Help me plzzzzzzzzzzzzzzzzzzz!!!!

- YumYum247

Help me plzzzzzzzzzzzzzzzzzzz!!!!

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- YumYum247

@tootzrll

- YumYum247

a= Vf/(Vi X t)

- tootzrll

i would say a. that's pretty much what theyre talking about, so i choose a.

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## More answers

- YumYum247

i got 4.4m/sec^2

- YumYum247

does it make sense??!?!?!?!

- tootzrll

what grade is this?

- YumYum247

11U

- tootzrll

11? okay, um im in 8th so idk how well thisll go over xDDD

- YumYum247

-___________________-

- tootzrll

:/// im so sorry

- YumYum247

@MAEMAEHOCKEY

- YumYum247

@Michele_Laino

- anonymous

a). \[\huge a_\text{average}=\frac{\Delta v}{t}\]

- YumYum247

@CShrix yes i got 4.4m/sec^2

- YumYum247

how about "B"? how faar will it travel?

- YumYum247

i got 79.4m

- YumYum247

for "B" i used the following formula.....d = Vf - Vi/2 X time

- YumYum247

@CShrix

- 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

i've done this question multiple ways, and i'm getting the same answer!!!! so i assume it's right.

- YumYum247

What about the last question...... wat kind of "POWER" was developing in this process!!!

- YumYum247

is it P = W/time????

- 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

It is probably a build up of kinetic energy over a time period of 6 seconds

- anonymous

In other words, mechanical power

- YumYum247

@q12157 yah but in joules.......

- anonymous

Yes kinetic energy is in joules, work is in joules. P=W/t

- YumYum247

and it'll be a potential energy right?????????????????

- YumYum247

since the car is constantly pushing for a period of 6 sec

- anonymous

Depends. Maybe the car uses gasoline, maybe the car uses electricity.

- anonymous

Whatever it is using, it is transferring chemical/electrical energy to kinetic.

- YumYum247

yah buddy, its potential......

- anonymous

So much sass LOL, but yeah its asking for an identification of power...which will probably be mechanical power.

- YumYum247

mechinical power is both the potential and kinetic energy right?????????

- anonymous

Potential and kinetic are not the only types of energy there are... so not necessarily, but if you insist...sure.

- 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

ok guys thanks alot, Love you guys to pieces. Stay cute and beast as always....Chao.

- anonymous

X) Ciao!

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