Answer:
The process in which heat flows by the mass movement of molecules from one place to another is C) convection.
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Answer:
1. Flexibility.
2. Muscular Fitness
3. Cardiovascular Fitness
4. Cardiovascular Fitness
5. Flexibility
6. Muscular Fitness
7. Cardiovascular Fitness
Answer:
1)
![v_{oy}=11.29\ m/s](https://tex.z-dn.net/?f=v_%7Boy%7D%3D11.29%5C%20m%2Fs)
2)
![y=7.39\ m](https://tex.z-dn.net/?f=y%3D7.39%5C%20m)
Explanation:
<u>Projectile Motion</u>
When an object is launched near the Earth's surface forming an angle
with the horizontal plane, it describes a well-known path called a parabola. The only force acting (neglecting the effects of the wind) is the gravity, which acts on the vertical axis.
The heigh of an object can be computed as
![\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20y%3Dy_o%2BV_%7Boy%7Dt-%5Cfrac%7Bgt%5E2%7D%7B2%7D)
Where
is the initial height above the ground level,
is the vertical component of the initial velocity and t is the time
The y-component of the speed is
![v_y=v_{oy}-gt](https://tex.z-dn.net/?f=v_y%3Dv_%7Boy%7D-gt)
1) We'll find the vertical component of the initial speed since we have not enough data to compute the magnitude of ![v_o](https://tex.z-dn.net/?f=v_o)
The object will reach the maximum height when
. It allows us to compute the time to reach that point
![v_{oy}-gt_m=0](https://tex.z-dn.net/?f=v_%7Boy%7D-gt_m%3D0)
Solving for ![t_m](https://tex.z-dn.net/?f=t_m)
![\displaystyle t_m=\frac{v_{oy}}{g}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20t_m%3D%5Cfrac%7Bv_%7Boy%7D%7D%7Bg%7D)
Thus, the maximum heigh is
![\displaystyle y_m=y_o+\frac{v_{oy}^2}{2g}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20y_m%3Dy_o%2B%5Cfrac%7Bv_%7Boy%7D%5E2%7D%7B2g%7D)
We know this value is 8 meters
![\displaystyle y_o+\frac{v_{oy}^2}{2g}=8](https://tex.z-dn.net/?f=%5Cdisplaystyle%20y_o%2B%5Cfrac%7Bv_%7Boy%7D%5E2%7D%7B2g%7D%3D8)
Solving for ![v_{oy}](https://tex.z-dn.net/?f=v_%7Boy%7D)
![\displaystyle v_{oy}=\sqrt{2g(8-y_o)}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20v_%7Boy%7D%3D%5Csqrt%7B2g%288-y_o%29%7D)
Replacing the known values
![\displaystyle v_{oy}=\sqrt{2(9.8)(8-1.5)}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20v_%7Boy%7D%3D%5Csqrt%7B2%289.8%29%288-1.5%29%7D)
![\displaystyle v_{oy}=11.29\ m/s](https://tex.z-dn.net/?f=%5Cdisplaystyle%20v_%7Boy%7D%3D11.29%5C%20m%2Fs)
2) We know at t=1.505 sec the ball is above Julie's head, we can compute
![\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20y%3Dy_o%2BV_%7Boy%7Dt-%5Cfrac%7Bgt%5E2%7D%7B2%7D)
![\displaystyle y=1.5+(11.29)(1.505)-\frac{9.8(1.505)^2}{2}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20y%3D1.5%2B%2811.29%29%281.505%29-%5Cfrac%7B9.8%281.505%29%5E2%7D%7B2%7D)
![\displaystyle y=1.5\ m+16,991\ m-11.098\ m](https://tex.z-dn.net/?f=%5Cdisplaystyle%20y%3D1.5%5C%20m%2B16%2C991%5C%20m-11.098%5C%20m)
![y=7.39\ m](https://tex.z-dn.net/?f=y%3D7.39%5C%20m)
Iron is considered a micro-nutrient because only small amounts are required to aid in normal plant growth. ... Plants can suffer iron deficiency with symptoms of chlorosis and stunted growth, but plants can also take in too much iron, especially under certain growing conditions.