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Ray Of Light [21]

2 years ago

6

Let A be the last two digits and let B be the sum of the last three digits of your 8-digit student ID. (Example: For 20245347, A

= 47 and B = 14) In a remote civilization, distance is measured in urks and an hour is divided into 125 time units named dorts. The length conversion is 1 urk = 58.0 m. Consider a speed of (25.0 + A + B) urks/dort. Convert this speed to meters per second (m/s). Round your final answer to 3 significant figures.

1 answer:

zheka24 [161]2 years ago

7 0

Answer:

The speed is 173 m/s.

Explanation:

Given that,

A = 47

B = 14

Length 1 urk = 58.0 m

An hour is divided into 125 time units named dorts.

3600 s = 125 dots

dorts = 28.8 s

Speed v= (25.0+A+B) urks/dort

We need to convert the speed into meters per second

Put the value of A and B into the speed

$v=25.0+47+14$

$v =86\ urk s/dort$

$v=86\times\dfrac{58.0}{28.8}$

$v=173.19\ m/s$

Hence, The speed is 173 m/s.

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Steam at 0.6 MPa, 200 oC, enters an insulated nozzle with a velocity of 50 m/s. It leaves at a pressure of 0.15 MPa and a veloci

Rudiy27

Answer:

x2 = 0.99

Explanation:

from superheated water table

at pressure p1 = 0.6MPa and temperature 200 degree celcius

h1 = 2850.6 kJ/kg

From energy equation we have following relation

$\dot m( h1+\frac{v1^2}{2}+ gz1 )+ Q = \dot m( h2+\frac{v2^2}{2}+ gz1) + W$

$\dot m( h1+\frac{v1^2}{2}) = \dot m( h2+\frac{v2^2}{2})$

$h1+\frac{v1^2}{2} = h2+\frac{v2^2}{2}$

$2850.6 + [\frac{50^2}{2} * \frac{1 kJ/kg}{1000 m^2/S^2}] = h2 +[ \frac{600^2}{2} * \frac{1 kJ/kg}{1000 m^2/S^2}]$

h2 = 2671.85 kJ/kg

from superheated water table

at pressure p2 = 0.15MPa

specific enthalpy of fluid hf = 467.13 kJ/kg

enthalpy change hfg = 2226.0 kJ/kg

specific enthalpy of the saturated gas hg = 2693.1 kJ/kg

as it can be seen from above value hf>h2>hg, so phase 2 is two phase region. so we have

quality of steam x2

h2 = hf + x2(hfg)

2671.85 = 467.13 +x2*2226.0

x2 = 0.99

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2 years ago

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Strike441 [17]

Answer:

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Explanation:

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2 years ago

A sample of an ideal gas is in a tank of constant volume. The sample absorbs heat energy so that its temperature changes from 38

oksano4ka [1.4K]

Answer:

the ratio is $\frac{V_2}{V_1}=\sqrt{2}$

Explanation:

Given

$Initial Temperature T_1=387 KFinal Temperature T_2=774 K$

The RMS velocity of molecules in a gas is given by

$V_{rms}=\sqrt{\dfrac{3k_bT}{m}}$

where T=temperature

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For T = 387K

$V_1=\sqrt{\frac{3k_b\cdot 387}{m}}----1$

For T = 774

$V_2=\sqrt{\frac{3k_b\cdot 774}{m}}----(2)$

dividing eqn 1 and eqn 2

$\frac{V_2}{V_1}=\sqrt{\frac{774}{387}}$

$\frac{V_2}{V_1}=\sqrt{2}$

Thus,the ratio is $\frac{V_2}{V_1}=\sqrt{2}$

5 0

2 years ago

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A nonuniform beam 4.50 m long and weighing 1.40 kN makes an angle of 25.0° below the horizontal. It is held in position by a fri

liubo4ka [24]

Answer:

$T = 7.64 kN$

$F_y = 0.52 kN$ (Downwards)

$F_x = 3.23 kN$ (Towards Left)

Explanation:

As we know that beam is in equilibrium

So here we can use torque balance as well as force balance for the beam

Now by torque balance equation at the pivot we can say

$F(4.50 cos\theta) + mg(2cos\theta) = T \times 3$

As we know that

mg = 1.40 kN

F = 5 kN

so we will have

$5 kN(4.50 cos25) + 1.40 kN(2 cos25) = 3 T$

$T = 7.64 kN$

Now force balance in vertical direction

$F + mg = Tsin65 + F_y$

$5 + 1.40 = 7.64 sin65 + F_y$

$F_y = 0.52 kN$ (Downwards)

Force balance in horizontal direction

$F_x = T cos65$

$F_x = 7.64 cos65$

$F_x = 3.23 kN$ (Towards Left)

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