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Solnce55 [7]
4 years ago
10

Suatu sistem gas berada didalam ruang yang fleksibel. Pada awalnya gas berada pada kondisi P1 = 1,5 × 10 pangkat 5 N/m2, T1 = 27

°C dan V1 = 12 L ketika gas menyerap kalor dari lingkungan secara isobarik, suhunya berubah menjadi 127°C. Volume gas sekarang adalah..
Physics
1 answer:
enot [183]4 years ago
6 0

Answer:

16.00L

Explanation:

First you calculate the number of moles in the system:

PV=nRT\\\\n=\frac{PV}{RT}\\\\n=\frac{(1.5*10^5N/m^2)(12L)}{(0.082L.atm/mol.K)(300.15K)}=73134.16\ mol

To find the new volume of the system you use the following formula for an isobaric procedure:

T_2-T_1=\frac{P}{nR}(V_2-V_1)\\\\V_2=\frac{nR}{P}(T_2-T_1)+V_1\\\\V_2=\frac{(73134.16\ mol)(0.082L.atm/mol.K)}{1.5*10^5N/m^2}(400.15-300.15)K+12L\\\\V_2=16.00L

hence, the new volume is 16.00L

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. The earth revolves around the sun in the counterclockwise direction, completing one full revolution about every 365 days. In r
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Explanation:

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3 years ago
A point charge, Q1 = -4.2 μC, is located at the origin. A rod of length L = 0.35 m is located along the x-axis with the near sid
igor_vitrenko [27]

Answer:

a) attractiva, b) dF = k \frac{Q_1 \ dQ_2}{dx}, c)  F = k Q_1 \frac{Q_2}{d \ (d+L)}, d) F = -1.09 N

Explanation:

a) q1 is negative and the charge of the bar is positive therefore the force is attractive

b) For this exercise we use Coulomb's law, where we assume a card dQ₂ at a distance x

           dF = k \frac{Q_1 \ dQ_2}{dx}

where k is a constant, Q₁ the charge at the origin, x the distance

c) To find the total force we must integrate from the beginning of the bar at x = d to the end point of the bar x = d + L

         ∫ dF = k \ Q_1 \int\limits^{d+L}_d     {\frac{1}{x^2} } \, dQ_2

as they indicate that the load on the bar is uniformly distributed, we use the concept of linear density

          λ = dQ₂ / dx

          DQ₂ = λ dx

we substitute

         F = k \ Q_1 \lambda \int\limits^{d+L}_d  \, \frac{dx}{x^2}

         F = k Q1 λ (-\frac{1}{x})  

we evaluate the integral

        F = k Q₁ λ (- \frac{1}{d+L} + \frac{1}{d} )

        F = k Q₁ λ  ( \frac{L}{d \ (d+L)})

we change the linear density by its value

      λ = Q2 / L

       F = k Q_1 \frac{Q_2}{d \ (d+L)}

d) we calculate the magnitude of F

       F =9 10⁹ (-4.2 10⁻⁶)   \frac{10.4 10x^{-6} }{0.45 ( 0.45 +0.35)}

       F = -1.09 N

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3 years ago
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A marble dropped in still water will create circular ripples or waves. the radius of each circular wave will increase 4 centimet
Gnom [1K]

The circumference of the circle after t seconds = 25.12 t

Given data,

The radius of each circular wave will increase by four centimeters per second (cm/s)

In 1 second of circular wave radius = 4cm

In 1*t second of circular wave radius = 4t cm

 

As a result, the radius of the circular wave after t seconds is 4t cm.

We already know that the circumference of a circle is given by = 2πr, where r is the circle's radius.

As a result, the radius of a circular wave after t seconds is = 2π * radius of a circular wave after t seconds = 2π * 4t

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( Taking π = 3.14 )

As a result, the circumference of the circle after t seconds = 8πt

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Find more on circumference at : brainly.in/question/49774764

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