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AleksAgata [21]
2 years ago
7

A train accelerated from rest to a speed of 60m/s in 200 seconds it continues with the speed reached for 40 seconds before comin

g to a rest in another 40 seconds. sketch a velocity time graph and find the acceleration, deceleration , total distance covered and its average speed.
Physics
1 answer:
Olenka [21]2 years ago
6 0

I don't know to be honest

Explanation:

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Which of the following is an example of a safe laboratory produced?
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3 years ago
Water, initially saturated vapor at 4 bar, fills a closed, rigid container. The water is heated until its temperature is 360°C.
salantis [7]

Explanation:

Using table A-3, we will obtain the properties of saturated water as follows.

Hence, pressure is given as p = 4 bar.

u_{1} = u_{g} = 2553.6 kJ/kg

v_{1} = v_{g} = 0.4625 m^{3}/kg

At state 2, we will obtain the properties. In a closed rigid container, the specific volume will remain constant.

Also, the specific volume saturated vapor at state 1 and 2 becomes equal. So, v_{2} = v_{g} = 0.4625 m^{3}/kg

According to the table A-4, properties of superheated water vapor will obtain the internal energy for state 2 at v_{2} = v_{g} = 0.4625 m^{3}/kg and temperature T_{2} = 360^{o}C so that it will fall in between range of pressure p = 5.0 bar and p = 7.0 bar.

Now, using interpolation we will find the internal energy as follows.

 u_{2} = u_{\text{at 5 bar, 400^{o}C}} + (\frac{v_{2} - v_{\text{at 5 bar, 400^{o}C}}}{v_{\text{at 7 bar, 400^{o}C - v_{at 5 bar, 400^{o}C}}}})(u_{at 7 bar, 400^{o}C - u_{at 5 bar, 400^{o}C}})

     u_{2} = 2963.2 + (\frac{0.4625 - 0.6173}{0.4397 - 0.6173})(2960.9 - 2963.2)

                   = 2963.2 - 2.005

                   = 2961.195 kJ/kg

Now, we will calculate the heat transfer in the system by applying the equation of energy balance as follows.

      Q - W = \Delta U + \Delta K.E + \Delta P.E ......... (1)

Since, the container is rigid so work will be equal to zero and the effects of both kinetic energy and potential energy can be ignored.

            \Delta K.E = \Delta P.E = 0

Now, equation will be as follows.

           Q - W = \Delta U + \Delta K.E + \Delta P.E

           Q - 0 = \Delta U + 0 + 0

           Q = \Delta U

Now, we will obtain the heat transfer per unit mass as follows.

          \frac{Q}{m} = \Delta u

         \frac{Q}{m} = u_{2} - u_{1}

                      = (2961.195 - 2553.6)

                      = 407.595 kJ/kg

Thus, we can conclude that the heat transfer is 407.595 kJ/kg.

4 0
3 years ago
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