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Radda [10]
3 years ago
12

Train levels a station and travels north at 60km/hr. Two hours later, a second train leaves on a parallel track and travels nort

h at 100km/hr. How far from the station will they meet
Engineering
1 answer:
Amanda [17]3 years ago
6 0

Answer: they would be 300 miles from the station.

Explanation:

At the point where both trains meet, they would have covered the same distance.

Let t represent the time spent by the first train in covering this distance.

Distance = speed × time

The first train leaves the station and travels north at 60km/hr.

Distance covered by the first train is

60 × t = 60t

Two hours later, a second train leaves on a parallel track and travels north at 100km/hr. Time spent by the second train in covering this distance is (t - 2) hours

Distance covered by the second train is

100(t - 2) = 100t - 200

Since both trains covered the same distance, then

100t - 200 = 60t

100t - 60t = 200

40t = 200

t = 200/40

t = 5 hours

The distance that they would be from the station is

60 × 5 = 300 miles

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Find the Rectangular form of the following phasors?
almond37 [142]

Answer:

The angles are missing in the question.

The angles are :

45,     30,    60,     90,    -34,     -56,      20,     -42,  -65,    -15

P=10, P=5,  P=25, P=54, P=65, P=95, P=250, P=8, P=35, P=150

Explanation:

1. P = 10,   θ = 45°  rectangular coordinates

x = r cosθ  ,   y = r sinθ

So, rectangular form is x + iy

x = P cosθ = 10 cos 45°

  = 7.07

y =P sinθ = 10 sin 45°

  = 7.07

Therefore, rectangular form

x + iy = 7.07 + i (7.07)

2. P = 5 , θ = 30°

x = 5 cos  30° = 4.33

y = 5 sin  30° = 2.5

So, (x+iy) = 4.33 + i (2.5)

3. P = 25 , θ = 60°

x = 25 cos  60° = 12.5

y = 25 sin  60° = 21.65

So, (x+iy) = 12.5 + i (21.65)

4. P = 54 , θ = 90°

x = 54 cos  90° = 0

y = 54 sin  90° = 54

So, (x+iy) = 0+ i (54)

5. P = 65 , θ = -34°

x = 65 cos  (-34°) = 53.88

y = 65 sin  (-34°) = -36.34

So, (x+iy) = 53.88 - i (36.34)

6. P = 95 , θ = -56°

x = 95 cos  (-56)° = 53.12

y = 95 sin  (-56)° = -78.75

So, (x+iy) = 53.12 - i (78.75)

7. P = 250 , θ = 20°

x = 250 cos  20° = 234.92

y = 250 sin 20° = 85.5

So, (x+iy) = 234.92 + i (85.5)

8. P = 8 , θ = (-42)°

x = 8 cos  (-42)° = 5.94

y = 8 sin  (-42)° = -5.353

So, (x+iy) = 5.94 - i (5.353)

9. P = 35 , θ = (-65)°

x = 35 cos  (-65)° = 14.79

y = 35 sin  (-65)° = -31.72

So, (x+iy) = 14.79 - i (31.72)

10. P = 150 , θ = (-15)°

x = 150 cos  (-15)° = 144.88

y = 150 sin  (-15)° = -38.82

So, (x+iy) = 144.88 - i (38.82)

6 0
2 years ago
Check the level of motor oil in your engine by ?
Usimov [2.4K]

Answer:

cpct gvxjjxjhdfjokjdzfjiyddzzsjhxf

6 0
1 year ago
Water vapor at 1.0 MPa, 300°C enters a turbine operating at steady state and expands to 15 kPa. The work developed by the turbin
Andre45 [30]

Answer:

a) isentropic efficiency = 84.905%

b) rate of entropy generation = .341 kj/(kg.k)

Please kindly see explaination and attachment.

Explanation:

a) isentropic efficiency = 84.905%

b) rate of entropy generation = .341 kj/(kg.k)

The Isentropic efficiency of a turbine is a comparison of the actual power output with the Isentropic case.

Entropy can be defined as the thermodynamic quantity representing the unavailability of a system's thermal energy for conversion into mechanical work, often interpreted as the degree of disorder or randomness in the system.

Please refer to attachment for step by step solution of the question.

5 0
3 years ago
The steel bracket is used to connect the ends of two cables. if the allowable normal stress for the steel is sallow = 30 ksi, de
garri49 [273]

The largest tensile force that can be applied to the cables given a rod with diameter 1.5 is 2013.15lb

<h3>The static equilibrium is given as:</h3>

F = P (Normal force)

Formula for moment at section

M = P(4 + 1.5/2)

= 4.75p

Solve for the cross sectional area

Area = \frac{\pi d^{2} }{4}

d = 1.5

\frac{\pi *1.5^{2} }{4}

= 1.767 inches²

<h3>Solve for inertia</h3>

\frac{\pi *0.75^4}{4}

= 0.2485inches⁴

Solve for the tensile force from here

\frac{F}{A} +\frac{Mc}{I}

30x10³ = \frac{P}{1.767} +\frac{4.75p*0.75}{0.2485} \\\\

30000 = 14.902 p

divide through by 14.902

2013.15 = P

The largest tensile force that can be applied to the cables given a rod with diameter 1.5 is 2013.15lb

Read more on tensile force here: brainly.com/question/25748369

4 0
2 years ago
A thick steel slab (rho= 7800 kg/m3 , cp= 480 J/kg K, k= 50 W/m K) is initially at 300 °C and is cooled by water jets impinging
dimaraw [331]

Answer:

t = 2244.3 sec

Explanation:

calculate the thermal diffusivity

\alpha = \frac{k}{\rho c}

           = \frac{50}{7800\times 480} = 1.34 \times 10^{-5} m^2/s

                   

Temperature at 28 mm distance after t time  = =  50 degree C

we know that

\frac[ T_{28} - T_s}{T_i -T_s} = erf(\frac{x}{2\sqrt{at}})

\frac{ 50 -25}{300-25} = erf [\frac{28\times 10^{-3}}{2\sqrt{1.34\times 10^{-5}\times t}}]

0.909 = erf{\frac{3.8245}{\sqrt{t}}}

from gaussian error function table , similarity variable w calculated as

erf w = 0.909

it is lie between erf w = 0.9008  and erf w = 0.11246 so by interpolation we have

w = 0.08073

erf 0.08073 = erf[\frac{3.8245}{\sqrt{t}}]

0.08073 = \frac{3.8245}{\sqrt{t}}

solving fot t we get

t = 2244.3 sec

3 0
2 years ago
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