<span>Well, since it's in the shape of a wheel and the person walks around the edge of it, they must have a centripetal acceleration. Since a=v^2/r you can solve for "v" using 2.20 as your "a" and 59.5 as your "r" (r=half of the diameter).
</span> a=v^2/r
v=(a*r)^(1/2)=((2.20)*(59.5))^(1/2)=<span>
<span>11.44 m/s.
</span></span><span> After you get "v," plugged that into T=2 pi r/ v. This will give you the 1rev per sec.
</span> T=2 pi r/ v= T=(2)*(pi)*(59.5)/(11.44)= <span>
<span>32.68 rev/s
</span></span> Use dimensional analysis to get rev per min (1rev / # sec) times (60 sec/min).
(32.68 rev/s)(60 s/min)=<span>
<span>1960.74 rev/min
</span></span>
D = distance between the cars at the start of time = 680 km
v₁ = speed of one car
v₂ = speed of other car = v₁ - 10
t = time taken to meet = 4 h
distance traveled by one car in time "t" + distance traveled by other car in time "t" = D
v₁ t + v₂ t = D
(v₁ + v₂) t = D
inserting the values
(v₁ + v₁ - 10) (4) = 680
v₁ = 90 km/h
rate of slower car is given as
v₂ = v₁ - 10
v₂ = 90 - 10 = 80 km/h
energy extracted out of liquids an atoms are left to come closer arrange themselves shorter distance and then they solidify
Answer:
F = ⅔ F₀
Explanation:
For this exercise we use Coulomb's law
F = k q₁q₂ / r²
let's use the subscript "o" for the initial conditions
F₀ = k q² / r²
now the charge changes q₁ = q₂ = 2q and the new distance is r = 3 r
we substitute
F = k 4q² / 9 r²
F = k q² r² 4/9
F = ⅔ F₀
