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2 years ago

A train travels 650 meters in 25 seconds. What is the train's velocity?

Engineering

1 answer:

frosja888 [35]2 years ago

3 0

The train is traveling 26 meters A second .

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thermal energy is being added to steam at 475.8 kPa and 75% quality. determine the amount of thermal energy to be added to creat

eduard

Answer:

$q_{in} = 528.6\,\frac{kJ}{kg}$

Explanation:

Let assume that heating process occurs at constant pressure, the phenomenon is modelled by the use of the First Law of Thermodynamics:

$q_{in} = h_{g} - h_{mix}$

The specific enthalpies are:

Liquid-Vapor Mixture:

$h_{mix} = 2217.2\,\frac{kJ}{kg}$

Saturated Vapor:

$h_{g} = 2745.8\,\frac{kJ}{kg}$

The thermal energy per unit mass required to heat the steam is:

$q_{in} = 2745.8\,\frac{kJ}{kg} - 2217.2\,\frac{kJ}{kg}$

$q_{in} = 528.6\,\frac{kJ}{kg}$

7 0

3 years ago

1. An automobile travels along a straight road at 15.65 m/s through a 11.18 m/s

djyliett [7]

Answer:

a.) Time = 17.13 seconds

b.) 31.88 m

c.) V = 11.18 m/s

d.) V = 7.1 m/s

Explanation:

The initial velocity U of the automobile is 15.65 m/s.

At the instant that the two vehicles are abreast of each other, the police car starts to pursue the automobile with initial velocity U = 0 at a constant acceleration of 1.96 m/s². Because the police is starting from rest.

For the automobile, let us use first equation of motion

V = U - at.

Acceleration a is negative since it is decelerating with a = 3.05 m/s² . And

V = 0.

Substitute U and a into the formula

0 = 15.65 - 3.05t

15.65 = 3.05t

t = 15.65/3.05

t = 5.13 seconds

But the motorist noticed the police car in his rear view mirror 12 s after the police car started the pursuit and applied his brakes and decelerates at 3.05 m/s².

The total time required for the police car to overtake the automobile will be

12 + 5.13 = 17.13 seconds.

b.) Using the third equation of motion formula for the police car at V = 11.18 m/s and a = 1.96 m/s²

V^2 = U^2 + 2aS

Where S = distance travelled.

Substitute V and a into the formula

11.18^2 = 0 + 2 × 1.96 ×S

124.99 = 3.92S

S = 124.99/3.92

S = 31.88 m

c.) The speed of the police car at the time it overtakes the automobile will be in line with the speed zone which is 11.18 m/s

d.) That will be the final velocity V of the automobile car.

We will use third equation of motion to solve that.

V^2 = U^2 + 2as

V^2 = 15.65^2 - 2 × 3.05 × 31.88

V^2 = 244.9225 - 194.468

V = sqrt( 50.4545)

V = 7.1 m/s

8 0

3 years ago

Consider a 1.5-m-high and 2.4-m-wide glass window whose thickness is 6 mm and thermal conductivity is k = 0.78 W/m⋅K. Determine

Bess [88]

Answer:

The steady rate of heat transfer through the glass window is 707.317 watts.

Explanation:

A figure describing the problem is included below as attachment. From First Law of Thermodynamics we get that steady rate of heat transfer through the glass window is the sum of thermal conductive and convective heat rates, all measured in watts:

$\dot Q_{total} = \dot Q_{cond} + \dot Q_{conv, in} + \dot Q_{conv, out}$ (Eq. 1)

Given that window is represented as a flat element, we can expand (Eq. 1) as follows:

$\dot Q_{total} = \frac{T_{i}-T_{o}}{R}$ (Eq. 2)

Where:

$T_{i}$ , $T_{o}$ - Indoor and outdoor temperatures, measured in Celsius.

$R$ - Overall thermal resistance, measured in Celsius per watt.

Now, we know that glass window is configurated in series and overall thermal resistance is:

$R = R_{cond} + R_{conv, in}+R_{conv, out}$ (Eq. 3)

Where:

$R_{cond}$ - Conductive thermal resistance, measured in Celsius per watt.

$R_{conv, in}$ , $R_{conv, out}$ - Indoor and outdoor convective thermal resistances, measured in Celsius per watt.

And we expand the expression as follows:

$R = \frac{l}{k\cdot w\cdot d} + \frac{1}{h_{i}\cdot w\cdot d} + \frac{1}{h_{i}\cdot w\cdot d}$

$R = \frac{1}{w\cdot d}\cdot \left(\frac{l}{k}+\frac{1}{h_{i}}+\frac{1}{h_{o}} \right)$ (Eq. 4)

Where:

$w$ - Width of the glass window, measured in meters.

$d$ - Length of the glass window, measured in meters.

$l$ - Thickness of the glass window, measured in meters.

$k$ - Thermal conductivity, measured in watts per meter-Celsius.

$h_{i}$ , $h_{o}$ - Indoor and outdoor convection coefficients, measured in watts per square meter-Celsius.

If we know that $w = 2.4\,m$ , $d = 1.5\,m$ , $l = 0.006\,m$ , $k = 0.78\,\frac{W}{m\cdot ^{\circ}C}$ , $h_{i} = 10\,\frac{W}{m^{2}\cdot ^{\circ}C}$ and $h_{o} = 25\,\frac{W}{m^{2}\cdot ^{\circ}C}$ , the overall thermal resistance is:

$R = \left[\frac{1}{(2.4\,m)\cdot (1.5\,m)}\right] \cdot \left(\frac{0.006\,m}{0.78\,\frac{W}{m\cdot ^{\circ}C} }+\frac{1}{10\,\frac{W}{m^{2}\cdot ^{\circ}C} }+\frac{1}{25\,\frac{W}{m^{2}\cdot ^{\circ}C} } \right)$

$R = 0.041\,\frac{^{\circ}C}{W}$

Now, we obtain the steady rate of heat transfer from (Eq. 2): ( $R = 0.041\,\frac{^{\circ}C}{W}$ , $T_{i} = -5\,^{\circ}C$ , $T_{o} = 24\,^{\circ}C$ )