By this equation: d = 4(32), we know what d equals, which means we have the value of d.
d= 4*32
d = 128
Answer: Decrease
Step-by-step explanation:
Answer:
a) ![v = \frac{[L]}{[T]} = LT^{-1}](https://tex.z-dn.net/?f=%20v%20%3D%20%5Cfrac%7B%5BL%5D%7D%7B%5BT%5D%7D%20%3D%20LT%5E%7B-1%7D)
b) ![a = \frac{[L}{T}^{-1}]}{{T}}= L T^{-1} T^{-1}= L T^{-2}](https://tex.z-dn.net/?f=%20a%20%3D%20%5Cfrac%7B%5BL%7D%7BT%7D%5E%7B-1%7D%5D%7D%7B%7BT%7D%7D%3D%20L%20T%5E%7B-1%7D%20T%5E%7B-1%7D%3D%20L%20T%5E%7B-2%7D)
c) ![\int v dt = s(t) = [L]=L](https://tex.z-dn.net/?f=%20%5Cint%20v%20dt%20%3D%20s%28t%29%20%3D%20%5BL%5D%3DL)
d) ![\int a dt = v(t) = [L][T]^{-1}=LT^{-1}](https://tex.z-dn.net/?f=%20%5Cint%20a%20dt%20%3D%20v%28t%29%20%3D%20%5BL%5D%5BT%5D%5E%7B-1%7D%3DLT%5E%7B-1%7D)
e) ![\frac{da}{dt}= \frac{[L][T]^{-2}}{T} = [L][T]^{-2} [T]^{-1} = LT^{-3}](https://tex.z-dn.net/?f=%20%5Cfrac%7Bda%7D%7Bdt%7D%3D%20%5Cfrac%7B%5BL%5D%5BT%5D%5E%7B-2%7D%7D%7BT%7D%20%3D%20%5BL%5D%5BT%5D%5E%7B-2%7D%20%5BT%5D%5E%7B-1%7D%20%3D%20LT%5E%7B-3%7D)
Step-by-step explanation:
Let define some notation:
[L]= represent longitude , [T] =represent time
And we have defined:
s(t) a position function


Part a
If we do the dimensional analysis for v we got:
![v = \frac{[L]}{[T]} = LT^{-1}](https://tex.z-dn.net/?f=%20v%20%3D%20%5Cfrac%7B%5BL%5D%7D%7B%5BT%5D%7D%20%3D%20LT%5E%7B-1%7D)
Part b
For the acceleration we can use the result obtained from part a and we got:
![a = \frac{[L}{T}^{-1}]}{{T}}= L T^{-1} T^{-1}= L T^{-2}](https://tex.z-dn.net/?f=%20a%20%3D%20%5Cfrac%7B%5BL%7D%7BT%7D%5E%7B-1%7D%5D%7D%7B%7BT%7D%7D%3D%20L%20T%5E%7B-1%7D%20T%5E%7B-1%7D%3D%20L%20T%5E%7B-2%7D)
Part c
From definition if we do the integral of the velocity respect to t we got the position:

And the dimensional analysis for the position is:
![\int v dt = s(t) = [L]=L](https://tex.z-dn.net/?f=%20%5Cint%20v%20dt%20%3D%20s%28t%29%20%3D%20%5BL%5D%3DL)
Part d
The integral for the acceleration respect to the time is the velocity:

And the dimensional analysis for the position is:
![\int a dt = v(t) = [L][T]^{-1}=LT^{-1}](https://tex.z-dn.net/?f=%20%5Cint%20a%20dt%20%3D%20v%28t%29%20%3D%20%5BL%5D%5BT%5D%5E%7B-1%7D%3DLT%5E%7B-1%7D)
Part e
If we take the derivate respect to the acceleration and we want to find the dimensional analysis for this case we got:
![\frac{da}{dt}= \frac{[L][T]^{-2}}{T} = [L][T]^{-2} [T]^{-1} = LT^{-3}](https://tex.z-dn.net/?f=%20%5Cfrac%7Bda%7D%7Bdt%7D%3D%20%5Cfrac%7B%5BL%5D%5BT%5D%5E%7B-2%7D%7D%7BT%7D%20%3D%20%5BL%5D%5BT%5D%5E%7B-2%7D%20%5BT%5D%5E%7B-1%7D%20%3D%20LT%5E%7B-3%7D)
Answer:
Step-by-step explanation:
First you need to write it down. Now, 1st step of the math, how many times will 7 go into 4? 0 times. So you have to do how many times 7 goes to 44. It goes in 6 times bc 7 times 7 is 49. Now, subtract 42 form 44 which is 2. Bring down the 2, you get 22. 7 goes into 22 3 times, subtract 21 from 22. One. So your answer is 63 remainder one!