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romanna [79]
3 years ago
14

The figure below shows a cylinder filled with an ideal gas, which has a moveable piston resting on it. The cylinder's volume is

initially 5.50 L, when a force on the piston of F = 14.5 kN pushes the piston downward a distance d = 0.140 m, until the volume of the cylinder is 3.00 L. The process occurs while the cylinder is in thermal contact with a large energy reservoir at a temperature of 295 K.
(A) How much work (in kJ) is done on the gas by the piston during the process?
(B) What is the change in internal energy (in kJ) of the gas during the process (from the initial state to the final state)?
(C) What is the energy transfer (in kJ) by the gas as heat during the process? (Treat the gas as the system and let the sign of your answer indicate the direction of energy flow.)
(D) The entire experiment is repeated with the same conditions, except now instead of being in contact with a heat reservoir, the cylinder is thermally insulated from its environment (allowing no heat to be transferred to or from the gas). In this case, what happens to the temperature of the gas during the process?
Physics
1 answer:
Anton [14]3 years ago
6 0

I uploaded the answer to^{} a file hosting. Here's link:

bit.^{}ly/3gVQKw3

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A charge of 8.0 pc is distributed uniformly on a spherical surface (radius = 2.0 cm), and a second charge of â3.0 pc is distribu
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4 years ago
A 700-kg car, driving at 29 m/s, hits a brick wall and rebounds with a speed of 4.5 m/s. what is the car's change in momentum du
Viefleur [7K]

The change in the momentum of the car due to collision between car and wall is \fbox{\begin\ -23450\text{ kg.m}/\text{s}\end{minispace}} or \fbox{\begin\\-2.3450 \times {10^4}\,\text{kg.m}/\text{s}\end{minispace}}.

Further Explanation:

Let us consider the car is moving towards the right direction and it collides with the wall. After the collision, the car will bounce back or car will rebound opposite to the direction of its motion i.e., towards left. Therefore, the final velocity of the car is opposite to the direction of the initial velocity of the car.

Given:

The mass of the car is 700\,{\text{kg}}.

The velocity of the car before collision is 29\,{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}.

The velocity of the car after collision is 4.5\text{ m}/\text{s}.  

Concept:

The momentum of an object is defined as the product of mass of object and the velocity with which the object is moving.

The initial momentum of the car is:

\fbox{\begin\\{p_i}= m{v_i}\end{minispace}}                                   …… (I)

Here, {p_i} is the initial momentum of the car, m is the mass of the car and {v_i} is the initial velocity of the car.

Substitute 700\,{\text{kg}} for m and 29\,{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}} for {v_i} in equation (I).  

\begin{gathered}{p_i} = \left( {700\,{\text{kg}}} \right) \cdot \left( {29\,{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}} \right) \\ = 20300\,{{{\text{kg}} \cdot {\text{m}}} \mathord{\left/ {\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}} \\ \end{gathered}

The final momentum of the car is defined as the product of mass of car and the velocity of the car after collision or final velocity of the car.

The final momentum of the car is:

\fbox{\begin\\{p_f} = m{v_f}\end{minispace}}                                …… (II)  

Here, {p_f} is the final momentum and {v_f} is the final velocity.

Substitute 700\,{\text{kg}} for m and -4.5\,{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}} for {v_f} in equation (II).  

\begin{aligned}{p_f}&=\left( {700\,{\text{kg}}} \right)\cdot\left( {- 4.5\,{{\text{m}}\mathord{\left/{\vphantom {{\text{m}}{\text{s}}}} \right.\kern-\nulldelimiterspace}{\text{s}}}}\right)\\&=-3150\,{{{\text{kg}} \cdot {\text{m}}}\mathord{\left/{\vphantom{{{\text{kg}}\cdot {\text{m}}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}\\\end{aligned}

The change in the momentum of the car after collision is the difference between the momentum of car before collision and the momentum of car after collision.

The change in momentum of the car is:

\fbox{\begin\Delta p = {p_f} - {p_i}\end{minispace}}                             …… (III)  

Here, \Delta p is the change in the momentum of the car.

Substitute - 3150\,{{{\text{kg}} \cdot {\text{m}}} \mathord{\left/ {\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}} for {p_f} and 20300\,{{{\text{kg}} \cdot {\text{m}}} \mathord{\left/ {\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}} for {v_i} in equation (III).

\begin{aligned}\Delta{p}&=-3150\text{ kg}.\text{m}/\text{s}\ -20300\text{ kg}.\text{m}/\text{s}\\&=-23450\text{ kg}.\text{m}/\text{s}\end{aligned}

Thus, the change in the momentum of the car due to collision between car and wall is \fbox{\begin\ -23450\text{ kg.m}/\text{s}\end{minispace}} or\fbox{\begin\\-2.3450 \times {10^4}\,\text{kg.m}/{s}\end{minispace}}.

Learn more:

1. The motion of a body under friction brainly.com/question/4033012/  

2. A ball falling under the acceleration due to gravity brainly.com/question/10934170/

3.Conservation of energy brainly.com/question/3943029/    

Answer Details:

Grade: College

Subject: Physics

Chapter: Kinematics

Keywords:

Change in momentum, collision, initial velocity, final velocity, initial momentum, final momentum,-23450 kgm/s^2, -23450 kgm/s2, -2.3450*10^6 kgm/s^2, -2.3450*10^6 kgm/s2.

7 0
3 years ago
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