Answer:
The horizontal component of the truck's velocity is: 23.70 m/s
The vertical component of the truck's velocity is: 3.13 m/s
Explanation:
You have to apply trigonometric identities for a right triangle (because the ramp can be seen as a right triangle where the speed is the hypotenuse), in order to obtain the components of the velocity vector.
The identities are:
Cosα= 
Senα= 
Where H is the hypotenuse, α is the angle, CA is the adjacent cathetus and CO is the opposite cathetus
The horizontal component of the truck's velocity is:
Let Vx represent it.
In this case, CA=Vx, H=24 and α=7.5 degrees
Vx=(24)Cos(7.5)
Vx=23.79 m/s
The vertical component of the truck's velocity is:
Let Vy represent it.
In this case, CO=Vy, H=24 and α=7.5 degrees
Vy=(24)Sen(7.5)
Vy=3.13 m/s
Answer:
I don't do physics , I'm sorry can't help you
Answer:

Explanation:
Given that
d= 1.5 in ( 1 in = 0.0254 m)
d= 0.0381 m
P= 75 hp ( 1 hp = 745.7 W)
P= 55927.5 W
N= 1800 rpm
We know that power P is given as

T=Torque
N=Speed

T=296.85 N.m
The maximum shear stress is given as



We know that 1 MPa =0.145 ksi

4A. PE = MxGxH. (You can consider g as 9.8 / 10m/s as well)
509 J = 12x10xH
509 J = 120xH
H = 509/120
H = 4.24 m
Hope u got the answer....pls rate the answer if it is helpful for u....and I'm sorry I could not understand B part so I didn't do it.
Thank you