YumYum247
  • YumYum247
Help me plzzzzzzzzzzzzzzzzzzz!!!!
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
YumYum247
  • YumYum247
@tootzrll
YumYum247
  • YumYum247
a= Vf/(Vi X t)
tootzrll
  • tootzrll
i would say a. that's pretty much what theyre talking about, so i choose a.

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More answers

YumYum247
  • YumYum247
i got 4.4m/sec^2
YumYum247
  • YumYum247
does it make sense??!?!?!?!
tootzrll
  • tootzrll
what grade is this?
YumYum247
  • YumYum247
11U
tootzrll
  • tootzrll
11? okay, um im in 8th so idk how well thisll go over xDDD
YumYum247
  • YumYum247
-___________________-
tootzrll
  • tootzrll
:/// im so sorry
YumYum247
  • YumYum247
@MAEMAEHOCKEY
YumYum247
  • YumYum247
@Michele_Laino
anonymous
  • anonymous
a). \[\huge a_\text{average}=\frac{\Delta v}{t}\]
YumYum247
  • YumYum247
@CShrix yes i got 4.4m/sec^2
YumYum247
  • YumYum247
how about "B"? how faar will it travel?
YumYum247
  • YumYum247
i got 79.4m
YumYum247
  • YumYum247
for "B" i used the following formula.....d = Vf - Vi/2 X time
YumYum247
  • YumYum247
@CShrix
anonymous
  • anonymous
\[\huge x=v_0t+\frac{ 1 }{ 2 }at^2 \implies x=\frac{ 1 }{ 2 }at^2\]might work for this. I'm honestly not that great at kinematics. Let's check with @IrishBoy123 or @Michele_Laino
YumYum247
  • YumYum247
i've done this question multiple ways, and i'm getting the same answer!!!! so i assume it's right.
YumYum247
  • YumYum247
What about the last question...... wat kind of "POWER" was developing in this process!!!
YumYum247
  • YumYum247
is it P = W/time????
anonymous
  • anonymous
\[\huge P=\frac{ \text{Work} }{ \text{Time} }=\frac{ Fd \cos(\theta) }{ t }\] Where \[\huge F=ma\] Additionally, since the force and displacement vectors are in the same direction, we can ignore the trig since \(\cos(0)=1\)
anonymous
  • anonymous
It is probably a build up of kinetic energy over a time period of 6 seconds
anonymous
  • anonymous
In other words, mechanical power
YumYum247
  • YumYum247
@q12157 yah but in joules.......
anonymous
  • anonymous
Yes kinetic energy is in joules, work is in joules. P=W/t
YumYum247
  • YumYum247
and it'll be a potential energy right?????????????????
YumYum247
  • YumYum247
since the car is constantly pushing for a period of 6 sec
anonymous
  • anonymous
Depends. Maybe the car uses gasoline, maybe the car uses electricity.
anonymous
  • anonymous
Whatever it is using, it is transferring chemical/electrical energy to kinetic.
YumYum247
  • YumYum247
yah buddy, its potential......
anonymous
  • anonymous
So much sass LOL, but yeah its asking for an identification of power...which will probably be mechanical power.
YumYum247
  • YumYum247
mechinical power is both the potential and kinetic energy right?????????
anonymous
  • anonymous
Potential and kinetic are not the only types of energy there are... so not necessarily, but if you insist...sure.
anonymous
  • anonymous
\[\huge P(t)=\vec{F} \cdot \vec{v}=|\vec{F}||\vec{v}| \cos(\theta)\] Again, we know that the angle in between is 0, so we can ignore the trig. We can also drop the magnitude symbols because we know that the values that we're calculating are already in magnitude. So we're just left with P=fv. Specifically this is mechanical power.
YumYum247
  • YumYum247
ok guys thanks alot, Love you guys to pieces. Stay cute and beast as always....Chao.
anonymous
  • anonymous
X) Ciao!

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