What Are the Units and Exponents in Basic Physics Equations?

AI Thread Summary
Understanding units and exponents in physics equations is crucial for grasping the subject. For the equation v = at^2 + bt^3, the units of a must be m/s^3 and b must be m/s^4 to ensure dimensional consistency. In the second equation A = B^n * C^m, the dimensions imply that n should equal 1 and m should equal 1 to maintain dimensional balance. It's important to remember that the units on both sides of an equation must match, and while you can multiply or divide different units, you can only add or subtract like units. Mastering these concepts is essential for solving basic physics problems effectively.
Catalyst333
Messages
2
Reaction score
0
I am very embarrassed to ask these questions because compared to other questions on these boards, these two seem very basic. Nonetheless, I have never taking physics and I am having a hard time grasping its concepts. So here are two simple questions that maybe someone could answer and explain for me. Thanks!

1st Q: During a short interval of time the speed v in m/s of an automobile is given by v = at^2 + bt^3, where the time t is in seconds. The units of a and b should be?

2nd Q: Suppose A = B^n*C^m, where A has dimensions LT, B has dimensions L^2T^–1, and C has dimensions LT^2. Then the exponents n and m have the values:
 
Physics news on Phys.org
You need to show your work before we can help-- forum rules. What do you know about the dimensions of an equation?
 
To be honest, I have no idea where to start with these two questions. I was hoping someone could get me started with either one. The only thing I know about dimensions is that acceleration is Length/Time^2 and speed is Length/time. My teacher zipped through this part and the book does not seem to talk about it. In other words, i don't know much about dimensions :(
 
The units on both sides of an equation must be the same.
You can multiply/divide different units but you can only add/subtract the same units ( a bit like canceling fractions).

so, area (m^2) = length(m) * width(m)
or, distance(m) = speed(m/s)*time(s) + 1/2 acceleration(m/s^2)*time(s)^2
 
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 '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