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:
Ice is water in solid phase, in this phase, the particles are very close together and relatively in fixed positions.
As the temperature starts to increase (thermal energy), also does the kinetic energy of the particles (so we have a change from thermal energy to kinetic energy), so they start to move "more", and the position of the particles starts to be less "fixed". There is a point where the particles have enough energy, and this point is where the phase of the water changes from solid to liquid phase (the fusion point). After this point the water can not hold his shape, and takes the shape of the container where it is.
The answer is 175184.08 joules
Answer:
The current in the rods is 171.26 A.
Explanation:
Given that,
Length of rod = 0.85 m
Mass of rod = 0.073 kg
Distance 
The rods carry the same current in the same direction.
We need to calculate the current
I is the current through each of the wires then the force per unit length on each of them is
Using formula of force
Where, m = mass of rod
l = length of rod
Put the value into the formula
Hence, The current in the rods is 171.26 A.
The options are;
a. V2 equals 2V1.
b. V2 equals (V1)/2.
c. V2 equals V1.
d. V2 equals (V1)/4.
e. V2 equals 4V1.
Answer:
Option A: V2 equals 2V1
Explanation:
Since the flow is steady, then we can say;
mass flow rate at input = mass flow rate at output.
Formula for mass flow rate is;
m' = ρVA
Thus;
At input;
m'1 = ρ1•V1•A1
At output;
m'2 = ρ2•V2•A2
So, m'1 = m'2
Now, we are told that the density of the fluid decreases to half its initial value.
Thus; ρ2 = (ρ1)/2
Since m'1 = m'2, then;
ρ1•V1•A1 = (ρ1)/2•V2•A2
Now, the pipe is uniform and thus the cross section doesn't change. Thus;
A1 = A2
We now have;
ρ1•V1•A1 = (ρ1)/2•V2•A1
A1 and ρ1 will cancel out to give;
V1 = (V2)/2
Thus, V2 = 2V1
