Answer:
(a) Height is 4.47 m
(b) Height is 4.37 m
Solution:
As per the question:
Initial velocity of teh ball, 
Angle made by the ramp, 
Distance traveled by the ball on the ramp, d = 5.00 m
Now,
(a) At any point on the projectile before attaining maximum height, the velocity can be given by the eqn-3 of motion:
where
H = 
g = 
= 19.06 m/s
Now, maximum height attained is given by:
Height from the ground = 
(b) now, considering the coefficient of friction bhetween ramp and the ball, 
:
velocity can be given by the eqn-3 of motion:
= 18.7 m/s
Now, maximum height attained is given by:
Height from the ground = 
Setting reference frame so that the x axis is along the incline and y is perpendicular to the incline
<span>X: mgsin65 - F = mAx </span>
<span>Y: N - mgcos65 = 0 (N is the normal force on the incline) N = mgcos65 (which we knew) </span>
<span>Moment about center of mass: </span>
<span>Fr = Iα </span>
<span>Now Ax = rα </span>
<span>and F = umgcos65 </span>
<span>mgsin65 - umgcos65 = mrα -------------> gsin65 - ugcos65 = rα (this is the X equation m's cancel) </span>
<span>umgcos65(r) = 0.4mr^2(α) -----------> ugcos65(r) = 0.4r(rα) (This is the moment equation m's cancel) </span>
<span>ugcos65(r) = 0.4r(gsin65 - ugcos65) ( moment equation subbing in X equation for rα) </span>
<span>ugcos65 = 0.4(gsin65 - ugcos65) </span>
<span>1.4ugcos65 = 0.4gsin65 </span>
<span>1.4ucos65 = 0.4sin65 </span>
<span>u = 0.4sin65/1.4cos65 </span>
<span>u = 0.613 </span>
Answer: so when a turbine converts the K.E and the potential of any moving fluid (more likely liquid or gas) to energy. once the proc is started the turbine generato, the fluid such as water, steam, combus gasses, or air pushes s big series of blades that have mounted on a shaft, which then will rotate the shaft that’s conn to the generator
Explanation: hope this helped plz mark brainest
