6. A meter stick whose mass is 0.200 kg is supported at the zero cm mark by a knifeedge and a force F at the 100 cm point. A mass of 700 grams is attached to the stickat the 40 cm mark. Find the magnitude of N and F in Newtons.Magnitude of NNNewtons, Magnitude of FNewtonsFitions for40 cm700 gm100 cm

Answered: Image /qna-images/answer/d1719765-1b5e-4df1-895c-9956f2526112.jpg

Answered: 6. A meter stick whose mass is 0.200 kg…

Glencoe Physics: Principles and Problems, Student Edition

1st Edition

ISBN:9780078807213

Author:Paul W. Zitzewitz

Publisher:Paul W. Zitzewitz

Chapter5: Displacement And Force In Two Dimensions

Section: Chapter Questions

Problem 6STP

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Similar questions

The tube, shown below, is supported by two cables (AB) and (AC). A force F = 600 lb is applied in cable AB. The component of the force Facting along cable (AC) is: |3 ft E 8 ft- В 12 ft 8 ft 15 ft Select one: a. 49.509 [lb] O b. 495.708 [lb] c. 600 [lb] O d. 459.708 [lb]
2/73 The bracket is fastened to the girder by means of the two rivets A and B and supports the 2-kN force. Re- place this force by a force acting along the centerline between the rivets and a couple. Then redistribute this force and couple by replacing it by two forces, one at A and the other at B, and ascertain the forces supported by the rivets. Ans. FA = 0.8 kN FB = 2.8 kN 250 mm 100 mm 2 kN 300 mm
The bar ABCD has a built-in support at A. Neglect the weight of the bar. T = 8.1 kN; M = 4.6 kNm. Calculate the resultant force (in kN) exerted by the support on the bar. B 270 500 T 450 9 KN Dimensions in mm M
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What is the y-component of the net Force (F = F+F2) measured in N? F=18N 50 山 300 F=30N X
A bridge of length 57 m and mass 80000 kg is supported at each end. A truck of mass 38200 kg is located 27 m from the left end. A A 27 m B 57 m What is the force on the bridge at the left point of support? The acceleration of gravity is 9.8 m/s². Answer in units of N. Your response... i Previous Responses X #1.412105.263 PALETTE
1. Calculate the weight in newtons and in lbf of an average person on the Earth. 2. During cell migration, fibroblasts can generate traction forces of approximately 2 Å~ 104μdynes. What is the equivalent force in KN?
The two planks are connected together by cable BC and a smooth spacer DE. Determine the reactions at the smooth supports A and F.and also find the force developed in the cable and spacer. Match the answers. 100 lb 200 lb -2 ft--2 ft- ft2 ft- 182 160 180 1. reaction at A= 168 2. reaction at F = 176 3. Foc- 128 4. FDE 5. 150 120 140
FORCE COMPONENTS: 1. Compute the X and Y Components of each of the forces. T=722 b P-200b. F-448lb. 2. Compute the X and Y Components of each of the forces. 200 lb 400 lb X 300 lb 300 lb
2. Twoforces,oneofmagnitude2Nandoneofmagnitude3N,areappliedto the ring of a force table. The directions of both forces are unknown. Which of the following best describes the magnitude of the equilibrium force? Justify your answer.a. FE ≤ 5N b. 2N≤FE≤3N c. FE ≥ 3Nd. 1N≤FE≤5N e. FE ≤ 2N
2. SOLVE THE FORCE ACTING ON THE 5 KN BLOCK IF IT IS IN EQUILIBRIUM. P 380 580
2. Bones and Muscles. A patient in therapy has a forearm that weighs -20.5Nj and that lifts a -112.0Nj weight. These two forces have direction vertically downward. the only other significant forces on his forearm come from the biceps muscle (which acts perpendicularly to the forearm) and the force at the elbow. If the biceps produces a pull of 232 N when the forearm is raised above 43° the horizontal, find the magnitude and direction of the force that the elbow exerts on the forearm. (The sum of the elbow force and the biceps force must balance the weight of the arm and the weight it is carrying, so their vector sum must be 132.5 N, upward.) Vector addition Given the following vectors: A = 10mi + 20m 2 4. P=Z+L 5. F=A+L+G+L 6. S=P-F G=55.2m, 45° South of West L = 10m, 85° East of South Draw the graphical representation and compute for the x- and y- components and the angle of the following resultant vectors: 3. R=A+G

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