Answer:
Explanation:
Range of projectile R = 20 m
formula of range
R = u² sin2θ / g
u is initial velocity , θ is angle of projectile
putting the values
20 = u² sin2x 40 / 9.8
u² = 199
u = 14.10 m /s
At the initial point
vertical component of u
= u sin40 = 14.1 x sin 40
= 9.06 m/s
Horizontal component
= u cos 30
At the final point where the ball strikes the ground after falling , its speed remains the same as that in the beginning .
Horizontal component of velocity
u cos 30
Vertical component
= - u sin 30
= - 9.06 m /s
So its horizontal component remains unchanged .
change in vertical component = 9.06 - ( - 9.06 )
= 18.12 m /s
change in momentum
mass x change in velocity
= .050 x 18.12
= .906 N.s
Impulse = change in momentum
= .906 N.s .
The maximum allowable torque must correspond to the allowable shear stress for maximization. To solve this, we use the torsion formula:
Max. Allowable Shear Stress = Maximum Torque ÷ Cross-Sectional Area
8 x 10^6 Pa = Maximum Torque ÷ pi*(d/2)²
Maximum Torque = 8 x 10^6 Pa * pi*(0.06/2)² m²
Maximum Torque = 22,619.47 J or
Maximum Torque = 22.62 kJ
As for the second question, I have no reference figure so I am unable to answer it. I hope I was still able to help you, though.
Hi there!
Use the following formula to solve:
KE = 1/2mv², where:
KE = kinetic energy
m = mass (kg)
v = velocity (m/s)
Therefore:
KE = 1/2(1500)(30)²
KE = 1/2(1500)(900)
KE = 675000 J
This is kinetic energy to heat energy
