Answers:
a) ![1.05 rad/s](https://tex.z-dn.net/?f=1.05%20rad%2Fs)
b) ![1.38 J](https://tex.z-dn.net/?f=1.38%20J)
Explanation:
a) Final angular velocity
:
Before solving this part, we have to stay clear that the angular momentum
is conserved, since we are dealing with circular motion, then:
Hence:
(1)
Where:
is the initial moment of inertia of the system
is the initial angular velocity
is the final moment of inertia of the system
is the final angular velocity
But first, we have to find
and
:
(2)
(3)
Where:
is the student's moment of inertia
is the mass of each object
is the initial radius
is the final radius
Then:
(4)
(5)
Substituting the results of (4) and (5) in (1):
(6)
Finding
:
(7) This is the final angular speed
b) Change in kinetic energy:
The rotational kinetic energy is defined as:
(8)
And the change in kinetic energy is:
(9)
Since we already calculated these values, we can solve (9):
(10)
Finally:
This is the change in kinetic energy
Depending on the height of the building they can break due to impact on the floor.
B would be the correct answer. All the other choices consist of physical changes
(a) The time the baseball spends in the air is 0.92 s.
(b) The horizontal distance from the roof edge to the point where the baseball lands on the ground is 3.1 m.
<h3>
Time spent in air by the baseball</h3>
h = vt - ¹/₂gt²
-2.1 = (4.05 x sin 34)t - ¹/₂(9.8)(t²)
-2.1 = 2.26t - 4.9t²
4.9t² - 2.26t - 2.1 = 0
t = 0.92 s
<h3>Horizontal distance traveled by the baseball</h3>
R = Vx(t)
R = (4.05 x cos 34)(0.92)
R = 3.1 m
Thus, the time the baseball spends in the air is 0.92 s.
The horizontal distance from the roof edge to the point where the baseball lands on the ground is 3.1 m.
Learn more about horizontal distance here: brainly.com/question/24784992
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Answer:
The speed of the water shoot out of the hole is 20 m/s.
(d) is correct option.
Explanation:
Given that,
Height = 20 m
We need to calculate the velocity
Using formula Bernoulli equation
![\dfrac{1}{2}\rho v_{1}^2+\rho gh_{1}=\dfrac{1}{2}\rho v_{2}^2+\rho gh_{2}](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B2%7D%5Crho%20v_%7B1%7D%5E2%2B%5Crho%20gh_%7B1%7D%3D%5Cdfrac%7B1%7D%7B2%7D%5Crho%20v_%7B2%7D%5E2%2B%5Crho%20gh_%7B2%7D)
Where,
v₁= initial velocity
v₂=final velocity
h₁=total height
h₂=height of the hole from the base
Put the value into the formula
![v_{1}^2=2g(h_{2}-h_{1})](https://tex.z-dn.net/?f=v_%7B1%7D%5E2%3D2g%28h_%7B2%7D-h_%7B1%7D%29)
![v_{1}=\sqrt{2g(h_{2}-h_{1})}](https://tex.z-dn.net/?f=v_%7B1%7D%3D%5Csqrt%7B2g%28h_%7B2%7D-h_%7B1%7D%29%7D)
![v_{1}=\sqrt{2\times9.8\times(20-0.005)}](https://tex.z-dn.net/?f=v_%7B1%7D%3D%5Csqrt%7B2%5Ctimes9.8%5Ctimes%2820-0.005%29%7D)
![v_{1}=19.7\ m/s= approximate\ 20\ m/s](https://tex.z-dn.net/?f=v_%7B1%7D%3D19.7%5C%20m%2Fs%3D%20approximate%5C%2020%5C%20m%2Fs)
Hence, The speed of the water shoot out of the hole is 20 m/s.