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Vanyuwa [196]
3 years ago
14

The motion map shows an object’s position and velocity at given times. A long black arrow right labeled x. Above that are 5 gree

n arrows pointing right end to end and are the same length. There are two more green arrows each a little shorter. Farther up from the line is a point. Much farther above that are 3 green arrows each longer than the previous. How can the map be changed so it shows constant acceleration after the object changes direction? by adding vectors that are all the same length and placing them above the current top row of vectors by adding vectors that gradually increase in length and placing them above the current top row of vectors by adding vectors that are all the same length and placing them below the current bottom row of vectors by adding vectors that gradually increase in length and placing them below the current bottom row of vectors

Physics
2 answers:
Digiron [165]3 years ago
5 0

The map be changed <u>by adding vectors that are all the same length and placing them above the current top row of vectors</u>

<h3>Further explanation</h3>

A motion map represents the position, velocity, or acceleration of an object at various time readings

a small point or dot describes the position of the object. the object's velocity / acceleration is represented by an arrow / vector.

In this motion map, The arrows (vectors) are all the same length (same speed) and point in the same direction (direction to the right) for t = 0 (origin) to t = 4s

At t = 5s, the object has slowed down, and at t = 7s and 8s the object stops

At t = 9s-11 s, the object accelerates to the left

So the position of the next moving object will be above the previous movement, and the length of the same vector shows the same speed / acceleration

So that motion maps show constant acceleration after the object changes direction, it can be done by adding vectors that are all the same length and placing them above the current top row of vectors

<h3>Learn more</h3>

The initial velocity of the bird before the gust of wind

brainly.com/question/13207873

Keywords: motion map,  constant acceleration, dot, arrows, object,length, vectors, row

#LearnwithBrainly

Ad libitum [116K]3 years ago
5 0

Answer:

The object starts away from the origin and then moves toward the origin at a constant velocity. Next, it stops for one second. Finally, it moves away from the origin at a greater constant velocity.v

Explanation:

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Allen Aby sets up an Atwood Machine and wants to find the acceleration and the tension in the string. Please help him. Two block
Vitek1552 [10]

<u>Answers:</u>

In order to solve this problem we will use Newton’s second Law, which is mathematically expressed after some simplifications as:

<h2>F=ma   (1) </h2>

This can be read as: The Net Force F of an object is equal to its mass m multiplied by its acceleration a.

We will also need to <u>draw the Free Body Diagram of each block</u> in order to know the direction of the acceleration in this system and find the Tension T of the string (<u>See figure attached).  </u>

We already know<u> m_{2} is greater than m_{1}</u>, this means the weight of the block 2 P_{2} is greater than the weight of the block 1 P_{1}; therefore <u>the acceleration of the system will be in the direction of P_{2}</u>, as shown in the figure attached.

We also know by the information given in the problem that <u>the pulley does not have friction and has negligible mass</u>, and <u>the string is massless</u>.

This means that the tension will be the same along the string regardless of the difference of mass of the blocks.

Now that we have the conditions clear, let’s begin with the calculations:

1) Firstly, we have to find the weight of each block, in order to verify that block 2 is heavier than block 1.

This is done using equation (1), where the force of the weight P is calculated using the <u>acceleration of gravity</u> g=9.8\frac{m}{s^{2}}  acting on the blocks:


<h2>P=mg   (2) </h2>

<u>For block 1: </u>

P_{1}=m_{1}g   (3)

P_{1}=1.5kg(9.8\frac{m}{s^{2}})    

<h2>P_{1}=14.7N   (4) </h2>

<u>For block 2: </u>

P_{2}=m_{2}g   (5)

P_{2}=2.4kg(9.8\frac{m}{s^{2}})    

<h2>P_{2}=23.52N      (6) </h2>

Then, we are going to <u>find the acceleration a of the whole system: </u>

F_{r}=P_{1}+P_{2}   (7)

<h2>P_{1}+P_{2}=(m_{1}+m_{2})a   (8) </h2>

Where the Resulting Force F_{r}  is equal to the sum of the weights P_{1} and P_{2}.  

In the figure attached, note that P_{1} is in opposite direction to the acceleration a, this means it must <u>have a negative sing</u>; while P_{2} is in the same direction of a.

Here we only have to isolate a from equation (8) and substitute the values according to the conditions of the system:

-14.7N+23.52N=(1.5kg+2.4kg)a  

8.82N=(3.9kg)a  

Then:

a=\frac{8.82N }{3.9kg}  

<h2>a=2.26\frac{m}{ s^{2}}  </h2><h2>This is the acceleration of the system. </h2>

2) For the second part of the problem, we have to find the tension T of the string.

We can choose either the Free Body Diagram of block A or block B to make the calculations, <u>the result will be the same</u>.  

Let’s prove it:

For m_{1}

we see in the free body diagram that the <u>acceleration is in the same direction of the tension of the string</u>, so:

F_{r}=T-P_{1}   (9)

T-P_{1}=m_{1}a   (10)

T-14.7N=(1.5kg)( 2.26\frac{m}{ s^{2}})    

Then;

<h2>T=18.09N   This is the tension of the string </h2><h2> </h2>

For m_{2}

we see in the free body diagram that the acceleration is in opposite direction of the tension of the string and must <u>have a negative sign,</u> so:

F_{r}=T-P_{2}   (9)

T-P_{2}=m_{2}a   (10)

T-23.52N=(2.4kg)(-2.26\frac{m}{ s^{2}})    

Then;

<h2>T=18.09N    This is the same tension of the string </h2>

6 0
3 years ago
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