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Natali [406]
2 years ago
13

Similarities and differences between distance Time graph and velocity time graph?

Physics
1 answer:
egoroff_w [7]2 years ago
4 0

Answer:

Similarity: >>Time is independent variable and such is on the x-axis. ... >>Distance time graph tells you how much distance you have travelled, while velocity time graph tells you your acceleration.  The difference between them is that the velocity-time graph reveals the speed of an object (and whether it is slowing down or speeding up), while the position-time graph describes the motion of an object over a period of time.

Explanation:

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Melting and frezzing are physical changes

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to know what that specif animal or thing does in the food chain

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An iron bar has more mass than a plastic bar of the same volume. so the iron bar will have greater inertia.
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The response is False, both bars, iron bars and plastic bars have de same inertia, this characteristic does not depend on the type of material, the inertia depends on his transverse section, since we can estimate in the following formula

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What is the minimum amount of data points an experiment should gather?
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2 years ago
Air at 400 kPa, 980 K enters a turbine operating at steady state and exits at 100 kPa, 670 K. Heat transfer from the turbine occ
Angelina_Jolie [31]

Answer:

a). \frac{\dot{W}}{m}= 311 kJ/kg

b). \frac{\dot{\sigma _{gen}}}{m}=0.9113 kJ/kg-K

Explanation:

a). The energy rate balance equation in the control volume is given by

\dot{Q} - \dot{W}+m(h_{1}-h_{2})=0

\frac{\dot{Q}}{m} = \frac{\dot{W}}{m}+m(h_{1}-h_{2})

\frac{\dot{W}}{m}= \frac{\dot{Q}}{m}+c_{p}(T_{1}-T_{2})

\frac{\dot{W}}{m}= -30+1.1(980-670)

\frac{\dot{W}}{m}= 311 kJ/kg

b). Entropy produced from the entropy balance equation in a control volume is given by

\frac{\dot{Q}}{T_{boundary}}+\dot{m}(s_{1}-s_{2})+\dot{\sigma _{gen}}=0

\frac{\dot{\sigma _{gen}}}{m}=\frac{-\frac{\dot{Q}}{m}}{T_{boundary}}+(s_{2}-s_{1})

\frac{\dot{\sigma _{gen}}}{m}=\frac{-\frac{\dot{Q}}{m}}{T_{boundary}}+c_{p}ln\frac{T_{2}}{T_{1}}-R.ln\frac{p_{2}}{p_{1}}

\frac{\dot{\sigma _{gen}}}{m}=\frac{-30}{315}+1.1ln\frac{670}{980}-0.287.ln\frac{100}{400}

\frac{\dot{\sigma _{gen}}}{m}=0.0952+0.4183+0.3978

\frac{\dot{\sigma _{gen}}}{m}=0.9113 kJ/kg-K

5 0
3 years ago
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