Answer:
Normal stress = 66/62.84 = 1.05kips/in²
shearing stress = T/2 = 0.952/2 = 0.476 kips/in²
Explanation:
A steel pipe of 12-in. outer diameter d₂ =12in d₁= 12 -4in = 8in
4 -in.-thick
angle of 25°
Axial force P = 66 kip axial force
determine the normal and shearing stresses
Normal stress б = force/area = P/A
= 66/ (П* (d₂²-d₁²)/4
=66/ (3.142* (12²-8²)/4
= 66/62.84 = 1.05kips/in²
Tangential stress T = force* cos ∅/area = P/A
= 66* cos 25/ (П* (d₂²-d₁²)/4
=59.82/ (3.142* (12²-8²)/4
= 59.82/62.84 = 0.952kips/in²
shearing stress = tangential stress /2
= T/2 = 0.952/2 = 0.476 kips/in²
Answer:
608kg
Explanation:
Formula : <u>Kinetic</u><u> </u><u>energy</u><u> </u>
½ ×mass x speed²
<u>47500</u>
½×12.5²
=608 Kg
<span>The 23.5 degree tilt is responsible for the seasons. If the earth had no tilt there would not be seasons. If the earth was tilted by 90 degrees the seasonal changes would be at the most extreme. The Earth's pole would point directly at the sun at a point on the track around the sun. As the Earth revolves around the Sun the pole would alternate twice each year between pointing directly at the sun and being perpendicular to the sun.
I hope this helps you!
xo, Leafling</span>
Using kinematics: 8.1 = 1/2(9.8)(t^2),
t = 1.2857 s.
Horizontal distance x travelled is x = vhorizontal * t, so 9.3 = v*1.2857, or v= 7.233 m/s horizontally.
Answer:
Fgparallel = 170N
Explanation:
Acting down the incline would be the paralell force. The Force of gravity on an incline for the parallel portion is mgsinθ.
Fg parallel = mgsinθ
mg is 289.9, as that is the weight. θ is 35.9
Fgpar = 289.9sin35.9
Fgparallel = 170N
