Answer: B. Concrete
Explanation:
Let N = reacting force pressing the bodies in context together (units in Newtons),
The question stated that the force pressing the two mounted/stacked objects together is equal to the weight of the object on top.
We need to start by finding the weight of the piece of wood.
friction is given by
f = μN
The value of f is 22.5,
and from the chart reference the coefficient of friction between wood and stone, μ is 0.30.
22.5 = 75. 0.30
Putting the values into the equation: 22.5 = 0.30N.
Divide both sides by 0.30 to find the value of N:
N= 22.5/0.3 = 75
Now that the piece of wood will be placed on another surface, its weight of 75 Newton is the force pressing the two bodies together.
To determine the new surface, you should find the new coefficient of friction by using the new value of the force of friction given 46.5:
46.5 = µ(75).
Divide both sides by 75 to isolate μ.
The refer chart also indicates that the coefficient of friction equals 0.62 between wood and concrete, so the new surface corresponding to 0.62 is the concrete, which is (B).
Answer:
c. The coefficient of kinetic friction is less than the coefficient of static friction
Explanation:
When the box finally does break loose. Then the component of the box weight which is parallel to the board weight parallel component, is equal to the
.
![w_{box}=\mu_gn](https://tex.z-dn.net/?f=w_%7Bbox%7D%3D%5Cmu_gn)
For the box to acce;erate thee must be non-zero net force acting on the box parallel to the board. Or we can say,
![w_1>f_g\\w_1>\mu_gn](https://tex.z-dn.net/?f=w_1%3Ef_g%5C%5Cw_1%3E%5Cmu_gn)
Therefore the force of kinetic friction must be less than the force of static friction. Thus,
![\mu_g](https://tex.z-dn.net/?f=%5Cmu_g%3C%5Cmu_s)
Answer:
Yes,earth has an electric field
ICD-10-CM codes are -S02.2XXA, W21.01XA, Y93.61, Y92.830
S02.2 for Fracture, Traumatic/Nasal (Bone(s)), ICD-10-CM Alphabetic Index. Both the open fracture code and the dislocation code are not reported. Only the fracture code is provided if a fracture and a dislocation happen at the same place. Search for "dislocation/with fracture" in the alphabetical index to be sent to a doctor. A closed fracture is a fracture with displacement. To report the conditions leading up to the injury, external cause codes are utilized. Look for Struck (accidentally) by/ball (struck) (thrown)/football W21.01 in the ICD-10-CM External Cause of Injuries Index. Seven characters are required in the Tabular List to finish the code. For the first encounter, X is utilized as a stand-in for character number six, and character number seven is given the letter A.
To learn more about ICD-10-CM please visit -brainly.com/question/27932590
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Answer:
5.9*10^{-4}m
Explanation:
to find the uncertainty of the displacement it is necessary to compute the uncertainty for the angular frequency:
![\omega=\frac{2\pi}{T}=\frac{2\pi}{0.40s}=15.707rad/s\\\\\frac{d \omega}{\omega}=\frac{dT}{T}\\\\d\omega=\omega \frac{dT}{T}=(15.707rad/s)\frac{0.020s}{0.40s}=0.785rad/s](https://tex.z-dn.net/?f=%5Comega%3D%5Cfrac%7B2%5Cpi%7D%7BT%7D%3D%5Cfrac%7B2%5Cpi%7D%7B0.40s%7D%3D15.707rad%2Fs%5C%5C%5C%5C%5Cfrac%7Bd%20%5Comega%7D%7B%5Comega%7D%3D%5Cfrac%7BdT%7D%7BT%7D%5C%5C%5C%5Cd%5Comega%3D%5Comega%20%5Cfrac%7BdT%7D%7BT%7D%3D%2815.707rad%2Fs%29%5Cfrac%7B0.020s%7D%7B0.40s%7D%3D0.785rad%2Fs)
then, you can calculate the uncertainty in angular displacement:
![\theta=\omega t\\\\\frac{d\theta}{\theta}=\sqrt{(\frac{d\omega}{\omega})^2+(\frac{dt}{t})^2}\\\\d\theta=\theta\sqrt{(\frac{d\omega}{\omega})^2+(\frac{dt}{t})^2}=0.0422](https://tex.z-dn.net/?f=%5Ctheta%3D%5Comega%20t%5C%5C%5C%5C%5Cfrac%7Bd%5Ctheta%7D%7B%5Ctheta%7D%3D%5Csqrt%7B%28%5Cfrac%7Bd%5Comega%7D%7B%5Comega%7D%29%5E2%2B%28%5Cfrac%7Bdt%7D%7Bt%7D%29%5E2%7D%5C%5C%5C%5Cd%5Ctheta%3D%5Ctheta%5Csqrt%7B%28%5Cfrac%7Bd%5Comega%7D%7B%5Comega%7D%29%5E2%2B%28%5Cfrac%7Bdt%7D%7Bt%7D%29%5E2%7D%3D0.0422)
finally, by using:
![y=Acos(\omega t)\\\\dy=dAcos(\omega t)d(\omega t)=(dA)cos(\theta)d\theta=(0.002m)cos(0.785)(0.0422)\\\\dy=5.9*10^{-4}m](https://tex.z-dn.net/?f=y%3DAcos%28%5Comega%20t%29%5C%5C%5C%5Cdy%3DdAcos%28%5Comega%20t%29d%28%5Comega%20t%29%3D%28dA%29cos%28%5Ctheta%29d%5Ctheta%3D%280.002m%29cos%280.785%29%280.0422%29%5C%5C%5C%5Cdy%3D5.9%2A10%5E%7B-4%7Dm)