Recent content by kza62

so Ff = 0.29 x 9.8 = 2.842 and then Fnet = 1400 - 2.842 = 1397.158 a = 1397.158/285 = 4.902308... ? what did i do wrong now?!

Homework Statement A car and driver weighing 5490 N passes a sign stating “Bridge Out 29.3 m Ahead.” She slams on the brakes, and the car decelerates at a constant rate of 10.9 m/s2. The acceleration of gravity is 9.8 m/s2 . What is the magnitude of the work done stopping the car if the...

oooo ok so KE = 1/2mv^2 alright now i can finish this thanks!

THAAATTSS what i needed ooohhh maaann thanks rl.bhat! so it would be (0.5)(43.5)(2.08333)^2 = 45.3124 J thanks again!

ooooo okokokok but how do u get Fn?? that's the only thing i need to plug in everything else :uhh:

1. Homework Statement Shawn and his bike have a total mass of 43.5 kg. Shawn rides his bike 1.3 km in 10.4 min at a constant velocity. The acceleration of gravity is 9.8 m/s^2 What is Shawn's kinetic energy? 2. Homework Equations F = ma Ff = u * Fn Fn = mg avgV = x/t x = 1/2(Vf +...

hey kcc aight so when ur doing projectile motion try using this equation to find t: t = √(2y/g) and now that u have t u can use the formula V=x/t to find x or arrange it so that its x=vt :D et voila'

not really dude, its moving horizontally, its not falling

Homework Statement Shawn and his bike have a total mass of 43.5 kg. Shawn rides his bike 1.3 km in 10.4 min at a constant velocity. The acceleration of gravity is 9.8 m/s^2 What is Shawn's kinetic energy? Homework Equations F = ma Ff = u * Fn Fn = mg avgV = x/t x = 1/2(Vf + Vi)t...

WOOHOO! thanks for ur help! i love physics when I get it :D

Homework Statement A cart with a mass of 285 kg is pushed horizontally with a force of 1400 N. The acceleration of gravity is 9.8 m/s/s. The coefficient of friction (u) with the crate and the surface is 0.29. Calculate the acceleration of the crate. Sooo given: Fa: 1400 N m: 285 kg...

*** vf^2 - vi^2 / 2a = x and a = -1.176

ok, i think i figured it out but I'm not sure how let me try to break it down again so if Fgravity is -9.8 then the Fnormal is 9.8? so is Ffriction = u * Fnormal? which is 1.176 is a? and then i used vf - vi / 2a which gave me 140.689 did i get it or was this a lucky guess?

A car is traveling at 40.7 mi/h on a horizontal highway. The acceleration of gravity is 9.8 m/s/s. If the coefficient of friction between road and tires on a rainy day is 0.12, what is the minimum distance in which the car will stop? (1 mi = 1.609 km) So given: vi(initial velocity): 40.7...