HELP URGENT WILL GIVE BRAINLIEST

seropon [69]

The slowest line is the solid line and the fastest is the dotted line that crosses the solid line
for future reference you just need to find the slope or the line which is traveling most vertical

8 0

3 years ago

What is the dimension of permeance

Dennis_Churaev [7]

Answer:

metric dimensions, i think

Explanation:

8 0

3 years ago

PLEASE HELP MEEE!!! I WILL MARK THE BRAINLIEST!!! ANYBODY PLEASE?!

disa [49]

Answer:

Gases: Oxygen and Carbon dioxide

Liquids: Water

Solid: Most Metals

6 0

3 years ago

Read 2 more answers

Neon is compressed from 100 kPa and 24°C to 500 kPa in an isothermal compressor. Determine the change in the specific volume and

trasher [3.6K]

Answer:

ΔV = -0.97 m³/ kg

ΔH = 0 kJ/ kg

Explanation:

To determine the change in the specific volume we need to use the ideal gas law:

$PV = RT$

where P: pressure of the gas V: volume of the gas, R: ideal gas constant= 0.4119 kJ/kg.K = 0.4119 kPa.m³/kg.K and T: temperature of the gas.

The V₁, at a compressed pressure is:

$V_{1}= \frac {RT}{P_{1}}$

$V_{1}= \frac {0.4119 \frac{kPa\cdot m^{3}}{kg\cdot K} \cdot (24 + 273 K)}{100 kPa}$

$V_{1}= 1.22 \frac{m^{3}}{kg}$

Similarly, the V₂ is:

$V_{2}= \frac {RT}{P_{2}}$

$V_{2}= \frac {0.4119 \frac{kPa\cdot m^{3}}{kg\cdot K} \cdot (24 + 273 K)}{500 kPa}$

$V_{2}= 0.25 \frac{m^{3}}{kg}$

Now, the change in the specific volume because the compressor is:

$V_{2} - V_{1} = 0.25 - 1.22 \frac{m^{3}}{kg}$

$V_{2} - V_{1} = -0.97 \frac{m^{3}}{kg}$

Finally, to calculate the change in the specific enthalpy, we need to remember that neon is an ideal gas and that is an isothermal process:

$\Delta H = C_{p} \cdot \Delta T$

$\Delta H = 1.0299 \frac{kJ}{kg \cdot K} \cdot 0$

$\Delta H = 0 \frac{kJ}{kg}$

Have a nice day!

7 0

4 years ago

The mass of Object 2 is double the mass of Object 5. The mass of Object 4 is half of the mass of Object 5 and the mass of Object

SVETLANKA909090 [29]

This is a great problem if you like getting tied up in knots
and making smoke come out of your brain.

I found that it makes the problem a lot easier if I give the objects some
numbers. I'm going to say that the mass of Object 5 is 20 clods.

Let the mass of Mass of Object 5 be 20 clods .

Then . . .

-- The mass of Object 2 is double the mass of Object 5 = 40 clods.

-- The mass of Object 4 is half of the mass of Object 5 = 10 clods.
and
-- the mass of Object 3 is half of the mass of Object 4 = 5 clods.

So now, here are the masses:

Object #1 . . . . . unknown
Object #2 . . . . . 40 clods
Object #3 . . . . . 5 clods
Object #4 . . . . . 10 clods
Object #5 . . . . . 20 clods .

Now let's check out the statements, and see how they stack up:

Choice-A:
Object 3 and Object 5 exert the same gravitational force on Object 1.
Can't be.
Objects #3 and #5 have different masses, so they can't both
exert the same force on the same mass.

Choice-B.
Object 2 and Object 4 exert the same gravitational force on Object 1.
Can't be.
Objects #2 and #4 have different masses, so they can't both
exert the same force on the same mass.

Choice-C.
The gravitational force between Object 1 and Object 2 is greater than
the gravitational force between Object 1 and Object 4.
Yes ! Yay !
Object-2 has more mass than Object-4 has, so it must exert more force on
ANYTHING than Object-4 does, (as long as the distances are the same).

Choice-D.
The gravitational force between Object 1 and Object 3 is greater than the gravitational force between Object 1 and Object 5.
Can't be.
Object-3 has less mass than Object-5 has, so it must exert less force on
ANYTHING than Object-4 does, (as long as the distances are the same).

Conclusion:
If the DISTANCE is the same for all the tests, then Choice-C is
the only one that can be true.

8 0

3 years ago