Answer:
The other angle is 75⁰
Explanation:
Given;
velocity of the projectile, v = 10 m/s
range of the projectile, R = 5.1 m
angle of projection, 15⁰
The range of a projectile is given as;
To find another angle of projection to give the same range;
<u>Check: </u>
sin(2θ) = sin(2 x 75) = sin(150) = 0.5
sin(2θ) = sin(2 x 15) = sin(30) = 0.5
It is potential energy because the band is not in movement, th band has the potential to move.
<span>Assume: neglect of the collar dimensions.
Ď_h=(P*r)/t=(5*125)/8=78.125 MPa ,Ď_a=Ď_h/2=39 MPa
τ=(S*Q)/(I*b)=(40*〖10〗^3*π(〖0.125〗^2-〖0.117〗^2 )*121*〖10〗^(-3))/(π/2 (〖0.125〗^4-〖0.117〗^4 )*8*〖10〗^(-3) )=41.277 MPa
@ Point K:
Ď_z=(+M*c)/I=(40*0.6*121*〖10〗^(-3))/(8.914*〖10〗^(-5) )=32.6 MPa
Using Mohr Circle:
Ď_max=(Ď_h+Ď_a)/2+âš(Ď„^2+((Ď_h-Ď_a)/2)^2 )
Ď_max=104.2 MPa, Ď„_max=45.62 MPa</span>
