If (and only if) by 'streched to a distance of 40cm' you mean that it's 40 cm from irs rest position, then the elastic potential energy can be written
Answer:
8.2 m/s
Explanation:
The horizontal component of the velocity is given by:
![v_x = v cos \theta](https://tex.z-dn.net/?f=v_x%20%3D%20v%20cos%20%5Ctheta)
where
v = 10 m/s is the magnitude of the velocity
is the angle at which the ball has been thrown, with respect to the horizontal
Substituting the values into the equation, we get
![v_x=(10 m/s)(cos 35^{\circ})=8.2 m/s](https://tex.z-dn.net/?f=v_x%3D%2810%20m%2Fs%29%28cos%2035%5E%7B%5Ccirc%7D%29%3D8.2%20m%2Fs)
Solution
<span>10 mg∙cm/s^2 = 10Kg 10^-6 .10^-2 m /s^2 = 1 Kg 10^1-6-2m / s^2
</span>10 mg∙cm/s^2 = <span>Kg 10^-7 m / s^2
</span>so <span>D. 1 x 10^-7 N</span>
Answer:
v = 6.95 m/s
Explanation:
Given that,
A diver is on a board 1.80 m above the water, s = 1.8 m
The initial speed of the diver, u = 3.62 m/s
Let v is the speed with which she hit the water. It will move under the action of gravity. Using the equation of motion as follows :
![v^2-u^2=2gs\\\\v=\sqrt{u^2+2gs} \\\\v=\sqrt{(3.62)^2+2(9.8)(1.8)} \\\\v=6.95\ m/s](https://tex.z-dn.net/?f=v%5E2-u%5E2%3D2gs%5C%5C%5C%5Cv%3D%5Csqrt%7Bu%5E2%2B2gs%7D%20%5C%5C%5C%5Cv%3D%5Csqrt%7B%283.62%29%5E2%2B2%289.8%29%281.8%29%7D%20%5C%5C%5C%5Cv%3D6.95%5C%20m%2Fs)
So, she will hit the water with a speed of 6.95 m/s.