How many seconds does it take for a car to deliver 4299.1N?

  • Thread starter Thread starter IDKPhysics101
  • Start date Start date
  • Tags Tags
    Car Seconds
AI Thread Summary
A 765.1 kg car accelerates from rest to 20.9 m/s under a force of 4299.1 N. The acceleration is calculated using the formula a = f/m, leading to a time calculation t = m * v / f. Substituting the values, the time taken for the car to reach the speed is determined to be approximately 3.71 seconds. This result is confirmed by participants in the discussion, who express appreciation for the solution. The conversation highlights the application of physics principles in calculating acceleration and time.
IDKPhysics101
Messages
31
Reaction score
0
A 765.1 kg car accelerates from rest to 20.9 m/s. If the engine can deliver 4299.1 N of force, how many seconds does it take the car to do this?
 
Physics news on Phys.org
v = a * t;

we know a = f/m;

t = m * v/f
 
t=m*v/f
t=765.1*20.9/4299.1
t=3.71seconds?
 
Yes, that's right.
 
awesome, thanks alot! i promote you and rock freak to be our class's teacher
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top