After rolling off the edge of the cliff and falling ' M ' meters down,
the speed of the boulder is
Square root of ( 19.6 M ) .
If M=111 meters, then the speed is <em>46.64 meters per second</em>.
We have known for roughly 500 years that if there's no air resistance,
the mass of the falling object makes no difference, and all objects fall
with the same acceleration, speed, time to splat, etc.
Answer:
a) True. There is dependence on the radius and moment of inertia, no data is given to calculate the moment of inertia
c) True. Information is missing to perform the calculation
Explanation:
Let's consider solving this exercise before seeing the final statements.
We use Newton's second law Rotational
τ = I α
T r = I α
T gR = I α
Alf = T R / I (1)
T = α I / R
Now let's use Newton's second law in the mass that descends
W- T = m a
a = (m g -T) / m
The two accelerations need related
a = R α
α = a / R
a = (m g - α I / R) / m
R α = g - α I /m R
α (R + I / mR) = g
α = g / R (1 + I / mR²)
We can see that the angular acceleration depends on the radius and the moments of inertia of the steering wheels, the mass is constant
Let's review the claims
a) True. There is dependence on the radius and moment of inertia, no data is given to calculate the moment of inertia
b) False. Missing data for calculation
c) True. Information is missing to perform the calculation
d) False. There is a dependency if the radius and moment of inertia increases angular acceleration decreases
Answer:
Option (2)
Explanation:
From the figure attached,
Horizontal component, 
![A_x=12[\text{Sin}(37)]](https://tex.z-dn.net/?f=A_x%3D12%5B%5Ctext%7BSin%7D%2837%29%5D)
= 7.22 m
Vertical component, ![A_y=A[\text{Cos}(37)]](https://tex.z-dn.net/?f=A_y%3DA%5B%5Ctext%7BCos%7D%2837%29%5D)
= 9.58 m
Similarly, Horizontal component of vector C,
= C[Cos(60)]
= 6[Cos(60)]
= 
= 3 m
![C_y=6[\text{Sin}(60)]](https://tex.z-dn.net/?f=C_y%3D6%5B%5Ctext%7BSin%7D%2860%29%5D)
= 5.20 m
Resultant Horizontal component of the vectors A + C,
m
= 4.38 m
Now magnitude of the resultant will be,
From ΔOBC,

= 
= 
= 6.1 m
Direction of the resultant will be towards vector A.
tan(∠COB) = 
= 
= 
m∠COB = 
= 46°
Therefore, magnitude of the resultant vector will be 6.1 m and direction will be 46°.
Option (2) will be the answer.
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