1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
natali 33 [55]
2 years ago
10

The density of a certain type of steel is 8.1 g/cm3. What is the mass of a 100 cm3 chunk of this steel

Engineering
1 answer:
irina1246 [14]2 years ago
4 0

Answer:

  810 g

Explanation:

Mass is the product of density and volume:

  m = ρV

  m = (8.1 g/cm³)(100 cm³) = 810 g

The mass of the chunk is 810 grams.

You might be interested in
Compressed Air In a piston-cylinder device, 10 gr of air is compressed isentropically. The air is initially at 27 °C and 110 kPa
Helen [10]

Answer:

(a) 2.39 MPa (b) 3.03 kJ (c) 3.035 kJ

Explanation:

Solution

Recall that:

A 10 gr of air is compressed isentropically

The initial air is at = 27 °C, 110 kPa

After compression air is at = a450 °C

For air,  R=287 J/kg.K

cv = 716.5 J/kg.K

y = 1.4

Now,

(a) W efind the pressure on [MPa]

Thus,

T₂/T₁ = (p₂/p₁)^r-1/r

=(450 + 273)/27 + 273) =

=(p₂/110) ^0.4/1.4

p₂ becomes  2390.3 kPa

So, p₂ = 2.39 MPa

(b) For the increase in total internal energy, is given below:

ΔU = mCv (T₂ - T₁)

=(10/100) (716.5) (450 -27)

ΔU =3030 J

ΔU =3.03 kJ

(c) The next step is to find the total work needed in kJ

ΔW = mR ( (T₂ - T₁) / k- 1

(10/100) (287) (450 -27)/1.4 -1

ΔW = 3035 J

Hence, the total work required is = 3.035 kJ

4 0
3 years ago
A small lake with volume of 160,000 m^3 receives agricultural drainage waters that contain 150 mg / L total dissolved solids (TD
Stels [109]

Answer:

Explanation:

Given that : -

The desirable limit is 500 mg / l , but

allowable upto 2000 mg / l.

The take volume is V = 160.000 m3

V = 160 , 000 x 103 l

The crainage gives 150 mg / l and lake has initialy 100 mg / l

Code of tpr frpm drawn = 150 x 60, 000 x 1000

Ci = 9000 kg / gr

Cl = 100 x 160,000 x 1000

Cl = 16, 000 kg

Since allowable limit = 2000 mg / l

Cn = ( 2000 x 160, 00 x 1000 )

= 320, 000 kg

so, each year the rate increases, by 9000 kg / yr

Read level = ( 320, 000 - 16,000 )

Li = 304, 000 kg

Tr=<u>304,000</u>

      900

=33.77

5 0
3 years ago
Read 2 more answers
On a average work day more than work place firs are reorted​
Basile [38]
What is the question? It looks like a statement...
8 0
3 years ago
The advantage of an interferometer is that
victus00 [196]

It can provide measurements of stars with a higher angular resolution than is possible with conventional telescopes.

5 0
1 year ago
An air-standard Carnot cycle is executed in a closed system between the temperature limits of 350 and 1200 K. The pressures befo
SVEN [57.7K]

This question is incomplete, the complete question is;

An air-standard Carnot cycle is executed in a closed system between the temperature limits of 350 and 1200 K. The pressures before and after the isothermal compression are 150 and 300 kPa, respectively. If the net work output per cycle is 0.5 kJ, determine (a) the maximum pressure in the cycle, (b) the heat transfer to air, and (c) the mass of air. Assume variable specific heats for air.

Answer:

a) the maximum pressure in the cycle is 30.01 Mpa

b) the heat transfer to air is 0.7058 KJ

c) mass of Air is 0.002957 kg

Explanation:

Given the data in the question;

We find the relative pressure of air at 1200 K (T1) and 350 K ( T4)

so from the "ideal gas properties of air table"

Pr1 = 238

Pr4 = 2.379

we know that Pressure P1 is only maximum at the beginning of the expansion process,

so

now we express the relative pressure and pressure relation for the process 4-1

P1 = (Pr2/Pr4)P4

so we substitute

P1 = (238/2.379)300 kPa

P1 = 30012.6 kPa = 30.01 Mpa

Therefore the maximum pressure in the cycle is 30.01 Mpa

b)

the Thermal heat efficiency of the Carnot cycle is expressed as;

ηth = 1 - (TL/TH)

we substitute

ηth = 1 - (350K/1200K)

ηth = 1 - 0.2916

ηth = 0.7084

now we find the heat transferred

Qin = W_net.out / ηth

given that the net work output per cycle is 0.5 kJ

we substitute

Qin = 0.5 / 0.7084

Qin = 0.7058 KJ

Therefore, the heat transfer to air is 0.7058 KJ

c)

first lets express the change in entropy for process 3 - 4

S4 - S3 = (S°4 - S°3) - R.In(P4/P3)

S4 - S3 = - (0.287 kJ/Kg.K) In(300/150)kPa

= -0.1989 Kj/Kg.K = S1 - S2

so that; S2 - S1 = 0.1989 Kj/Kg.K

Next we find the net work output per unit mass for the Carnot cycle

W"_netout = (S2 - S1)(TH - TL)

we substitute

W"_netout = ( 0.1989 Kj/Kg.K )( 1200 - 350)K

= 169.065 kJ/kg

Finally we find the mass

mass m = W_ net.out /  W"_netout

we substitute

m = 0.5 / 169.065

m = 0.002957 kg

Therefore, mass of Air is 0.002957 kg

5 0
3 years ago
Other questions:
  • The advantages of solar cells include all of the following, except a.moderate net energy yield b.little or no direct emissions o
    11·1 answer
  • A saturated 1.5 ft3 clay sample has a natural water content of 25%, shrinkage limit (SL) of 12% and a specific gravity (GS) of 2
    11·1 answer
  • What are atomic bombs made out of <br> Just wondering
    10·1 answer
  • Please answwr the above question screenshot.​
    15·1 answer
  • PLEASE HELP ASAP!!! Thanks
    11·1 answer
  • How will the proposed study contribute to your career?*<br>(quantity Surveying​
    11·1 answer
  • Imagine waking up every day and no one is home besides your dog and you how would you feel?
    12·2 answers
  • PROBLEM IN PICTURE HELP ME DEAR GODDDDDD UGHHH NONONO I HAVE 2 MINUTES TO FINISH THIS ❕❗️❕❗️❗️❕❕❕❕❗️❕❕❗️❕❗️❗️❗️❕‼️‼️‼️‼️❗️‼️❗️
    11·2 answers
  • An unknown relative passes away and bequeaths upon you a small tract of land in Amherst. You decide to build a two-story storage
    14·1 answer
  • A thin aluminum sheet is placed between two very large parallel plates that are maintained at uniform temperatures T1 = 900 K, T
    6·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!