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igomit [66]
3 years ago
7

Examples of drawing packages

Physics
1 answer:
Marat540 [252]3 years ago
5 0

Answer:

The answer are given above in attachment.

You might be interested in
5.54 Two kilograms of air within a piston–cylinder assembly WP execute a Carnot power cycle with maximum and minimum temper atur
Step2247 [10]

Answer:

A) 60%

B) p2 = 1237.2 kPa

   v2 = 0.348 m^3

C) w1-2 = w3-4 = 1615.5 kJ

   Q2-3 = 60 kJ

Explanation:

A) calculate thermal efficiency

  Л = 1 - \frac{Tl}{Th}  

where Tl = 300 k

            Th = 750 k

hence thermal efficiency ( Л ) = [1 - ( 300 / 750 )] * 100 = 60%

B) calculate the pressure and volume at the beginning of the isothermal expansion

calculate pressure ( P2 )  :

= P3v3 = mRT3  ----- (1)

v3 = 0.4m , mR = 2* 0.287, T3 = 750

hence P3 = 1076.25

next equation to determine P2

Qex = p3v3 ln( p2/p3 )

60 = 1076.25 * 0.4 ln(p2/p3)

hence ; P2 = 1237.2 kpa

calculate volume ( V2 )

p2v2 = p3v3

v2 = p3v3 / p2

   = (1076.25 * 0.4 ) /  1237.2  

  = 0.348 m^3

C) calculate the work and heat transfer for each four processes

work :

W1-2 = mCv( T2 - T1 )

        = 2*0.718 ( 750 - 300 ) = 1615.5 kJ

W3-4 = 1615.5 kJ

heat transfer

Q2-3 = W2-3 = 60KJ

Q3-4 = 0

D ) sketch of the cycle on p-V coordinates

attached below  

6 0
3 years ago
What should we do to be a computer engineer ‍ ‍?​
ycow [4]

Answer:

to be an eginere u would have to go to college and study hard

Explanation:

6 0
3 years ago
Read 2 more answers
which statement is a Hypothesis? A: Most of the earthworms moved to the shaded area during experiment. B: If an earthworm is giv
True [87]

Answer:

B

Explanation:

It's saying what you think

3 0
3 years ago
Starting from zero, the electric current takes 2 seconds to reach half its maximum possible value in an RL circuit with a resist
Leno4ka [110]

Answer:

time=4s

Explanation:

we know that in a RL circuit with a resistance R, an inductance L and a battery of emf E, the current (i) will vary in following fashion

i(t)=\frac{E}{R}(1-e^\frac{-t}{\frac{L}{R}}), where imax=\frac{E}{R}

Given that, at i(2)=\frac{imax}{2} =\frac{E}{2R}

⇒\frac{E}{2R}=\frac{E}{R}(1-e^\frac{-2}{\frac{L}{R}})

⇒\frac{1}{2}=1-e^\frac{-2}{\frac{L}{R}}

⇒\frac{1}{2}=e^\frac{-2}{\frac{L}{R}}

Applying logarithm on both sides,

⇒log(\frac{1}{2})=\frac{-2}{\frac{L}{R}}

⇒log(2)=\frac{2}{\frac{L}{R}}

⇒\frac{L}{R}=\frac{2}{log2}

Now substitute i(t)=\frac{3}{4}imax=\frac{3E}{4R}

⇒\frac{3E}{4R}=\frac{E}{R}(1-e^\frac{-t}{\frac{L}{R}})

⇒\frac{3}{4}=1-e^\frac{-t}{\frac{L}{R}}

⇒\frac{1}{4}=e^\frac{-t}{\frac{L}{R}}

Applying logarithm on both sides,

⇒log(\frac{1}{4})=\frac{-t}{\frac{L}{R}}

⇒log(4)=\frac{t}{\frac{L}{R}}

⇒t=log4\frac{L}{R}

now subs. \frac{L}{R}=\frac{2}{log2}

⇒t=log4\frac{2}{log2}

also log4=log2^{2}=2log2

⇒t=2log2\frac{2}{log2}

⇒t=4

5 0
3 years ago
the maximum intensity levels of a trumpet, trombone, and a bass drum, each at a distance of 3m are 94 dB, 107dB, and 113dB respe
Gwar [14]

Answer:

β = 114 db

Explanation:

The intensity of sound in decibles is

          β = 10 log \frac{I}{I_{o}}

in most cases Io is the hearing threshold 1 10-12 W / cm²

let's calculate the intensity of each instrument

            I / I₀ = 10 (β / 10)

            I = I₀ 10 (β / 10)

trumpet

            I1 = 1 10⁻¹² 10 (94/10)

            I1 = 2.51 10⁻³ / cm²

Thrombus

           I2 = 1 10⁻¹² 10 (107/10)

           I2 = 5.01 10-2 W / cm²

low

           I3 =1 1-12    (113/10) W/cm²

            I3 = 1,995 10-1 W / cm²

when we place the three instruments together their sounds reinforce

           I_total = I₁ + I₂ + I₃

           I_ttoal = 2.51 10-3 + 5.01 10-2 + ​​1.995 10-1

           I_total = 0.00251 + 0.0501 + 0.1995

           I_total = 0.25211 W / cm²

let's bring this amount to the SI system

         β = 10 log (0.25211 / 1 10⁻¹²)

           β = 114 db

7 0
3 years ago
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