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Len [333]
3 years ago
13

Kuhygtfchvgbjkniljuhygutyfvghbjkljhbvghuijohbiuljkhbjiuoljkhftgyjhvgjk

Physics
1 answer:
Lisa [10]3 years ago
5 0

Answer:

....

Explanation:

trrkytl8f5vob779y

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A piston–cylinder device with a set of stops initially contains 0.6 kg of steam at 1.0 MPa and 400°C. The location of the stops
Ilya [14]

Answer:

(a) Compression work at the final state with a pressure of 1(MPa) is: 44.32(KJ), (b) Compression work at the final state with a pressure of 500(KPa): 110.37(KJ) and (c) temperaure of the final state in part b: T=151.83(°C).  

Explanation:

Remember that the substance is steam so it's water (H2O) and the initial conditions are P_{1} =1MPa, T_{1}=400^{0}C, m=0.6Kg andv_{2} =0.4v_{1} from a saturated water table and the initial conditions we can determine that the state phase is superheated (see Table 1 attached) because the T_{sat}=179.88^{0} C \leq T_{1} from the table 1 we get:v_{1} =0.30661(m^{3}/Kg). Now we have second conditions as: P_{2}=1(MPa), T_{2}=250^{0}C so from the same table we can see the state still superheated and we getv_{2}=0.23275(m^{3}/Kg), knowing that it's a isobaric process we can find the compression's work as:W_{b}=m*P(v_{2}-v_{1})=0.6*1000*(0.23275-0.30661)=-44.32(KJ) so the compressor's work is: 44.32(KJ). (b) Then the piston reaches the stop and there are two processes in this stage, so Process 1 is isobaric and:W_{1}=m*P*(v_{2}-v_{1}) =0.6*1000*(0.4*0.30661-0.30661)=-110.38(KJ) and the second process is isochoric:W_{2}=zero,nowW_{b}=W_{1}+ W_{2} =110.38+0=110.38(KJ). Finally to get the temperarure at the final state in part (b) we get:v_{2} =0.4v_{1} =0.4*0.30661=0.122644(m^{3}/Kg), P_{2}=500(KPa) from table 2 (see attached) we comparev_{f} andv_{g} at the saturated water table and find the following:v_{f}=0.001093(m^{3}/Kg), so we know that the final state phase is a satured mixture and we get the temperature at the final state as:T_{2} =T_{sat} =151.83^{0}C.

3 0
4 years ago
Researchers recorded the forces over a short period of time (0.065 s) for three different airbag designs as well as one crash wi
kotegsom [21]

Answer:

Design 2

Explanation:

I had this same question

My answer:

"Design 2 is the well-designed one, because the air molecules are the most compact and could protect the individual better than 1,3,and no air bag."

6 0
4 years ago
PLZZZ HELP 100 POINTS WILL MARK BRAINELST
Kitty [74]

Explanation:

Crust...molten

a. Oceanic, iron

b. Continental, silicates

c. less

3. Mantle, Denser

a. Lithosphere

b. Asthenosphere

4. Core

a. elements, rocks

b. liquid, magnetic

(I guess the liquid should come after the is)

Couldn't answer all but wanted to help

3 0
3 years ago
Read 2 more answers
What is, by far, the most abundant component of Earth’s atmosphere?
Serjik [45]

Answer:

Molecular Nitrogen

Explanation:

Earth's atmosphere is a mixture of several gases in their molecular form . The most abundant of them is the element nitrogen. 78% of it is nitrogen and the rest that follow are Oxygen, Carbon dioxide and a mixture of other gases like Neon, Helium, Methane, Water vapor. hen the air is dry, argon is the third most abundant element in the earth's atmosphere.

As the height above the earth's surface increases, the air molecules decrease.

5 0
3 years ago
The specific heat of soil is 0.20 kcal/kg*C and the specific heat of water is 1.00 kcal/kg*C. This means that if 1 kg of soil an
stiv31 [10]

Answer: The soil will be 4\°C warmer than the water.

Explanation:

The heat (thermal energy) absorbed can be found using the following equation:

Q=m.C.\Delta T

Where:

Q is the heat  

m is the mass of the element

C is the specific heat capacity of the material.

\Delta T is the variation in temperature

<u>In the case of soil we have:</u>

Q_{soil}=m_{soil}.C_{soil}.\Delta T_{soil} (1)

Where:

Q_{soil}=1 kcal

m_{soil}=1 kg

C_{soil}=0.2 kcal/kg \°C

\Delta T_{soil}

<u>In the case of water we have:</u>

Q_{water}=m_{water}.C_{water}.\Delta T_{water} (2)

Where:

Q_{water}=1 kcal

m_{water}=1 kg

C_{water}=1 kcal/kg \°C

\Delta T_{water}

Isolating \Delta T from both equations:

\Delta T_{soil}=\frac{Q_{soil}}{m_{soil}.C_{soil}} (3)

\Delta T_{soil}=\frac{1 kcal}{1 kg(0.2 kcal/kg \°C)}

\Delta T_{soil}=5\°C (4)

\Delta T_{water}=\frac{Q_{water}}{m_{water}.C_{water}} (5)

\Delta T_{water}=\frac{1 kcal}{1 kg(1 kcal/kg \°C)}

\Delta T_{water}=1\°C (6)

Comparing (4) and (6) we can find the soil will be 4\°C warmer than the water.

8 0
3 years ago
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