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4 years ago

7.There is no gain in mechanical advantage in a single fixed pulley. En

Physics

1 answer:

AlexFokin [52]4 years ago

7 0

Answer:

The answer to this question can be defined as follows:

Explanation:

Advantage:

The load can be raised with one fixed pulley by adjusting the shifting pressure path.
Its a fixed input pulley doesn't even have a mechanical advantage, the movement is larger than for the load.
It is preferable to use because the pulley must not be moved up or down.
It gains the measurement of just an increase in output power-to-input force ratio.

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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.

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3 years ago

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OverLord2011 [107]

The correct answer for this question is this one:

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Explanation:

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