Answer:
D
Explanation:
The greater the mass, the greater the inertia, and vice versa.
Remark: This means that a more massive object has a greater tendency to resist a change in its state of rest or motion.
Answer:
Cruising at 35,000 feet in an airliner, straight toward the east,
at 500 miles per hour
Explanation:
Answer:
50 N
Explanation:
Efficiency of a machine can't be more than 1, so I assume you mean 40%. (Remember, efficiency and mechanical advantage are not the same).
Efficiency is the ratio of work out of a system to the work in to the system.
e = Wout / Win
Work is force times distance, so:
e = (Fout × Dout) / (Fin × Din)
Rearranging:
Fin = (Fout × Dout) / (e × Din)
Fin = (Fout / e) × (Dout / Din)
Fin = (Fout / e) / (Din / Dout)
We know that e = 0.40, and Fout = 120 N. Since there are 6 pulleys, we also know that Din/Dout = 6.
F = (120 N / 0.4) / 6
F = 50 N
Answer:
0.02 m/s^2
Explanation:
change in velocity= 4.5m/s - 2.3m/s = 2.2 m/s
acceleration= change in velocity/change in time
acceleration= 2.2/120= 0.0183
= 0.02 (to 2 significant figures)
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
