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zavuch27 [327]
2 years ago
8

A sample of gas occupies a volume of 57.8 mL 57.8 mL . As it expands, it does 110.8 J 110.8 J of work on its surroundings at a c

onstant pressure of 783 Torr 783 Torr . What is the final volume of the gas?
Physics
1 answer:
Vinvika [58]2 years ago
7 0

Answer: The final volume of the gas is 1.12 L

Explanation:

W = work done on or by the system

w = work done by the system = P\Delta V  

P = pressure = 783 torr = 1.03 atm     (1atm=760 torr)

w= 110.8 J = 1.094 J       (1Latm=101.3J)

V_1 = initial volume = 57.8 ml = 0.0578L   (1L=1000ml)

V_2 = final volume = ?

1.094Latm=1.03atm\times (V_2-0.0578)L

(V_2-0.0578)=1.06

V_2=1.12L

The final volume of the gas is 1.12 L

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The units of G must be C. m³ / ( kg s² )

<h3>Further explanation</h3>

Newton's gravitational law states that the force of attraction between two objects can be formulated as follows:

\large {\boxed {F = G \frac{m_1 ~ m_2}{R^2}} }

<em>F = Gravitational Force ( Newton )</em>

<em>G = Gravitational Constant ( 6.67 × 10⁻¹¹ Nm² / kg² )</em>

<em>m = Object's Mass ( kg )</em>

<em>R = Distance Between Objects ( m )</em>

Let us now tackle the problem !

To find unit of Gravitational Constant can be carried out in the following way:

F = G \frac{m_1 ~ m_2}{R^2}

{[N]}= G\frac{{[kg]}{[kg]}}{{[m^2]}}

{[kg ~ m / s^2]}= G \frac{{[kg^2]}}{{[m^2]}}

G = \frac{{[kg ~ m / s^2]}{[m^2]}} {{[kg^2]} }

G = \frac{{[kg ~ m^3 / s^2]}} {{[kg^2]} }

G = \frac{{[m^3 / s^2]}} {{[kg]} }

\boxed {G = \frac{{[m^3]}} {{[kg ~ s^2]} }}

The unit of G must be \large {\boxed {\frac{m^3} {kg ~ s^2 }}}

<h3>Learn more</h3>
  • Impacts of Gravity : brainly.com/question/5330244
  • Effect of Earth’s Gravity on Objects : brainly.com/question/8844454
  • The Acceleration Due To Gravity : brainly.com/question/4189441

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Gravitational Fields

Keywords: Gravity , Unit , Magnitude , Attraction , Distance , Mass , Newton , Law , Gravitational , Constant

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Two blocks joined by a string have masses of 6 and 9 kg. They rest on a frictionless horizontal surface. A 2nd string, attached
Tom [10]

Answer:

12N

Explanation:

Suppose the string mass is negligible, the total mass of the 2 block system is 6 + 9 = 15 kg

So the acceleration of the system when subjected to 30N force is

a = F / M = 30 / 15 = 2 m/s2

So both blocks would have the same acceleration, however, the force acting on the 6kg block would have a magnitude of

f = am = 2 * 6 = 12N

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2 years ago
Find the torque required for the shaft to transmit 40 kW when (a) The shaft speed is 2500 rev/min. (b) The shaft speed is 250 re
Luba_88 [7]

Answer:

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(b) 1528.5 Nm

Explanation:

According to the formula of power

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So, 40 x 1000 = τ x 2 x 3.14 x 41.67

τ = 152.85 Nm

(b) f = 250 rpm = 250 / 60 = 4.167 rps

So, 40 x 1000 = τ x 2 x 3.14 x 4.167

τ = 1528.5 Nm

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Calculate the Potential Energy of an object that has a mass of 14-kg and is located at a height of 24-m?
inessss [21]
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