10. A racecar travels 40 m/s around a banked (45° with the horizontal) circular (radius = 0.20 km) track. What is the magnitude of the resultant force on the 80-kg driver of this car? a. 0.68 kN b. 0.64 kN c. 0.72 kN d. 0.76 kN e. 0.52 kN
Added by Andre R.
Close
Step 1
In this case, $m = 80 kg$, $v = 40 m/s$, and $r = 0.20 km = 200 m$. $F_c = \frac{(80 kg)(40 m/s)^2}{200 m} = \frac{80 \times 1600}{200} = 640 N = 0.64 kN$ Show more…
Show all steps
Your feedback will help us improve your experience
Ankur S and 101 other Physics 101 Mechanics educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
A racing car travels at 40 m / s around a circular track (radius = 0.20 km) with a cant (45 ° to horizontal). What is the magnitude of the resultant force on the 80-kg driver of this car? TO. 0.52 kN B. 0.72 kN C. 0.76 kN D. 0.64 kN AND. 0.68 kN
Ankur S.
A car travels along the perimeter of a vertical circle (radius = 0.16 km) at a constant speed of 27 m/s. What is the magnitude of the resultant force on the 72-kg driver of the car at the lowest point on this circular path? a. 0.24 kN b. 0.17 kN c. 0.33 kN d. 0.39 kN e. 0.65 kN
Sri K.
Mahendra K.
Recommended Textbooks
University Physics with Modern Physics
Physics: Principles with Applications
Fundamentals of Physics
Transcript
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD