Answer:
a) 93.852 kN
b) 128.043 mm
Explanation:
Stress is load over section:
σ = P / A
If plastic deformation begins with a stress of 297 MPa, the maximum load before plastic deformation will be:
P = σ * A
316 mm^2 = 3.16*10^-4
P = 297*10^6 * 3.16*10^-4 = 93852 N = 93.852 kN
The stiffness of the specimen is:
k = E * A / l
k = 113*10^9 * 3.16*10^-4 / 0.128 = 279 MN/m
Hooke's law:
x' = x0 * (1 + P/k)
x' = 0.128 * (1 + 93.852*10^3 / 279*10^6) = 0.128043 m = 128.043 mm
Answer:
Technician B
Explanation:
Both AWG and metric are sized by cross-sectional area.
Technician A is wrong: 12 gauge wire is larger diameter rated for 20 amps in free air. 18 awg is smaller diameter and typically used for speaker wiring, Class II or low voltage and sub-circuits within appliances.
Answer:
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Explanation:
Answer:
work=281.4KJ/kg
Power=4Kw
Explanation:
Hi!
To solve follow the steps below!
1. Find the density of the air at the entrance using the equation for ideal gases

where
P=pressure=120kPa
T=20C=293k
R= 0.287 kJ/(kg*K)=
gas constant ideal for air

2.find the mass flow by finding the product between the flow rate and the density
m=(density)(flow rate)
flow rate=10L/s=0.01m^3/s
m=(1.43kg/m^3)(0.01m^3/s)=0.0143kg/s
3. Please use the equation the first law of thermodynamics that states that the energy that enters is the same as the one that must come out, we infer the following equation, note = remember that power is the product of work and mass flow
Work
w=Cp(T1-T2)
Where
Cp= specific heat for air=1.005KJ/kgK
w=work
T1=inlet temperature=20C
T2=outlet temperature=300C
w=1.005(300-20)=281.4KJ/kg
Power
W=mw
W=(0.0143)(281.4KJ/kg)=4Kw
Answer:
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Explanation: