Recent content by L_0611

Homework Statement A bullet is shot vertically into the air at a speed of 512 m/s. a) to what maximum height does the bullet go? b) how much time passes before the bullet stops rising? c) what is the velocity of the bullet after 60s? Homework Equations a) V2²=V1²+2ad b) a=...

yeah I wasn't sure how to write them on the computer, basically I am following my book. Fg - force of gravity FgI - Fg(cosθ) FgII = Fg(sinθ) Fn - normal force Fnet - net force Fk - force of kinetic friction m - mass a - acceleration g - gravity μk - coefficient of kinetic friction please let me...

Homework Statement A child is tobogganing down a hillside. The child and the toboggan together have a mass of 50.0kg. The slope is at an angle of 30.0° to the horizontal. Assume that the positive y-direction is pointing in the direction of the normal force. Assume that the positive...

Alright then, funny how sometimes I can miss the simplest things, thanks a lot.

ok so that's basically all I have to do? put it into words...?

ok perfect, thank you. As part of my solution I'm asked to find the x and y components but I'm not sure how to do so, can anyone help me? they want me to use these five steps to solve the problem, the steps are as follows: Step 1: select the object to be studied. Step 2: draw a "free-body...

ok so I made a drawing like ehild recommended and I think I figured out angle θ. This is what I got for an answer. tanθ = 5/0.2 = 25 θ = tan-1*25 = 87.7 then then formula I used to solve the problem would be 2Tcos87.7=mg 2Tcos87.7=40*9.8 then i rearranged the formula to read...

Homework Statement A clothesline is attached to two fixed ends which are 10.0m apart. A pulley of mass 40.0kg hangs freely in the middle of the line. The sag at the centre is 0.20m. Find the tension in each half of the clothesline. Homework Equations The Attempt at a Solution...

Homework Statement A ferryboat, whose speed in still water is 4.00m/s, must cross a river whose current is 3.00m/s. The river runs from west to east and is 128m wide. The boat is pointed north. a) If the boat does not compensate for the flow of the river water and allows itself to be...

so I just did my homework question and this are the answers I got: Part a: V2²=V1²+2ad a=(V2²-V1²)/2d a=(2.0x10^7²m/s-0m/s²)/2(0.10m) a=(4x10^14m/s)/0.20m a= 2x10^15m/s² therefore the electron's acceleration is 2x10^15m/s² Part b: a=ΔV/Δt Δt=ΔV/a Δt=(2.0x10^7m/s-0m/s)/(2x10^15m/s²)...

ok perfect, thanks for the explanation, I think I get it now. I really appreciate you taking your time to help me out with this question and obviously I was right out to lunch, I am taking this course through correspondence so I don`t have the privilege of asking a teacher, so I really...

so are you saying that time should equal change in velocity divided by change in distance, not acceleration? but because both my initial velocity and initial distance is 0, I can just use time= velocity/acceleration? I think I'm somewhat starting to understand it, however still very confused.

wait a second I just redid that and realized it should be t=V/a not t=aV. Thanks for making me realize my mistake.

wait what? I thought the original formula is a=V/t and if I rearrange it to solve for t wouldn't I end up with time=acceleration x velocity. If that is wrong like you are saying haruspex, can you please explain why, thank you.

ok perfect, I re did it using the same formula and squaring both V1 and V2. Thanks for your help, I really appreciate it.