Answer:
Clockwise
Explanation:
All of the planets rotate the same way around the sun.
Answer:
h = 9.57 seconds
Explanation:
It is given that,
Initial speed of Kalea, u = 13.7 m/s
At maximum height, v = 0
Let t is the time taken by the ball to reach its maximum point. It cane be calculated as :
![v=u-gt](https://tex.z-dn.net/?f=v%3Du-gt)
![u=gt](https://tex.z-dn.net/?f=u%3Dgt)
![t=\dfrac{u}{g}](https://tex.z-dn.net/?f=t%3D%5Cdfrac%7Bu%7D%7Bg%7D)
![t=\dfrac{13.7}{9.8}](https://tex.z-dn.net/?f=t%3D%5Cdfrac%7B13.7%7D%7B9.8%7D)
t = 1.39 s
Let h is the height reached by the ball above its release point. It can be calculated using second equation of motion as :
![h=ut+\dfrac{1}{2}at^2](https://tex.z-dn.net/?f=h%3Dut%2B%5Cdfrac%7B1%7D%7B2%7Dat%5E2)
Here, a = -g
![h=ut-\dfrac{1}{2}gt^2](https://tex.z-dn.net/?f=h%3Dut-%5Cdfrac%7B1%7D%7B2%7Dgt%5E2)
![h=13.7\times 1.39-\dfrac{1}{2}\times 9.8\times (1.39)^2](https://tex.z-dn.net/?f=h%3D13.7%5Ctimes%201.39-%5Cdfrac%7B1%7D%7B2%7D%5Ctimes%209.8%5Ctimes%20%281.39%29%5E2)
h = 9.57 meters
So, the height attained by the ball above its release point is 9.57 meters. Hence, this is the required solution.
Take the missile's starting position to be the origin. Assuming the angles given are taken to be counterclockwise from the positive horizontal axis, the missile has position vector with components
![x=v_0\cos20.0^\circ t+\dfrac12a_xt^2](https://tex.z-dn.net/?f=x%3Dv_0%5Ccos20.0%5E%5Ccirc%20t%2B%5Cdfrac12a_xt%5E2)
![y=v_0\sin20.0^\circ t+\dfrac12a_yt^2](https://tex.z-dn.net/?f=y%3Dv_0%5Csin20.0%5E%5Ccirc%20t%2B%5Cdfrac12a_yt%5E2)
The missile's final position after 9.20 s has to be a vector whose distance from the origin is 19,500 m and situated 32.0 deg relative the positive horizontal axis. This means the final position should have components
![x_{9.20\,\mathrm s}=(19,500\,\mathrm m)\cos32.0^\circ](https://tex.z-dn.net/?f=x_%7B9.20%5C%2C%5Cmathrm%20s%7D%3D%2819%2C500%5C%2C%5Cmathrm%20m%29%5Ccos32.0%5E%5Ccirc)
![y_{9.20\,\mathrm s}=(19,500\,\mathrm m)\sin32.0^\circ](https://tex.z-dn.net/?f=y_%7B9.20%5C%2C%5Cmathrm%20s%7D%3D%2819%2C500%5C%2C%5Cmathrm%20m%29%5Csin32.0%5E%5Ccirc)
So we have enough information to solve for the components of the acceleration vector,
and
:
![x_{9.20\,\mathrm s}=\left(1810\,\dfrac{\mathrm m}{\mathrm s}\right)\cos20.0^\circ(9.20\,\mathrm s)+\dfrac12a_x(9.20\,\mathrm s)^2\implies a_x=21.0\,\dfrac{\mathrm m}{\mathrm s^2}](https://tex.z-dn.net/?f=x_%7B9.20%5C%2C%5Cmathrm%20s%7D%3D%5Cleft%281810%5C%2C%5Cdfrac%7B%5Cmathrm%20m%7D%7B%5Cmathrm%20s%7D%5Cright%29%5Ccos20.0%5E%5Ccirc%289.20%5C%2C%5Cmathrm%20s%29%2B%5Cdfrac12a_x%289.20%5C%2C%5Cmathrm%20s%29%5E2%5Cimplies%20a_x%3D21.0%5C%2C%5Cdfrac%7B%5Cmathrm%20m%7D%7B%5Cmathrm%20s%5E2%7D)
![y_{9.20\,\mathrm s}=\left(1810\,\dfrac{\mathrm m}{\mathrm s}\right)\sin20.0^\circ(9.20\,\mathrm s)+\dfrac12a_y(9.20\,\mathrm s)^2\implies a_y=110\,\dfrac{\mathrm m}{\mathrm s^2}](https://tex.z-dn.net/?f=y_%7B9.20%5C%2C%5Cmathrm%20s%7D%3D%5Cleft%281810%5C%2C%5Cdfrac%7B%5Cmathrm%20m%7D%7B%5Cmathrm%20s%7D%5Cright%29%5Csin20.0%5E%5Ccirc%289.20%5C%2C%5Cmathrm%20s%29%2B%5Cdfrac12a_y%289.20%5C%2C%5Cmathrm%20s%29%5E2%5Cimplies%20a_y%3D110%5C%2C%5Cdfrac%7B%5Cmathrm%20m%7D%7B%5Cmathrm%20s%5E2%7D)
The acceleration vector then has direction
where
![\tan\theta=\dfrac{a_y}{a_x}\implies\theta=79.2^\circ](https://tex.z-dn.net/?f=%5Ctan%5Ctheta%3D%5Cdfrac%7Ba_y%7D%7Ba_x%7D%5Cimplies%5Ctheta%3D79.2%5E%5Ccirc)
<span>Crust. The thin solid outermost layer of Earth. ...Asthenosphere. The lower layer of the crust. ...Lithosphere.Plasticity: is solid but still being able to. flow without being a liquid.The cool, rigid outermost layer of the Earth. ...<span>the solid part of the earth consisting of the crust and outer mantle.</span></span>