1. A force P = 800 N is shown in Fig. a. Find the y-component of P with respect to x and y axis. b. Find the y'-component of P with respect to x' and y' axis. c. Find the y-component of P with respect tox' and y axis. d. Find the y'-component of P with respect to x and y' axis. P = 800 N 30 X'

1. A force P = 800 N is shown in Fig. a. Find the y-component of P with respect to x and y axis. b. Find the y'-component of P with respect to x' and y' axis. c. Find the y-component of P with respect tox' and y axis. d. Find the y'-component of P with respect to x and y' axis. P = 800 N 30 X'

Name: ( 45 aM > 1:0YHW1.pdf >Chapter 1HW1Exercises1. A force P = 800 N is s
Uploaded: 2024-03-20T06:56:46.000Z
Channel: Bartleby
Description: ( 45 aM > 1:0YHW1.pdf >Chapter 1HW1Exercises1. A force P = 800 N is shown in Fig.a. Find the y-component of P with respect to x and y axis.b. Find the y'-component of P with respect to x' and y' axis.c. Find the y-component of P with respect to x' a

Principles of Physics: A Calculus-Based Text

5th Edition

ISBN:9781133104261

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter1: Introduction And Vectors

Section: Chapter Questions

Problem 59P

See similar textbooks

Similar questions

My question is why did we used sine and cosine 45 degree? Is not is supposed to be 315 degree since 45 degree is in quadrant four? Why can't we use 315 degree?
Services Academic Departments ETC CIM 1. Based on the values provided, complete the following table. Test No Weight (N) H, (cm) H2 (cm) P.E. (initial) (J) P.E. (final) (J) Loss in P.E. (J) 1 1.0 50 30.5 1.2 50 31.7 1.4 50 32.9 4 1.6 50 34.2 1.8 50 35.5 3.
If e = 60° and the radius of the quarter and semi – circle is both 5 m. Determine the following: a.) The x - component if F = 7 N. b.) The y - component if F = 9 N. y F Answer: (a) Fx = 3.130 N (b) Fy = 8.050 N
Consider the force shown in the figure below. If the y-component of the force is Fy = 16.0 N and the angle is θ = 22.0 ◦ , then the x-component and magnitude of the force are A. Fx = 0 and |F~ | = 16.0 N B. Fx = −16.0 N and |F~ | = 22.6 N C. Fx = −39.6 N and |F~ | = 42.7 N D. Fx = −6.46 N and |F~ | = 17.3 N
1. Based on the values provided, complete the following table. Eof Test No Weight (N) H, (cm) H2 (cm) P.E. (initial) (J) P.E. (final) (J) Loss in P.E. (J) estion 1 1.0 50 30.5 1.2 50 31.7 1.4 50 32.9 4 1.6 50 34.2 1.8 50 35.5
Either all parts or none... I vll definitely upvote
STATICS OF RIGID BODIES PRODUCTS OF VECTOR VECTOR CROSS AND DOT PRODUCT
1. Based on the values provided, complete the following table. Test No Weight (N) H1 (cm) H2 (cm) P.E. (initial) (J) P.E. (final) (J) Loss in P.E. (J) 1 1.0 50 30.5 1.2 50 31.7 1.4 50 32.9 4 1.6 50 34.2 1.8 50 35.5 2.
PART A) What is the maximum force that can be applied to a bone whose minimum cross-sectional area is 3.0cm^2? (This is approximately the cross-sectional area of a tibia, or shin bone, at its narrowest point.) Express your answer using two significant figures. PART B) Estimate the maximum height from which a 70kg man could jump and not fracture the tibia. Take the time between when he first touches the floor and when he has stopped to be 0.006s, and assume that the stress is distributed equally between his legs. Express your answer using two significant figures. Thank you!!!
A. Draw fbd B. Determine value of force
18. Replace the two forces shown by a force-couple system with the force acting at O. B 360 mm 50 N 270 mm 30° A 90 N Ý Ans. R = 135.62N, C = 6.3N-m, CCW
Two forces of magnitude 2F and the square root of F act such that the resultant force is square root of 10F. What is the angle between the forces given in the problem? O o deg 120 deg 90 deg 45 deg 30 deg