.
Try solving on your own before revealing the answer!
Final Answer: (1) 7.4 m/s
Time up = 0.75 s; m/s. This matches the answer 7.4 m/s.
Q6. Vector A is 3.6 cm long at 14° above the x-axis. Vector B has x = -1.2 cm, y = +2.4 cm. What is the magnitude of C = A - 2B?
Background
Topic: Vectors – Components and Vector Operations
This question tests your ability to break vectors into components, perform vector addition/subtraction, and find the magnitude of the result.
Key Terms and Formulas
Component form: ,
Vector subtraction:
Magnitude:
Step-by-Step Guidance
Find the x and y components of vector A using its magnitude and angle.
Multiply the components of vector B by 2 to get 2B.
Subtract 2B from A component-wise to get the components of C.
Set up the formula for the magnitude of C using its components.
Try solving on your own before revealing the answer!
Final Answer: (1) 7.1 cm
After calculating components and combining, cm.
Q7. You throw a ball at 10 m/s, 50° above the horizontal. It reaches a max height of 0.66 m. How long until you catch it?
Background
Topic: Projectile Motion
This question tests your ability to analyze projectile motion, using initial velocity, angle, and maximum height to find total time of flight.
Key Terms and Formulas
Vertical component:
Max height:
Time to max height:
Total time in air:
Step-by-Step Guidance
Calculate the vertical component of the initial velocity.
Use the max height formula to check if the vertical component matches the given height.
Find the time to reach max height using .
Double this time to get the total time in the air.
Try solving on your own before revealing the answer!
Final Answer: (2) 0.78 s
Using the vertical component and the formulas, the total time in the air is approximately 0.78 s.
Q8. When you are initially pushing a heavy cabinet and it is accelerating, what is true about the forces between you and the cabinet?
Background
Topic: Newton's Third Law
This question tests your understanding of action-reaction force pairs and how they apply when objects are accelerating.
Key Terms and Concepts
Newton's Third Law: For every action, there is an equal and opposite reaction.
Action-reaction pairs act on different objects.
Step-by-Step Guidance
Recall Newton's Third Law and what it says about forces between two interacting objects.
Consider whether the acceleration of the cabinet changes the magnitude of the action-reaction pair.
Identify which answer choice correctly describes the relationship between the force you exert and the force the cabinet exerts on you.
Try solving on your own before revealing the answer!
Final Answer: (1) The force you exert on the cabinet is equal to the force of the cabinet pushing back on you.
Newton's Third Law always applies, regardless of acceleration.
Q9. Ignoring air resistance, which free body diagram is correct for a golf ball in flight?
Background
Topic: Free Body Diagrams – Projectile Motion
This question tests your ability to identify the correct forces acting on a projectile in flight, neglecting air resistance.
Key Terms and Concepts
Gravity is the only force acting on the ball in flight (ignoring air resistance).
Free body diagram should show only the weight force acting downward.
Step-by-Step Guidance
Recall what forces act on a projectile after it leaves the club and is in the air.
Identify which diagram shows only the force of gravity acting downward.
Eliminate any diagrams that show additional or incorrect forces.
Try solving on your own before revealing the answer!
Final Answer: (3)
The correct diagram shows only a downward force (gravity) acting on the ball.
Q10. You pull a 4.1 kg box with a 5.2 N force at 37° above the horizontal, accelerating it at 0.4 m/s². What is the coefficient of kinetic friction?
Background
Topic: Newton's Second Law, Friction
This question tests your ability to analyze forces on an object on a surface, resolve components, and use Newton's laws to find the coefficient of kinetic friction.
Key Terms and Formulas
Newton's Second Law:
Friction force:
Normal force: (if pulling up at an angle)
Horizontal component of pull:
Step-by-Step Guidance
Resolve the pulling force into horizontal and vertical components.
Write the equation for the normal force, considering the upward component of the pull.
Set up Newton's Second Law in the horizontal direction, including friction and the horizontal pull.
Express the friction force in terms of the coefficient of kinetic friction and the normal force.
Set up the equation to solve for but stop before plugging in all the numbers.
Try solving on your own before revealing the answer!
Final Answer: (1) 0.07
After resolving forces and applying Newton's laws, .