<span>...a concordant intrusion.
In geology, "concordant" means the same as "sill" -- or, an intrusion that has gotten in between older layers of rock (or even beds of volcanic lava). An intrusion with boundaries parallel to layering in surrounding rocks suggests this, meaning it is considered to be a concordant intrusion.</span>
m = mass of the car moving in horizontal circle = 1750 kg
v = Constant speed of the car moving in the horizontal circle = 15 m/s
r = radius of the horizontal circular track traced by the car = 45.0 m
F = magnitude of the centripetal force acting on the car
To move in a circle . centripetal force is required which is given as
F = m v²/r
inserting the above values in the formula
F = (1750) (15)²/(45)
F = (1750) (225)/(45)
F = 1750 x 5
F = 8750 N
Explanation:
... in every interaction, there is a pair of forces acting on the two interacting objects. The size of the force on the first object equals the size of the force on the second object. The direction of the force on the first object is opposite to the direction of the force on the second object. Forces always come in pairs - equal and opposite action-reaction force pairs.
Answer:
A(3.56m)
Explanation:
We have a conservation of energy problem here as well. Potential energy is being converted into linear kinetic energy and rotational kinetic energy.
We are given ω= 4.27rad/s, so v = ωr, which is 6.832 m/s. Place your coordinate system at top of the hill so E initial is 0.
Ef= Ug+Klin+Krot= -mgh+1/2mv^2+1/2Iω^2
Since it is a solid uniform disk I= 1/2MR^2, so Krot will be 1/4Mv^2(r^2ω^2= v^2).
Ef= -mgh+3/4mv^2
Since Ef=Ei=0
Mgh=3/4mv^2
gh=3/4v^2
h=0.75v^2/g
plug in givens to get h= 3.57m
Here is the full question:
The rotational inertia I of any given body of mass M about any given axis is equal to the rotational inertia of an equivalent hoop about that axis, if the hoop has the same mass M and a radius k given by:
The radius k of the equivalent hoop is called the radius of gyration of the given body. Using this formula, find the radius of gyration of (a) a cylinder of radius 1.20 m, (b) a thin spherical shell of radius 1.20 m, and (c) a solid sphere of radius 1.20 m, all rotating about their central axes.
Answer:
a) 0.85 m
b) 0.98 m
c) 0.76 m
Explanation:
Given that: the radius of gyration
So, moment of rotational inertia (I) of a cylinder about it axis = 
k = 0.8455 m
k ≅ 0.85 m
For the spherical shell of radius
(I) = 
k = 0.9797 m
k ≅ 0.98 m
For the solid sphere of radius
(I) = 
k = 0.7560
k ≅ 0.76 m
