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:
• They depend solely on the load that generates it
• Two or more electrical charges interact, which can be positive or negative
• The energy source is based on the electrical voltage
V^2= 2* g * s
V^2 = 2 * 9.8 * 50 = 980
v = square root (980)=...... m/s
The answer should be B. a stable isotope to a decaying isotope.
Answer:
a) 3.43 m/s
Explanation:
Due to the law of conservation of momentum, the total momentum of the bullet - rifle system must be conserved.
The total momentum before the bullet is shot is zero, because they are both at rest, so:

Instead the total momentum of the system after the shot is:

where:
m = 0.006 kg is the mass of the bullet
M = 1.4 kg is the mass of the rifle
v = 800 m/s is the velocity of the bullet
V is the recoil velocity of the rifle
The total momentum is conserved, therefore we can write:

Which means:

Solving for V, we can find the recoil velocity of the rifle:

where the negative sign indicates that the velocity is opposite to direction of the bullet: so the recoil speed is
a) 3.43 m/s