20. Use the results from Example Problem 2 to answer questions abouta scale in an elevator on Earth. What force would be exerted bythe scale on a person in the following situations?a. The elevator moves at constant speed.b. It slows at 2.00 m/s² while moving upward.c. It speeds up at 2.00 m/s² while moving downward.d. It moves downward at constant speed.e. It slows to a stop at a constant magnitude of acceleration. EXAMPLE Problem 2Real and Apparent Weight Your mass is 75.0 kg, and you arestanding on a bathroom scale in an elevator. Starting from rest,the elevator accelerates upward at 2.00 m/s² for 2.00 s and thencontinues at a constant speed. Is the scale reading during accelerationgreater than, equal to, or less than the scale reading when theelevator is at rest?1 Analyze and Sketch the Problem• Sketch the situation.F.scaleSystemChoose a coordinate system with the positivedirection as upward.F.netscale+yDraw the motion diagram. Label v and a.• Draw the free-body diagram. The net forceis in the same direction as the acceleration,so the upward force is greater than thedownward force.Known:Unknown:m= 75.0 kgE.scale?a = 2.00 m/s²t = 2.00 s= 9.80 N2 Solve for the UnknownFF.= manet+ (-F)F, is negative because it is in the negative direction defined bythe coordinate system.netscaleSolve for FscaleFnet + Fg= EscaleElevator at rest:F.= Eret + E.gThe elevator is not accelerating. Thus, FnetSubstitute Fet = 0.00 Nscalenet= 0.00 N.%3DgSubstitute F, = mgSubstitute m = 75.0 kg, g = 9.80 m/s²mg= (75.0 kg)(9.80 m/s²)= 735 NMath HandbookOperations withSignificant DigitsAcceleration of the elevator:pages 835–836Fecale = Fnet + g= ma + mg%3DSubstitute Fnet= ma, F, = mgSubstitute m = 75.0 kg, a = 2.00 m/s², g = 9.80 m/s²= m(a + g)(75.0 kg)(2.00 m/s² + 9.80 m/s²)= 885 NThe scale reading when the elevator is accelerating (885 N) is larger than the scalereading at rest (735 N).

Name: 20. Use the results from Example Problem 2 to answer questions abouta
Uploaded: 2024-03-20T06:57:37.000Z
Channel: Bartleby
Description: 20. Use the results from Example Problem 2 to answer questions abouta scale in an elevator on Earth. What force would be exerted bythe scale on a person in the following situations?a. The elevator moves at constant speed.b. It slows at 2.00 m/s² whi

Given:- the mass of the person m = 75 kg Find:- The force exerted on the scale a) When the…

Science

Physics

- Use the results from Example Problem 2 to ansv a scale in an elevator on Earth. What force woul the scale on a person in the following situations a. The elevator moves at constant speed. b. It slows at 2.00 m/s² while moving upward. c. It speeds up at 2.00 m/s² while moving dowr movee downuword ot oonotont onoed

- Use the results from Example Problem 2 to ansv a scale in an elevator on Earth. What force woul the scale on a person in the following situations a. The elevator moves at constant speed. b. It slows at 2.00 m/s² while moving upward. c. It speeds up at 2.00 m/s² while moving dowr movee downuword ot oonotont onoed

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

See similar textbooks

Similar questions

SD6) What is the correct solution fo this question.
A kind of vector that is required to occupy a unique position in space. A. Fixed B. Free C. Sliding D. Equal
Two identical boxes with mass m are placed on inclined surfaces with coefficient of kinetic friction and connected by a light rope passing through a frictionless pulley as shown in the figure. Which of the following gives the correct expression for the system's accelera- tion? A. a = (sin 3-sina) - B. a (sin 3-sina) - C. a = (sin a-sin 3) - D. a (sin a sin 3) - (cosa = - = (cosa + cos 3) (cosa - cos (3) (cosa + cos 3) cos 3) α |6|× m B
mass. You place the mass on the ramp at an angle high enough that the mass starts sliding. You measure the time it takes to fall down a known distance, The time it takes to fall down the 45 degree ramp You have devised an experìment to measure the kinetic coefficient of friction between a ramp and a starting from a standstill is At = 0.55 sec, m-1 kg, and the distance it falls, Az, is 0.5 m. What is ? 7. Ax in At Ug
Need correct ans wrong ans gives u many downvote
Which of the following does not describe a vector quantity?a. a 900 kg massb. a ball projected upwardc. Wind moving at 130 km/hr NEd. A car traveling 100 km/hr along South superhighway
A 60 kg block is pushed by a horizontal forcecausing the block to accelerate by 1 m/s2. The block has an initial velocity of 5 m/s. The coefficient of friction between the block andthe horizontal floor is 0.40. Find the distance travelled by the blockafter 10 seconds.a. 90 m b. 100 mc. 70 m d. 120 m
Help
What happens to the object when you apply a constant force on it in space? A. It will not move B. It will continuously move C. It will move faster D. None of the above
A Roller Coaster train and its passengers have a combined mass of 1375 kg. The train comes over the top of the first hill on a roller coaster, 45 m above the ground with a velocity of 10.5 m/s. a) The train goes down the hill on to another hill 18 m above the ground. What is its velocity on top of the second hill. b) What will be its velocity when it reaches the ground level from the top of the second hill.
I have absolutely no clue what she's asking on these problems. Please help!
A 10.0-kg box is pulled along a horizontal surface by a force of 40.0 N applied at a 30.0° angle with vertical. Coefficient of kinetic friction is 0.30. Calculate the acceleration. a. 0.52 meter per second squared b. 3.76 meter per second squared c. 1.12 meter per second squared d. 3.16 meter per second squared e. None of the above