Explanation:
The moment of inertia of each disk is:
Idisk = 1/2 MR²
Using parallel axis theorem, the moment of inertia of each rod is:
Irod = 1/2 mr² + m (R − r)²
The total moment of inertia is:
I = 2Idisk + 5Irod
I = 2 (1/2 MR²) + 5 [1/2 mr² + m (R − r)²]
I = MR² + 5/2 mr² + 5m (R − r)²
Plugging in values:
I = (125 g) (5 cm)² + 5/2 (250 g) (1 cm)² + 5 (250 g) (5 cm − 1 cm)²
I = 23,750 g cm²
Answer:
15
Explanation:
mass, M = 5Kg
horizontal force, F_h = 40N
acceleration, a =5 m/s^2
frictional force, F_f =?
net force = ma
net force = F_h - F_f = 40N - F_f
40 - F_f = 5 x 5
- F_f = 25 - 40
multiply both side by -1
F_f = 40 - 25 = 15
the frictional force is 15N
Answer:
a) r = 6122 m and b) v = 32.5 m / s
Explanation:
a) The train in the curve is subject to centripetal acceleration
a = v2 / r
Where v is The speed and r the radius of the curve
They indicate that the maximum acceleration of the person is 0.060g,
a = 0.060 g
a = 0.060 9.8
a = 0.588 m /s²
Let's calculate the radius
v = 216 km / h (1000m / 1km) (1 h / 3600 s =
v = 60 m / s
r = v² / a
r = 60² /0.588
r = 6122 m
b) Let's calculate the speed, for a radius curve 1.80 km = 1800 m
v = √a r
v = √( 0.588 1800)
v = 32.5 m / s
Answer:
The speed of the roller coaster at this point is 18.74 m/s.
Explanation:
Given that,
Weight of the student, W = 655 kg
Weight of the roller coaster, 
Radius of the roller coaster, r = 18 m
At the bottom of the loop, the weight of the roller coaster us given by :

If m is the mass of the roller coaster,



m = 66.83 kg
So,



v = 18.74 m/s
So, the speed of the roller coaster at this point is 18.74 m/s. Hence, this is the required solution.
(a) The lowest frequency (called fundamental frequency) of a wire stretched under a tension T is given by

where
L is the wire length
T is the tension
m is the wire mass
In our problem, L=10.9 m, m=55.8 g=0.0558 kg and T=253 N, therefore the fundamental frequency of the wire is

b) The frequency of the nth-harmonic for a standing wave in a wire is given by

where n is the order of the harmonic and f1 is the fundamental frequency. If we use n=2, we find the second lowest frequency of the wire:

c) Similarly, the third lowest frequency (third harmonic) is given by