Which of these elements is unlikely to have a reaction with any element or compound?

Air as an ideal gas enters a diffuser operating at steady state at 5 bar, 280 K with a velocity of 510 m/s. The exit velocity is

Nataly [62]

Answer:

Explanation:

Calculating the exit temperature for K = 1.4

The value of $c_p$ is determined via the expression:

$c_p = \frac{KR}{K_1}$

where ;

R = universal gas constant = $\frac{8.314 \ J}{28.97 \ kg.K}$

k = constant = 1.4

$c_p = \frac{1.4(\frac{8.314}{28.97} )}{1.4 -1}$

$c_p= 1.004 \ kJ/kg.K$

The derived expression from mass and energy rate balances reduce for the isothermal process of ideal gas is :

$0=(h_1-h_2)+\frac{(v_1^2-v_2^2)}{2}$ ------ equation(1)

we can rewrite the above equation as :

$0 = c_p(T_1-T_2)+ \frac{(v_1^2-v_2^2)}{2}$

$T_2 =T_1+ \frac{(v_1^2-v_2^2)}{2 c_p}$

where:

$T_1 = 280 K \\ \\ v_1 = 510 m/s \\ \\ v_2 = 120 m/s \\ \\c_p = 1.0004 \ kJ/kg.K$

$T_2= 280+\frac{((510)^2-(120)^2)}{2(1.004)} *\frac{1}{10^3}$

$T_2 = 402.36 \ K$

Thus, the exit temperature = 402.36 K

The exit pressure is determined by using the relation: $\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{k}{k-1}$

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

$P_2 = 5 (\frac{402.36}{280} )^\frac{1.4}{1.4-1}$

$P_2 = 17.79 \ bar$

Therefore, the exit pressure is 17.79 bar

7 0

3 years ago

what is the angular speed w of the system immediately after the collision in terms f the sstem parameters and I

denis23 [38]

Answer: hello some part of your question is missing attached below is the missing detail

answer :

<em>w</em>f = M( v cos∅ )D / I

Explanation:

The Angular speed <em>wf </em>of the system after collision in terms of the system parameters and I can be expressed as

considering angular momentum conservation

Li = Lf

M( v cos∅ ) D = ( ML^2 / 3 + mD^2 ) <em>w</em>f

where ; ( ML^2 / 3 + mD^2 ) = I ( Inertia )

In terms of system parameters and I

<em>w</em>f = M( v cos∅ )D / I

5 0

3 years ago

To balance a chemical equation we use______. <br> -coefficients <br> -subscripts

Vitek1552 [10]

Answer:

Coefficients

Explanation:

Coefficients are used to balance chemical equations.
According to the law of conservation of mass, the mass of the reactants in a chemical equation should be equivalent to the mass of the products.
The conservation of mass is achieved in chemical equations is achieved through balancing chemical equations.
<em><u>Balancing chemical equations is a try and error of putting appropriate coefficients to the reactants or products to ensure that the number of atoms of each element are equal on the side of the reactants and that of products. </u></em>

3 0

3 years ago

David rowed a boat upstream for three miles and then returned to point he started from. The entire journey took four hours. The

frez [133]

Upstream speed = S - 1
Downstream speed = S + 1

Average speed = total distance / total time

Average speed = (S - 1) + (S + 1) / 2
= S

S = 6 miles / 4 hours
S = 1.5 miles per hour

4 0

3 years ago

A block of mass 1.5 hangs at the of end of a weight cord suspended from the ceiling.what is the tension in the cord, and with wh

Len [333]

The tension in the cord is 14.7 N and the force of pull of the cord is 14.7 N, assuming the block is stationary.

<h3>What is the tension in the cord?</h3>

The tension in the cord is calculated as follows;

T = ma + mg

where;

a is the acceleration of the block
g is acceleration due to gravity
m is mass of the block

T = m(a + g)

T = 1.5(a + 9.8)

T = 1.5a + 14.7

Thus, the tension in the cord is (1.5a + 14.7) N.

If the block is at rest, the tension is 14.7 N.

<h3>Force of the force</h3>

The force with which the cord pulls is equal to the tension in the cord

F = T = m(a + g)

F = (1.5a + 14.7) N

If the block is stationary, a = 0, the tension and force of pull of the cord = 14.7 N.

Thus, the tension in the cord is 14.7 N and the force of pull of the cord is 14.7 N, assuming the block is stationary.

Learn more about tension here: brainly.com/question/187404

#SPJ1

4 0

1 year ago