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Stells [14]
3 years ago
12

An insulated piston-cylinder device initially contains 0.16 m2 of CO2 at 150 kPa and 41 °C. Electric resistance heater supplied

heat for 10 mins. During this procedure the volume has doubled while pressure stayed the same. Considering electric resistance heater running on 110 V, calculate the needed current in A ((Give your answer with three decimals, and do NOT enter units!!!).

Engineering
1 answer:
lyudmila [28]3 years ago
4 0

Answer:

I=0.3636

Explanation:

See the attached picture for explanation.

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5. A typical paper clip weighs 0.59 g and consists of BCC iron. Calculate (a) the number of
marta [7]

Answer:

(a) 3.185*10^{21} cells

(b) 6.37*10^{21} atoms

Explanation:

(a)

Volume, V of unit cell

V=(2.866*10^{-8})^{3}=2.354*10^{-23}

Number of unit cells, N

N=\frac {W_{mat}}{V\rho_{mat}} Where W_{mat} is weight of material and \rho_{mat} is density of material

N=\frac{0.59}{7.87*(2.354*10^{-23}}=3.185*10^{21} cells

(b)

Number of atoms in paper clip

This is a product of number of unit cells and number of atoms per cell

Since iron has 2 atoms per cell

Number of atoms of iron=3.185*10^{21} cells*2 atoms/cell=6.37*10^{21} atoms

8 0
3 years ago
A solid circular rod that is 600 mm long and 20 mm in diameter is subjected to an axial force of P = 50 kN The elongation of the
Kay [80]

Answer:

a) V = 0.354

b)  G = 25.34 GPA

Explanation:

Solution:

We first determine Modulus of Elasticity and Modulus of rigidity

Elongation of rod ΔL = 1.4 mm

Normal stress, δ = P/A

Where P = Force acting on the cross-section

A = Area of the cross-section

Using Area, A = π/4 · d²

= π/4 · (0.0020)²  = 3.14 × 10⁻⁴m²

δ = 50/3.14 × 10⁻⁴    = 159.155 MPA

E(long) = Δl/l  = 1.4/600 = 2.33 × 10⁻³mm/mm

Modulus of Elasticity Е = δ/ε

= 159.155 × 10⁶/2.33 × 10⁻³    = 68.306 GPA

Also final diameter d(f) = 19.9837 mm

Initial diameter d(i) = 20 mm

Poisson said that V = Е(elasticity)/Е(long)

= -  <u>( 19.9837 - 20 /20)</u>

        2.33 × 10⁻³                  

= 0.354,

∴ v = 0.354

Also G = Е/2. (1+V)

=  68.306 × 10⁹/ 2.(1+ 0.354)

= 25.34 GPA

⇒ G = 25.34 GPA

5 0
3 years ago
Biblical studies of john​
nevsk [136]

Answer:

<h2>the answer of sols brother is correct</h2><h3>hope it helps you have a good day</h3><h2 />

5 0
3 years ago
How should you move your board through the planer? (Pick two choices.)
iragen [17]

Answer:

I would say do it at an even pace

Explanation:

Doing it a slow pace takes time quickly will probably not to good gor you and doing it at an irregular pace is just way to fast

4 0
3 years ago
An Ideal gas is being heated in a circular duct as while flowing over an electric heater of 130 kW. The diameter of duct is 500
max2010maxim [7]

Answer: The exit temperature of the gas in deg C is 32^{o}C.

Explanation:

The given data is as follows.

C_{p} = 1000 J/kg K,   R = 500 J/kg K = 0.5 kJ/kg K (as 1 kJ = 1000 J)

P_{1} = 100 kPa,     V_{1} = 15 m^{3}/s

T_{1} = 27^{o}C = (27 + 273) K = 300 K

We know that for an ideal gas the mass flow rate will be calculated as follows.

     P_{1}V_{1} = mRT_{1}

or,         m = \frac{P_{1}V_{1}}{RT_{1}}

                = \frac{100 \times 15}{0.5 \times 300}

                = 10 kg/s

Now, according to the steady flow energy equation:

mh_{1} + Q = mh_{2} + W

h_{1} + \frac{Q}{m} = h_{2} + \frac{W}{m}

C_{p}T_{1} - \frac{80}{10} = C_{p}T_{2} - \frac{130}{10}

(T_{2} - T_{1})C_{p} = \frac{130 - 80}{10}

(T_{2} - T_{1}) = 5 K

T_{2} = 5 K + 300 K

T_{2} = 305 K

           = (305 K - 273 K)

           = 32^{o}C

Therefore, we can conclude that the exit temperature of the gas in deg C is 32^{o}C.

7 0
3 years ago
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