Physics!

Hi everyone :)

Today in class we went over the booklet, Work and Energy. The notes basically introduce different concepts of "work." We also corrected Chapter 10 Study Guide: 10.1 Work and Energy. Here are the answers:
Work
force, distance, direction
W=FxD
work, force, distance
scalar
joule
Newton, meter, joule
moves
force, displacement
Work and Direction of Force
in the direction of
perpendicular
in the direction of
cosine, direction of motion
negative
energy
energy, motion
positive, negative
Power
rate, rate, energy
P=W/T
power, work, time
watt
joule, one second
kilowatts

Corrections were also done for p.330 of the Physics book, answers are as follows..
1.a) W=FD
=(40N)(0.15m)
=6J

b) W=FD
=(50kg)(9.8)(195m)
=956J

c) W=F∆D(cosΘ)
=(120)(4.0)(cos25)
=435J

2) W=FD
=(5000)[1/2(25)5^2]
=156 250J

3) W=F∆D(cosΘ)
=(78N)(10m)(cos35)
=447J

I wrote down the answers for questions 4-5 but, I wasn't sure of the answers! Sorry =/

6) D=W/F
=480J/30N
=16m

Ms.K also handed back our Centripetal Force Labs back, here are the answers to the questions in case you didn't get them. Thanks to group mates Dave and Camy for getting the correct answers..nerds❤ ..or i'm just dumb =S
2) The relationship revealed is a power relationship between the two variables.
4) A straight line is obtained, representing a linear relationship between F&V^2.
5) F=mv^2/R... m(mass) and R(radius) were kept constant throughout the investigation. Therefore yes, the results of our investigation do support the equation because F over V^2, representing a power relationship as displayed by the results.

Homework:

• Chapter 11 Study Guide: 11.1 Energy in its Many Forms
• 8-1, 8-2 Concept-Development Practice Page

Zaiza to scribe next.