0+ A 3. Knight P10.19 As a 15,000 kg jet p an aircraft carrier, its tail hook sna slow it down. The cable is attached with spring constant 60,000 N/m. stretches 30 m to stop the plane, w plane's landing speed? Answers to one significant figure 1. 1 m 2. 8 m/s 3. 60 m/s
0+ A 3. Knight P10.19 As a 15,000 kg jet p an aircraft carrier, its tail hook sna slow it down. The cable is attached with spring constant 60,000 N/m. stretches 30 m to stop the plane, w plane's landing speed? Answers to one significant figure 1. 1 m 2. 8 m/s 3. 60 m/s
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)...
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Help with the following question:
The second photo is not a question, it’s just to help you solve
![1.
2.
3.
0+
A
Answers to one significant figure
1 m
8 m/s
60 m/s
B
3. Knight P10.19 As a 15,000 kg jet plane lands on
an aircraft carrier, its tail hook snags a cable to
slow it down. The cable is attached to a spring
with spring constant 60,000 N/m. If the spring
stretches 30 m to stop the plane, what was the
plane's landing speed?
X](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F760b950f-bf86-4f0e-8f61-09da7fb76390%2F00454320-1d89-41b5-ab7e-d5e5aee340b3%2Fxsx2ss3_processed.jpeg&w=3840&q=75)
Transcribed Image Text:1.
2.
3.
0+
A
Answers to one significant figure
1 m
8 m/s
60 m/s
B
3. Knight P10.19 As a 15,000 kg jet plane lands on
an aircraft carrier, its tail hook snags a cable to
slow it down. The cable is attached to a spring
with spring constant 60,000 N/m. If the spring
stretches 30 m to stop the plane, what was the
plane's landing speed?
X
![Useful information
gravitational acceleration g = 9.8 m/s²
General energy theorem
Ko + Uo+Wext = Kf+Uf+AEth
Isolated and dissipationless system
Kinetic energy.
Spring relations
Gravity relations
Ko+Uo= Kf+Uf
KE=
Fsp=
Usp
mv²
= -kAx
= ½k(Ar)²
Fg = -mg
Ug = mgAy](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F760b950f-bf86-4f0e-8f61-09da7fb76390%2F00454320-1d89-41b5-ab7e-d5e5aee340b3%2Fd4utmvm_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Useful information
gravitational acceleration g = 9.8 m/s²
General energy theorem
Ko + Uo+Wext = Kf+Uf+AEth
Isolated and dissipationless system
Kinetic energy.
Spring relations
Gravity relations
Ko+Uo= Kf+Uf
KE=
Fsp=
Usp
mv²
= -kAx
= ½k(Ar)²
Fg = -mg
Ug = mgAy
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