Answer:
Number of revolution made by tire is 1.57 x 10⁷
Explanation:
Radius of tire, r = 0.220 m
Circumference of tire, C = 2πr
Substitute the value of r in the above equation.
C = 2 x π x 0.220 m = 1.38 m
Total distance covered by tire in a year, D = 13500 miles
But 1 mile = 1609.34 m
So, D = 13500 x 1609.34 m
Number of revolutions take by tire, N = ![\frac{D}{C}](https://tex.z-dn.net/?f=%5Cfrac%7BD%7D%7BC%7D)
![N=\frac{13500\times1609.34}{1.38}](https://tex.z-dn.net/?f=N%3D%5Cfrac%7B13500%5Ctimes1609.34%7D%7B1.38%7D)
N = 15743543
For this problem the figure below shows the representation of a student who pulls on a 20kg box. We know this variables:
Weight of the box = 20kg
Force used by the student to pull on the box = 50N (This is the tension T)
Angle relative to the horizontal = 45 degrees
Aceleration of the box =
![1.5m/s^{2}](https://tex.z-dn.net/?f=1.5m%2Fs%5E%7B2%7D%20)
The figure also shows the Free-Body diagram, Applying Newton's Second Law we can find the equation for this diagram, related to the x-axis as:
![Tcos(45)-f_{k}=ma_{x}](https://tex.z-dn.net/?f=Tcos%2845%29-f_%7Bk%7D%3Dma_%7Bx%7D)
Isolating
![f_{k}](https://tex.z-dn.net/?f=f_%7Bk%7D)
:
![f_{k}=Tcos(45)-ma_{x} = 50cos(45)-20(1.5)=5.355N](https://tex.z-dn.net/?f=f_%7Bk%7D%3DTcos%2845%29-ma_%7Bx%7D%20%3D%2050cos%2845%29-20%281.5%29%3D5.355N)
<span>That is the friction force on the box.</span>
Explanation:
to provide a stable base load of energy