Answer: 76.3 ° C
Explanation:
Outside Distance = 10 m
Outside diameter, d = 2.2 m
R=D/2 = 2.2/2= 1.1 m
inner radius, Rs = = 1.1 m
Surface Temperature Ts = 20°C
Given that the thermal conductivity of soil, K = 0.52 W/mK
We can find the outside temperature of the container by finding the thermal temperature of conductor, T, and then summing up with the surface temperature
That is (T + Ts)
we know amount of heat, Q is given by :
Q = [(T-Ts) / (R- Rs)/4πKRRs
500 = T-20 [4π×0.52×1.1×10] / (10-1.1)
500= T-20 (71.88/8.9)
500 = T-20 (8.08)
500 = 8.08T - 161.6
8.08T = 500 + 161.6
8.08T = 661.6
T = 76.33 ° C
Answer:
The complete question is:
Use a stem and leaf display to organize the following distribution of scores. Use six stems with each stem corresponding to a 10-point interval.
Scores: 16, 42, 53, 41, 69
34, 33, 40, 61, 23
53, 49, 55, 29, 10
44, 64, 51, 21, 39
36, 58, 60, 27, 47
14, 44, 38, 31, 56
The answer is:
The decimal point is 1 digit(s) to the right of the |
1 | 046
2 | 1379
3 | 134689
4 | 0124479
5 | 133568
6 | 0149
Explanation:
Answer:
Viscoelastic stress relaxation
Viscoelastic creep
Explanation:
- Viscoelastic stress refers to the reduction of tensile stress with the relaxation time that occurs when the body is kept at a certain length under tensile stress. The purpose of this study was to demonstrate viscoelastic stress relaxation
- Therefore, when the viscous tension stress occurs, the pressure decreases due to the steady-state stress of the phases.
- When under constant pressure, the viscoelastic material experiences a time-dependent increase in pressure. This phenomenon is called viscoelastic creep.
- Therefore the phase constant pressure decreases when there is relaxation in the viscoelastic stress