I'm really bad at measurements so I don't understand this.

What is the volume of the rectangular prism shown.

Likurg_2 [28]

It would be 72cm bc u need to add up all the line in the back to

7 0

2 years ago

A spring-loaded piston-cylinder contains 1 kg of carbon dioxide. This system is heated from 104 kPa and 25 °C to 1,068 kPa and 3

labwork [276]

Answer:

Q = -68.859 kJ

Explanation:

given details

mass $co_2 = 1 kg$

initial pressure P_1 = 104 kPa

Temperature T_1 = 25 Degree C = 25+ 273 K = 298 K

final pressure P_2 = 1068 kPa

Temperature T_2 = 311 Degree C = 311+ 273 K = 584 K

we know that

molecular mass of $co_2 = 44$

R = 8.314/44 = 0.189 kJ/kg K

c_v = 0.657 kJ/kgK

from ideal gas equation

PV =mRT

$V_1 = \frac{m RT_1}{P_1}$

$=\frac{1*0.189*298}{104}$

$V_1 = 0.5415 m3$

$V_2 = \frac{m RT_2}{P_2}$

$=\frac{1*0.189*584}{1068}$

$V_1 = 0.1033 m3$

WORK DONE

$W =P_{avg}*{V_2-V_1}$

w = 586*(0.1033 -0.514)

W =256.76 kJ

INTERNAL ENERGY IS

$\Delta U = m *c_v*{V_2-V_1}$

$\Delta U = 1*0.657*(584-298)$

$\Delta U =187.902 kJ$

HEAT TRANSFER

$Q = \Delta U +W$

= 187.902 +(-256.46)

Q = -68.859 kJ

7 0

3 years ago

Which task best fits the role of a planning engineer?

Nonamiya [84]

Answer:

D

Explanation:

ensuring project end on time through carefully planning and organizing

8 0

2 years ago

6) A deep underground cavern Contains 980 cuft

Elza [17]

Answer:

15625 moles of methane is present in this gas deposit

Explanation:

As we know,

PV = nRT

P = Pressure = 230 psia = 1585.79 kPA

V = Volume = 980 cuft = 27750.5 Liters

n = number of moles

R = ideal gas constant = 8.315

T = Temperature = 150°F = 338.706 Kelvin

Substituting the given values, we get -

1585.79 kPA * 27750.5 Liters = n * 8.315 * 338.706 Kelvin

n = (1585.79*27750.5)/(8.315 * 338.706) = 15625

3 0

2 years ago

A student takes 60 voltages readings across a resistor and finds a mean voltage of 2.501V with a sample standard deviation of 0.

Lena [83]

Answer:

There are 2 expected readings greater than 2.70 V

Solution:

As per the question:

Total no. of readings, n = 60 V

Mean of the voltage, $\mu = 2.501 V$

standard deviation, $\sigma = 0.113 V$

Now, to find the no. of readings greater than 2.70 V, we find:

The probability of the readings less than 2.70 V, $P(X\leq 2.70)$ :

$z = \frac{x - \mu}{\sigma} = \frac{2.70 - 2.501}{0.113} = 1.761$

Now, from the Probability table of standard normal distribution:

$P(z\leq 1.761) = 0.9608$

Now,

$P(X\geq 2.70) = 1 - P(X\leq 2.70) = 1 - 0.9608 = 0.0392 = 3.92%$

Now, for the expected no. of readings greater than 2.70 V:

$P(X\geq 2.70) = \frac{No.\ of\ readings\ expected\ to\ be\ greater\ than\ 2.70\ V}{Total\ no.\ of\ readings}$

No. of readings expected to be greater than 2.70 V = $P(X\geq 2.70)\times Total\ no.\ of\ readings$

No. of readings expected to be greater than 2.70 V = $0.0392\times 60 = 2.352$ ≈ 2

7 0

3 years ago