(a) The average speed from A to B would be 1.76 metre per second and the average velocity from A to B would also be 1.76 metre per second

(b) The average speed from A to C would be 1.73 metre per second and the average velocity from A to C would be 0.87 metre per second

3 0

3 years ago

What must be the acceleration of a train in order for it to stop 12 m/s in a distance if 541 m

earnstyle [38]

Answer:

The acceleration of the train must be - 0.133 m/s²

Explanation:

Lets explain how to solve the problem

A train in order for it to stop 12 m/s in a distance if 541 m

That means the initial velocity of the train is 12 m/s

Its final velocity is zero (stop)

The distance it covers is 541 m

We want to find its acceleration

The acceleration will be negative quantity because the train reduced its

velocity from 12 m/s to zero

We need rule contains velocity, acceleration and distance

So we will use ⇒ v² = u² + 2as, where v is the final velocity, u is the

initial velocity, a is the acceleration and s is the distance

v = 0, u = 12 m/s, s = 541 m

Substitute these values in the rule

(0)² = (12)² + 2(a)(541)

0 = 144 + 1082 a

Subtract 144 from both sides

-144 = 1082 a

Divide both sides by 1082

- 0.133 = a

The acceleration of the train must be - 0.133 m/s²

3 0

3 years ago

Air enters a turbine operating at steady state at 8 bar, 1600 K and expands to 0.8 bar. The turbine is well insulated, and kinet

kobusy [5.1K]

Answer:

the maximum theoretical work that could be developed by the turbine is 775.140kJ/kg

Explanation:

To solve this problem it is necessary to apply the concepts related to the adiabatic process that relate the temperature and pressure variables

Mathematically this can be determined as

$\frac{T_2}{T_1} = (\frac{P_2}{P_1})^{(\frac{\gamma-1}{\gamma})}$

Where

Temperature at inlet of turbine

Temperature at exit of turbine

Pressure at exit of turbine

The steady flow Energy equation for an open system is given as follows:

$m_i = m_0 = mm(h_i+\frac{V_i^2}{2}+gZ_i)+Q = m(h_0+\frac{V_0^2}{2}+gZ_0)+W$

Where,

m = mass

m(i) = mass at inlet

m(o)= Mass at outlet

h(i)= Enthalpy at inlet

h(o)= Enthalpy at outlet

W = Work done

Q = Heat transferred

v(i) = Velocity at inlet

v(o)= Velocity at outlet

Z(i)= Height at inlet

Z(o)= Height at outlet

For the insulated system with neglecting kinetic and potential energy effects

$h_i = h_0 + WW = h_i -h_0$

Using the relation T-P we can find the final temperature:

$\frac{T_2}{T_1} = (\frac{P_2}{P_1})^{(\frac{\gamma-1}{\gamma})}\\$

$\frac{T_2}{1600K} = (\frac{0.8bar}{8nar})^{(\frac{1.4-1}{1.4})}\\ = 828.716K$

From this point we can find the work done using the value of the specific heat of the air that is 1,005kJ / kgK

$W = h_i -h_0W = C_p (T_1-T_2)W = 1.005(1600 - 828.716)W = 775.140kJ/Kg$

the maximum theoretical work that could be developed by the turbine is 775.140kJ/kg

4 0

2 years ago

Snow and sleet when they fall to the ground is solar energy or gravitational force?

Nikolay [14]

Gravitational energy

3 0

2 years ago