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2 years ago

Calculate the pressure in a 212 liter tank containing 23.3 kg of argon gas at 25°c

1 answer:

7 0

First of all, let's find the number of moles of the gas.

The molar mass of argon is $M_m=40 g/mol=0.40 kg/mol$ . Since we have $m=23.3 kg$ of gas, the number of moles is
$n= \frac{m}{M_m}= \frac{23.3 kg}{0.40 kg/mol}=58.3 mol$

Now we can use the ideal gas law to calculate the pressure of the gas:
$pV=nRT$
where
p is the pressure
$V=212 L=0.212 m^3$ is the volume
$n=58.3 mol$ is the number of moles
$R=8.31 J/mol K$ is the gas constant
$T=25^{\circ}+273=298 K$ is the absolute temperature

Rearranging the equation, we find
$p= \frac{nRT}{V}= \frac{(58.3 mol)(8.31 J/mol K)(298 K)}{0.212 m^3}=6.81 \cdot 10^5 Pa$

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In a circus act, an acrobat rebounds upward from the surface of a trampoline at the exact moment that another acrobat, perched 9

slega [8]

Answer:

1.6 secs

Explanation:

In a circus act, an acrobat upwards from the surface of a trampoline

At that same moment another acrobat perched 9.0m above him

A ball is released from rest

While still in motion the acrobat catches the ball

He left the ball with a trampoline of 5.6m/s

Since the ball is falling downwards from a distance then acceleration will be negative

a= -9.8

s= d

s= 1/2at^2

= 1/2 × (-9.8)t^2

= 0.5× (-9.8)t^2

d = -4.9t^2

Therefore the time the acrobat stays in the air before catching the ball can be calculated as follows

9 - 4.9t^2= 5.6t + 1/2(-9.8)t^2

9 - 4.9t^2= 5.6t + (-4.9)t^2

9 - 4.9t^2= 5.6t - 4.9t^2

9= 5.6t

t= 9/5.6

t= 1.6 secs

6 0

2 years ago

Lilli suggests that they explore the simulation starting with varying only a single parameter in order to understand the role of

mrs_skeptik [129]

Answer:

Explanation:

One of the ways to address this issue is through the options given by the statement. The concepts related to the continuity equation and the Bernoulli equation.

Through these two equations it is possible to observe the behavior of the fluid, specifically the velocity at a constant height.

By definition the equation of continuity is,

$A_1V_1=A_2V_2$

In the problem $A_2$ is $2A_1$ , then

$A_1V_1=2A_1V_2$

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

<em>Here we can conclude that by means of the continuity when increasing the Area, a decrease will be obtained - in the diminished times in the area - in the speed.</em>

For the particular case of Bernoulli we have to

$P_1 + \frac{1}{2}\rho V_1^2 = P_2 +\frac{1}{2}\rho V_2^2$

$P_2-P_1 = \frac{1}{2} \rho (V_1^2-V_2^2)$

For the previous definition we can now replace,

$P_2-P_1 = \frac{1}{2} \rho (V_1^2-(\frac{V_1}{2})^2)$

$\Delta P = \frac{3}{8} \rho V_1^2$

<em>Expressed from Bernoulli's equation we can identify that the greater the change that exists in pressure, fluid velocity will tend to decrease</em>

The correct answer is B: "If we increase A2 then by the continuity equation the speed of the fluid should decrease. Bernoulli's equation then shows that if the velocity of the fluid decreases (at constant height conditions) then the pressure of the fluid should increase"

4 0

2 years ago

refrigerant 134a enters a compressor operating at steady state as saturated vapor at 0.12 MPa and exits at 1.2 MPa and 70 C at a

Afina-wow [57]

Answer:

the power input to the compressor is 7.19Kw

Explanation:

Hello!

To solve this problem follow the steps below.

1. We will call 1 the refrigerant state at the compressor inlet and 2 at the outlet.

2. We use thermodynamic tables to determine enthalpies in states 1 and 2.

(note: Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)

through prior knowledge of two other properties such as pressure and temperature. )

h1[quality=1, P=0.12Mpa)=237KJ/Kg

h2(P=1.2Mpa, t=70C)=300.6KJ/kg

3. uses the first law of thermodynamics in the compressor that states that the energy that enters a system is the same that must come out

Q=heat=0.32kJ/s

W=power input to the compressor

m=mass flow=0.108kg/S

m(h1)+W=Q+m(h2)

solving for W

W=Q+m(h2-h1)

W=0.32+0.108(300.6-237)=7.19Kw

the power input to the compressor is 7.19Kw

7 0

2 years ago

A girl rolls a ball up an incline and allows it to re- turn to her. For the angle and ball involved, the acceleration of the bal

zalisa [80]

Answer:

3.28 m

3.28 s

Explanation:

We can adopt a system of reference with an axis along the incline, the origin being at the position of the girl and the positive X axis going up slope.

Then we know that the ball is subject to a constant acceleration of 0.25*g (2.45 m/s^2) pointing down slope. Since the acceleration is constant we can use the equation for constant acceleration:

X(t) = X0 + V0 * t + 1/2 * a * t^2

X0 = 0

V0 = 4 m/s

a = -2.45 m/s^2 (because the acceleration is down slope)

Then:

X(t) = 4*t - 1.22*t^2

And the equation for speed is:

V(t) = V0 + a * t

V(t) = 4 - 2.45 * t

If we equate this to zero we can find the moment where it stops and begins rolling down, that will be the highest point:

0 = 4 - 2.45 * t

4 = 2.45 * t

t = 1.63 s

Replacing that time on the position equation:

X(1.63) = 4 * 1.63 - 1.22 * 1.63^2 = 3.28 m

To find the time it will take to return we equate the position equation to zero:

0 = 4 * t - 1.22 * t^2

Since this is a quadratic equation it will have to answers, one will be the moment the ball was released (t = 0), the other will eb the moment when it returns:

0 = t * (4 - 1.22*t)

t1 = 0

0 = 4 - 1.22*t2

1.22 * t2 = 4

t2 = 3.28 s

7 0

2 years ago

A diver shines light up to the surface of a flat glass-bottomed boat at an angle of 30° relative to the normal. If the index of

son4ous [18]

Answer:

Explanation:

According to snell's law which states that the ratio of the sin of incidence (i) to the angle of refraction(n) is a constant for a given pair of media.

sini/sinr = n

n is the constant = refractive index

Since the diver shines light up to the surface of a flat glass-bottomed boat, the refractive index n = nw/ng

nw is the refractive index of water and ng is that of glass

sini/sinr = nw/ng

given i = 30°, nw = 1.33, ng = 1.5, r = angle the light leave the glass

On substitution;

sin 30/sinr = 1.33/1.5

1.5sin30 = 1.33sinr

sinr = 1.5sin30/1.33

sinr = 0.75/1.33

sinr = 0.5639

r = arcsin0.5639

r ≈35°

angle the light leave the glass is 35°

7 0

2 years ago