Answer:
1561.84 MPa
Explanation:
L=20 cm
d1=0.21 cm
d2=0.25 cm
F=5500 N
a) σ= F/A1= 5000/(π/4×(0.0025)^2)= 1018.5916 MPa
lateral strain= Δd/d1= (0.0021-0.0025)/0.0025= -0.16
longitudinal strain (ε_l)= -lateral strain/ν = -(-0.16)/0.3
(assuming a poisson's ration of 0.3)
ε_l =0.16/0.3 = 0.5333
b) σ_true= σ(1+ ε_l)= 1018.5916( 1+0.5333)
σ_true = 1561.84 MPa
ε_true = ln( 1+ε_l)= ln(1+0.5333)
ε_true= 0.4274222
The engineering stress on the rod when it is loaded with a 5500 N weight is 1561.84 MPa.
Answer:
The question has some details missing : The 35-kg block A is released from rest. Determine the velocity of the 13-kgkg block BB in 4 ss . Express your answer to three significant figures and include the appropriate units. Enter positive value if the velocity is upward and negative value if the velocity is downward.
Explanation:
The detailed steps and appropriate calculation is as shown in the attached file.
Answer:
for 5.6V 9 turns, for 12.0V 19 turns, for 480V 755 turns
Explanation:
Vp/Vs= Np/Ns
Vp: Primary voltage
Vs: Secondary Voltage
Np: number of turns on primary side
Ns: number of turns on secondary side
for output 5.6V
140/5.6= 220/Ns
Ns= 8.8 or 9 Turns
for output 12.0V
140/12= 220/Ns
Ns= 18.9 or 19 turns
for output 480V
140/480= 220/Ns
Ns= 754.3 or 755 turns
Answer:
Check the v(t) signal referred to in the question and the solution to each part in the files attached
Explanation:
The detailed solutions of parts a to d are clearly expressed in the second file attached.
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