In the simulation above, as the projectile travels upward, how does the vertical velocity change? Vertical velocity is zero. Ver
2 answers:
Answer:
Vertical velocity decreases.
Explanation:
The figure of the simulation is missing, however we can still answer. In fact, the problem tells us that the projectile is moving upward. We are only interested in the vertical velocity of the projectile, so we can consider only its vertical motion.
The vertical motion of the projectile is an accelerated motion, with constant acceleration g = 9.8 m/s^2, directed downward (acceleration due to gravity). The problem says that the projectile is still moving upward, so its velocity is directed upward. This means that the vertical acceleration is in the opposite direction of the vertical velocity: therefore, the vertical velocity must be decreasing, according to the equation
where v0 is the initial vertical velocity of the projectile, and t the time. Due to the negative sign in the formula, we see that as t increases, v(t) decreases.
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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
When you draw an illustration for this problem, you would come up with the same drawing as shown in the picture. As the hot-air balloon travels upwards, there is a slight time when the bag of sand rises up until it reaches the maximum height. Then, it goes back down to the ground. The total time would be t₁ + t₂. The solution is as follows:
H = v₀²/2g = (2.45)²/2(9.81) = 0.306 m
t₁ = H/v₀ = 0.306 m/2.45 m/s = 0.125 s
t₂ = √2(H + 98.8)/g = √2(0.306+ 98.8)/9.81
t₂ = 4.495 s
Total time = 0.125 s + 4.495 s = 4.62 seconds
H = v₀²/2g = (2.45)²/2(9.81) = 0.306 m
t₁ = H/v₀ = 0.306 m/2.45 m/s = 0.125 s
t₂ = √2(H + 98.8)/g = √2(0.306+ 98.8)/9.81
t₂ = 4.495 s
Total time = 0.125 s + 4.495 s = 4.62 seconds