Calculating Velocity from Kinetic Energy: A Homework Problem Solution"

AI Thread Summary
To calculate the velocity of a car with a mass of 1500 kg and a kinetic energy of 1,400,000 Joules, the equation K.E. = 1/2 x m x v² is used. Rearranging the equation to solve for velocity (v) is necessary. The correct approach involves isolating v by multiplying both sides by 2 and then dividing by the mass. This leads to the formula v = √(2 * K.E. / m). Ultimately, applying these steps will yield the car's velocity.
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Homework Statement


A car witha mass of 1500 kg has a Kinetic Energy of 1,400,000 Joules. What is its velocity?


Homework Equations


I would assume that (K.E.=1/2 x m x v 2) could be used.


The Attempt at a Solution



I am not exactly sure how to attempt this question. Do we try to make velocity by itself in the above equation? I am not quite sure how to go about that...
 
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Yes, that's how you solve for v, isn't it?
 

Thread 'Errors and significant figures'

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...
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Thread 'A cylinder connected to a hanging mass'

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...
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