Recent content by Butterfly30

Oh, but do u enter my (a) as negative since the car is slowing down?

A 2319kg car is moving down a road with a slope grade of 11% and slowing down at a rate of 3.8m/s^2.Find the direction and magnitude of the frictional force ( define positive in the forward direction ie down the slope) So the equation I have is f= ma+ mgsinθ. For my angle I get...

So basically given this problem: a 1892kg car is moving down a road with slope (grade) of 12% while slowing down at a rate of 3.7m/s^2 .What is the direction and magnitude of the frictional force?(define positive in the forward direction) Would my force be the tangent of the angle 12...

Sorry these were the numbers, I accidently mixed two probs together:/ a 1892kg car is moving down a road with slope (grade) of 12% while slowing down at a rate of 3.7m/s^2 .What is the direction and magnitude of the frictional force?(define positive in the forward direction) Using...

a 1000 kg car is moving down a road with slope (grade) of 15% while slowing down at a rate of 3.7m/s^2 .What is the direction and magnitude of the frictional force?(define positive in the forward direction) Using f=ma I have the mass 1000 kg, the acceleration 3.7m/s^2, and slope 15%...

Would it be equal to the mass?

Zero? The frictional force would be zero. So, then the grade (the incline) was given just to make us think and try to picture it, I guess. Ok, how about if we're given the grade and told that the the car was speeding up at 3m/s^2 I know the mass (1000kg) and I think theta would be the...

fictional force problem...given a slope(grade)? A 1000kg car is moving down a road with a slope(grade) of 15% at a constant speed of 15m/s. What is the direction and magnitude of the frictional force? So... V= 15m/s a= 0 (constant) The slope really throws me off..I don't know where...

Yay, actually I got it! Thanks sobmuch!

Yes this is what I had, but I mentioned the #'s changed when i entered my answer so 7.25 became 5.16. It is late, thank you for your help... I wouldn't have gotten this far so thank u and good night :)

Ok, the #'s change each time but same question so, given y0=4ft(1.21m) v0y=5.16m/s v0x=2.41. Quad eqtn 1.21+5.16t+(-4.91)t^2 End up with -5.16+-sqrt50.3/2(-4.9) T=1.25. So when I plug this into the x eqtn x=v0t. X=2.41•1.25. X= 3.01m I don't know I this is right, once I enter it...

Hmmm... I'm still getting it wrong. When I put into the quadratic it's 1.21m + 7.25 t + -4.9t2. Btw it's 1.21m from the 4ft given... I get a time that is a decimal so when I put it into the x eqtn it's very low... I'm stuck

So the quadratic eqtn would give me the time it would take ... So I would then have to plug this into the eqtn x=x0+v0t+1/2at^2 but it would become x=x0+v0t since a for x equals zero... Right?

Oops yes it is thrown from a height of 4ft. So Would I still make my y and y0 = 0 and them handle the 4 ft afterwards?

I'm still confused...if I try to solve for t what is my change in y? It says 4ft from the ground...but what about what it travels when it's is thrown up? Also wouldn't I have to make a= 9.81?