El sol es una bola de gas así como las estrellas:)
A body accelerates due to net force acting on it. As newton's second law of motion states:
when a net force acts on a body, it produces acceleration in the body in the direction of the net force
Answer:
94.13 ft/s
Explanation:
<u>Given:</u>
= time interval in which the rock hits the opponent = 10 s - 5 s = 5 s
= distance to be moved by the rock long the horizontal = 98 yards
= displacement to be moved by the rock during the time of flight along the vertical = 0 yard
<u>Assume:</u>
= magnitude of initial velocity of the rock
= angle of the initial velocity with the horizontal.
For the motion of the rock along the vertical during the time of flight, the rock has a constant acceleration in the vertically downward direction.
![\therefore y = u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow 0 = u\sin \theta 5 +\dfrac{1}{2}(-9.8)\times 5^2\\\Rightarrow u\sin \theta 5 =\dfrac{1}{2}(9.8)\times 5^2......(1)\\](https://tex.z-dn.net/?f=%5Ctherefore%20y%20%3D%20u%5Csin%20%5Ctheta%20t%20%2B%5Cdfrac%7B1%7D%7B2%7D%28-g%29t%5E2%5C%5C%5CRightarrow%200%20%3D%20u%5Csin%20%5Ctheta%205%20%2B%5Cdfrac%7B1%7D%7B2%7D%28-9.8%29%5Ctimes%205%5E2%5C%5C%5CRightarrow%20u%5Csin%20%5Ctheta%205%20%3D%5Cdfrac%7B1%7D%7B2%7D%289.8%29%5Ctimes%205%5E2......%281%29%5C%5C)
Now the rock has zero acceleration along the horizontal. This means it has a constant velocity along the horizontal during the time of flight.
![\therefore u\cos \theta t = s\\\Rightarrow u\cos \theta 5 = 98.....(2)\\](https://tex.z-dn.net/?f=%5Ctherefore%20u%5Ccos%20%5Ctheta%20t%20%3D%20s%5C%5C%5CRightarrow%20u%5Ccos%20%5Ctheta%205%20%3D%2098.....%282%29%5C%5C)
On dividing equation (1) by (2), we have
![\tan \theta = \dfrac{25}{20}\\\Rightarrow \tan \theta = 1.25\\\Rightarrow \theta = \tan^{-1}1.25\\\Rightarrow \theta = 51.34^\circ](https://tex.z-dn.net/?f=%5Ctan%20%5Ctheta%20%3D%20%5Cdfrac%7B25%7D%7B20%7D%5C%5C%5CRightarrow%20%5Ctan%20%5Ctheta%20%3D%201.25%5C%5C%5CRightarrow%20%5Ctheta%20%3D%20%5Ctan%5E%7B-1%7D1.25%5C%5C%5CRightarrow%20%5Ctheta%20%3D%2051.34%5E%5Ccirc)
Now, putting this value in equation (2), we have
![u\cos 51.34^\circ\times 5 = 98\\\Rightarrow u = \dfrac{98}{5\cos 51.34^\circ}\\\Rightarrow u =31.38\ yard/s\\\Rightarrow u =31.38\times 3\ ft/s\\\Rightarrow u =94.13\ ft/s](https://tex.z-dn.net/?f=u%5Ccos%2051.34%5E%5Ccirc%5Ctimes%20%205%20%3D%2098%5C%5C%5CRightarrow%20u%20%3D%20%5Cdfrac%7B98%7D%7B5%5Ccos%2051.34%5E%5Ccirc%7D%5C%5C%5CRightarrow%20u%20%3D31.38%5C%20yard%2Fs%5C%5C%5CRightarrow%20u%20%3D31.38%5Ctimes%203%5C%20ft%2Fs%5C%5C%5CRightarrow%20u%20%3D94.13%5C%20ft%2Fs)
Hence, the initial velocity of the rock must a magnitude of 94.13 ft/s to hit the opponent exactly at 98 yards.
Answer: 62.7
Explanation:
The specific heat capacity of water is 4.186 J you multiply 4.186 and 15.0 and you will get 62.7
Its like the weather man. They are the people you see on the news that use radars ect. To predict the weather.
Hope that helps?