Question

calculate the kinetic energy of an 8.0x10^4 airliner flying at 600km/h

Answer 1

Do you remember the formula for kinetic energy of an object? ^_^

Answer 2

@noobslyr101 ?

Answer 3

KE=1/2MV^2?

Answer 4

@AllTehMaffs

Answer 5

yup yup! And you have both mass and velocity - the only problem now is converting your velocity into m/s from km/h in order to get units of joules.

\[ \left( \frac{600 \ km}{h} \right) \left( \frac{1000 m}{1 km} \right) \left( \frac{1 h}{ 60 min} \right) \left( \frac{1 min}{60 s} \right) \]

Answer 6

so
v=?

Answer 7

v=60x60x600?

Answer 8

close ^_^

v = ((600•1000)\(3600)) m/s

Answer 9

ooohhh so i have to change to kilometers to meters then the seconds

Answer 10

so the bottom is 60*60 and the top is 600*1000 cause im converting kilomoters to meters
!

Answer 11

Correct!! ^_^

Answer 12

so i got 166.6

Answer 13

m/s

Answer 14

yup yup yup ^_^

Also notice that the order itself doesn't matter, just the fact that you're canceling out km and hours (and minutes in between seconds)

\[ \left( \frac{600 \ \cancel{km}}{cancel{h}} \right) \left( \frac{1000 m}{1 \cancel{km}} \right) \left( \frac{1 \cancel{h}}{ 60 \cancel{min}} \right) \left( \frac{1 \cancel{min}}{60 s} \right) \]
=
\[ \left( \frac{600 \cancel{km}}{\cancel{h}} \right) \left( \frac{1 \cancel{h}}{ 60 \cancel{min}} \right) \left( \frac{1 \cancel{min}}{60 s} \right) \left( \frac{1000 m}{1 \cancel{km}} \right) \]
= 166.67 m/s

Answer 15

OMG thank you soo much so all i have to do left is 166.67^2*(80000/2)

Answer 16

so i get about 1.1 x 10^9

Answer 17

Yeah, you got it! ^_^

Answer 18

And that's with units of Joules

Answer 19

Jeeze you are the best! I cant believe im learning from this site! I have a huge test tomorrow on work and energy
My only problem with physics is taking out the word problem and changing it into an equation. Picturing the problem is one of my main problem with physics, but once i can draw the picture, the math part comes easy to me

Answer 20

Thanks ^_^ And that's a good approach - drawing pictures is definitely the best way to approach pretty much every physics problem I think. Energy is difficult in that regard because it's often just a static piece of information - but in those instances, often remembering the formula and comparing your knowns and unknowns to the variables in it (and looking to see what units the answer needs to have)is the best way to chip away towards an answer. ^_^ Good luck studying!!

Answer 21

could you help me out with 1 more question?

Answer 22

of course!

Answer 23

two 3g bullets are fired with velocities of 40m/s and 80m/s respectively/ what are their kinetic energies?

What i did was \[KE=\frac{ 1 }{ 2 } 0.003*40\]

Answer 24

omg i forgot to square

Answer 25

ahh i solved my own question hahaha i keep leaving out that detail in the kinetic equation

Answer 26

Other than the squaring, that's exactly right ^_^ - the one traveling at 80m/s should have 4 times the kinetic energy of the slower one!!

Answer 27

And solving your own question just means that you understand it well after forming into words the thing you're unsure of ^_^ it's always a good thing!

Answer 28

how long will you be on for? cause id like to ask some more questions later if im suck if that is ok with you.

Answer 29

For sure - I'll be on for a while, so if you need something just tag me in the question or message me (although the messages haven't been working on my end for a while, so tagging might yield better results).

Answer 30

alright thanks alot!!

Answer 31

Welcome, and good luck studyin' ! ^_^

Answer 32

@AllTehMaffs A 40kg child is in a swing that is attached to ropes 2.00 m long. find the PEg associated with the child relative to the child's lowest position under the following conditions. When the rope makes a 30 degree angle with the vertical

Answer 33

PEg=785j

Answer 34

work (like, show work, not what's the work ^_^)?

Answer 35

ok umm

Answer 36

40(9.81)2

Answer 37

2m is the rope. 9.81 force gravity and 40 is my mass

Answer 38

So that's the Potential energy of the system when the rope is parallel to the ground.