The so-called "terminal velocity" is the fastest that something can fall
through a fluid. Even though there's a constant force pulling it through,
the friction or resistance of plowing through the surrounding substance
gets bigger as the speed grows, so there's some speed where the resistance
is equal to the pulling force, and then the falling object can't go any faster.
A few examples:
-- the terminal velocity of a sky-diver falling through air,
-- the terminal velocity of a pecan falling through honey,
-- the terminal velocity of a stone falling through water.
It's not possible to say that "the terminal velocity is ----- miles per hour".
If any of these things changes, then the terminal velocity changes too:
-- weight of the falling object
-- shape of the object
-- surface texture (smoothness) of the object
-- density of the surrounding fluid
-- viscosity of the surrounding fluid .
Now the vertical velocity of the ball thrown at an angle 10° is given as
Voy(initial vertical velocity)= 30m/s x sin 10
Voy(initial vertical velocity)= 5.2m/s
Now the ball is decelerating with an acceleration due to gravity equivalent to 9.8m/s^2.
Let Vy be the final velocity and that is equal to zero in this case.
Now
Vy= Voy- tx9.8
Where t is the time at which the vertical velocity becomes 0.
Substituting the values we get
0= 5.2-tx9.8
9.8t=5.2
t=0.53 secs
Answer:
the magnitude of the total angular momentum of the blades is <em>743.71 kg·m²</em>
Explanation:
Converting the angular speed into radians per second:
ω = 334 rpm · (2π rad / 1 rev) · (1 min / 60 s)
ω = 34.98 rad/s
The rotational kinetic energy of the blades is given by:
EK = 1/2 I ω²
where
- I is the moment of inertia
- ω is the angular speed
Therefore, rearranging the above equation, we get:
1/2 I ω² = EK
I ω² = 2 EK
I = 2(EK) / ω²
I = 2(4.55 × 10⁵ J) / (34.98 rad/s)²
<em>I = 743.71 kg·m²</em>
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Therefore, the magnitude of the total angular momentum of the blades is <em>743.71 kg·m²</em>.
