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Nady [450]
3 years ago
6

Matthew drives 10 meters to go to the park at her the park he goes to the movie theatre which is 15 meters away from the park. H

e then comes back home which is 10 meters away from the park. Find the displacement Mathew travels
Physics
1 answer:
Step2247 [10]3 years ago
4 0

Answer: 0 m

Explanation:

Let's begin by stating clear that movement is the change of position of a body at a certain time. So, during this movement, the body will have a trajectory and a displacement, being both different:

The trajectory is the <u>path followed by the body</u> (is a scalar quantity).  

The displacement is <u>the distance in a straight line between the initial and final position</u> (is a vector quantity).

According to this, in the description Matthew's home is placed at 0 on a number line, then he moves 10 m to the park (this is the distance between the park and Mattew's home), then 15 m to the movie theatre until he finally comes back to his home (position 0). So, in this case we are talking about the <u>path followed by Matthew</u>, hence <u>his trajectory</u>.

However, if we talk about Matthew's displacement, we have to draw a straight line between Matthew's initial position (point 0) to his final position (also point 0).

Now, being this an unidimensional problem, the displacement vector for Matthew is 0 meters.

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Students perform a set of experiments by placing a block of mass m against a spring, compressing the spring a distance x along a
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Increasing the angle of inclination of the plane decreases the velocity of the block as it leaves the spring.

  • The statement that indicates how the relationship between <em>v</em> and <em>x</em> changes is;<u> As </u><u><em>x</em></u><u> increases, </u><u><em>v</em></u><u> increases, but the relationship is no longer linear and the values of </u><u><em>v</em></u><u> will be less for the same value of </u><u><em>x</em></u><u>.</u>

Reasons:

The energy given  to the block by the spring = \mathbf{0.5  \cdot k  \cdot x^2}

According to the principle of conservation of energy, we have;

On a flat plane, energy given to the block = 0.5  \cdot k  \cdot x^2 = kinetic energy of

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The energy of the spring = \mathbf{0.5  \cdot k  \cdot x^2}

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The energy given to the block = 0.5 \cdot k \cdot x^2 - m \cdot g  \cdot sin(\theta) = The kinetic energy of block as it leaves the spring = \mathbf{0.5  \cdot m  \cdot v^2}

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a, b, and <em>c</em> are constants

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Therefore;

  • As <em>x</em> increases, <em>v</em> increases, however, the value of <em>v</em> obtained will be lesser than the same value of <em>x</em> as when the block is on a flat plane.

<em>Please find attached a drawing related to the question obtained from a similar question online</em>

<em>The possible question options are;</em>

  • <em>As x increases, v increases, but the relationship is no longer linear and the values of v will be less for the same value of x</em>
  • <em>The relationship is no longer linear and v will be more for the same value of x</em>
  • <em>The relationship is still linear, with lesser value of v</em>
  • <em>The relationship is still linear, with higher value of v</em>
  • <em>The relationship is still linear, but vary inversely, such that as x increases, v decreases</em>

<em />

Learn more here:

brainly.com/question/9134528

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