Explanation:
the object will begin to move
Answer: µ=0.205
Explanation:
The horizontal forces acting on the ladder are the friction(f) at the floor and the normal force (Fw) at the wall. For horizontal equilibrium,
f=Fw
The sum of the moments about the base of the ladder Is 0
ΣM = 0 = Fw*L*sin74.3º - (25.8kg*(L/2) + 67.08kg*0.82L)*cos74.3º*9.8m/s²
Note that it doesn't matter WHAT the length of the ladder is -- it cancels.
Solve this for Fw.
0= 0.9637FwL - (67.91L)2.652
Fw=180.1/0.9637
Fw=186.87N
f=186.81N
Since Fw=f
We know Fw, so we know f.
But f = µ*Fn
where Fn is the normal force at the floor --
Fn = (25.8 + 67.08)kg * 9.8m/s² =
910.22N
so
µ = f / Fn
186.81/910.22
µ= 0.205
From the information given,
diameter of ornament = 8
radius = diameter/2 = 8/2
radius of curvature, r = 4
Recall,
focal length, f = radius of curvature/2 = 4/2
f = 2
Recall,
magnification = image d
Answer:
m³/(kg⋅s²)
Explanation:
Hello.
In this case, since the involved formula is:
By writing a dimensional analysis with the proper algebra handling, we obtain:
![N[=]G*\frac{kg*kg}{m^2}\\ \\kg*\frac{m}{s^2}[=]G *\frac{kg*kg}{m^2}\\\\G[=]\frac{kg*m*m^2}{kg^2*s^2}\\ \\G[=]\frac{m^3}{kg*s^2}](https://tex.z-dn.net/?f=N%5B%3D%5DG%2A%5Cfrac%7Bkg%2Akg%7D%7Bm%5E2%7D%5C%5C%20%5C%5Ckg%2A%5Cfrac%7Bm%7D%7Bs%5E2%7D%5B%3D%5DG%20%2A%5Cfrac%7Bkg%2Akg%7D%7Bm%5E2%7D%5C%5C%5C%5CG%5B%3D%5D%5Cfrac%7Bkg%2Am%2Am%5E2%7D%7Bkg%5E2%2As%5E2%7D%5C%5C%20%5C%5CG%5B%3D%5D%5Cfrac%7Bm%5E3%7D%7Bkg%2As%5E2%7D)
Thus, answer is:
m³/(kg⋅s²)
Note that the [=] is used to indicate the units of G.
Best regards
