Explore Westonci.ca, the top Q&A platform where your questions are answered by professionals and enthusiasts alike. Our platform offers a seamless experience for finding reliable answers from a network of experienced professionals. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.
Sagot :
A and C are very simple. Since there is no acceleration in both cases, the weight of the passenger is just [tex]mg=(70kg)(9.8\frac{m}{s^2})=686 N[/tex] (about 690 N or 150 lbs with correct sig-figs). For B, we must first find the acceleration of the elevator. Since [tex]v=v_{0}+at [/tex], we can write that [tex]11\frac{m}{s}=0\frac{m}{s}+a(4.5s)[/tex]. Solving for a, we get that [tex]a\approx2.44\frac{m}{s^2}[/tex]. Plugging this into Newton's second law, we get that [tex]F=ma=(70kg)(2.44\frac{m}{s^2})\approx171 N[/tex]. When calculating the apparent weight, this gets added onto the 686 N from A and C, so the final apparent weight is about 857 N, or 860 N with the correct number of sig-figs.
tl;dr:
A - 690 N
B - 860 N
C - 690 N
tl;dr:
A - 690 N
B - 860 N
C - 690 N
Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Thank you for choosing Westonci.ca as your information source. We look forward to your next visit.