Answer:
a)Yes will deform plastically
b) Will NOT experience necking
Explanation:
Given:
- Applied Force F = 850 lb
- Diameter of wire D = 0.15 in
- Yield Strength Y=45,000 psi
- Ultimate Tensile strength U = 55,000 psi
Find:
a) Whether there will be plastic deformation
b) Whether there will be necking.
Solution:
Assuming a constant Force F, the stress in the wire will be:
stress = F / Area
Area = pi*D^2 / 4
Area = pi*0.15^2 / 4 = 0.0176715 in^2
stress = 850 / 0.0176715
stress = 48,100.16 psi
Yield Strength < Applied stress > Ultimate Tensile strength
45,000 < 48,100 < 55,000
Hence, stress applied is greater than Yield strength beyond which the wire will deform plasticly but insufficient enough to reach UTS responsible for the necking to initiate. Hence, wire deforms plastically but does not experience necking.
Answer:
Part a)

Part b)
t = 12 s
Explanation:
Part a)
Tension in the rope at a distance x from the lower end is given as

so the speed of the wave at that position is given as

here we know that

now we have


Part b)
time taken by the wave to reach the top is given as




Answer:
7800kg/m³
Explanation:
Density of iron in CGS unit is 7.8 g/cm3. Its density is SI is
Given the density of iron = 7.8 g/cm3.
The SI units must be in kg/m³
7.8g = 7.8/1000 kg
7.8g = 0.0078kg
1cm³ = 0.000001m³
7.8g/cm³
= 0.0078/0.000001 kg/m³
= 7800kg/m³
Hence the density in SI unit is 7800kg/m³
Explanation:
It varies with altitude, but at sea level, it's 9.8 m/s².