Answer:
Explanation:
Given
mass of sled =26 kg
coefficient of static friction 
coefficient of kinetic friction 
In order to move sled from rest we need to provide a force greater than static friction which is given by
After Moving Sled kinetic friction comes in to play which is less than static friction
therefore minimum force to keep moving sledge at constant velocity is 18.34 N
Answer:
The answer are given above in attachment.
Answer:
1 N
Explanation:
First the equation is momentum = Force / distance
20 cm = 0.2 m
5 N/m = F / 0.2 m
F = 1 N
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The moment of the resultant of these two forces with respect to O 376 lb-ft CCW which is <span>about moment center point O.</span>
Explanation:
The expression is :
A =[LT], B=[L²T⁻¹], C=[LT²]
Using dimensional of A, B and C in above formula. So,
![A=B^nC^m\\\\\ [LT]=[L^2T^{-1}]^n[LT^2}]^m\\\\\ [LT]=L^{2n}T^{-n}L^mT^{2m}\\\\\ [LT]=L^{2n+m}T^{2m-n}](https://tex.z-dn.net/?f=A%3DB%5EnC%5Em%5C%5C%5C%5C%5C%20%5BLT%5D%3D%5BL%5E2T%5E%7B-1%7D%5D%5En%5BLT%5E2%7D%5D%5Em%5C%5C%5C%5C%5C%20%5BLT%5D%3DL%5E%7B2n%7DT%5E%7B-n%7DL%5EmT%5E%7B2m%7D%5C%5C%5C%5C%5C%20%5BLT%5D%3DL%5E%7B2n%2Bm%7DT%5E%7B2m-n%7D)
Comparing the powers both sides,
2n+m=1 ...(1)
2m-n=1 ...(2)
Now, solving equation (1) and (2) we get :
Hence, the correct option is (E).
