Question

Help me plzzzzzzzzzzzzzzzzzzz!!!!

Answer 1

@tootzrll

Answer 2

a= Vf/(Vi X t)

Answer 3

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

Answer 4

i got 4.4m/sec^2

Answer 5

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

Answer 6

what grade is this?

Answer 7

11U

Answer 8

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

Answer 9

-___________________-

Answer 10

:/// im so sorry

Answer 11

@MAEMAEHOCKEY

Answer 12

@Michele_Laino

Answer 13

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

Answer 14

@CShrix yes i got 4.4m/sec^2

Answer 15

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

Answer 16

i got 79.4m

Answer 17

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

Answer 18

@CShrix

Answer 19

\[\huge x=v_0t+\frac{ 1 }{ 2 }at^2 \implies x=\frac{ 1 }{ 2 }at^2\]might work for this. I'm honestly not that great at kinematics. Let's check with @IrishBoy123 or @Michele_Laino

Answer 20

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

Answer 21

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

Answer 22

is it P = W/time????

Answer 23

\[\huge P=\frac{ \text{Work} }{ \text{Time} }=\frac{ Fd \cos(\theta) }{ t }\]
Where
\[\huge F=ma\]
Additionally, since the force and displacement vectors are in the same direction, we can ignore the trig since \(\cos(0)=1\)

Answer 24

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

Answer 25

In other words, mechanical power

Answer 26

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

Answer 27

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

Answer 28

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

Answer 29

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

Answer 30

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

Answer 31

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

Answer 32

yah buddy, its potential......

Answer 33

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

Answer 34

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

Answer 35

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

Answer 36

\[\huge P(t)=\vec{F} \cdot \vec{v}=|\vec{F}||\vec{v}| \cos(\theta)\]
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.

Answer 37

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

Answer 38

X) Ciao!