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Natali5045456 [20]
1 year ago
13

similar to the case in the experiment you use two rubber bands, side to side first, to hold a weight of 3 reference objects. the

n, as in the experiment, the bands are pulled apart in their free ends until they form an angle of 30 degrees with respect to the horizontal. what are the deformations observed in these bands? ...
Physics
1 answer:
Sever21 [200]1 year ago
7 0

Three units deformations observed in these bands.

<h3>What forces do a rubber band encounter?</h3>

Elastic force is the force that permits some materials to regain their former shape after being stretched or crushed. Thus, a stretched rubber band is subject to an elastic force.

The rubber band experiment uses a straightforward rubber band to show entropic force and a refrigeration cycle. The rubber band experiment involves stretching and then releasing a rubber band while measuring its temperature.

Always acting in the opposite direction of motion is friction. This indicates that if friction is there, it cancels out some of the force driving the motion (if the object is being accelerated). This results in a decreased acceleration and a smaller net force.

learn more about Elastic force refer

brainly.com/question/5055063

#SPJ14

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posledela

Answer:

a)  y₂ = 49.1 m ,    t = 1.02 s , b)   y = 49.1 m , t= 1.02 s

Explanation:

a) We will solve this problem with the missile launch kinematic equations, to find the maximum height, at this point the vertical speed is zero

            v_{y}² = v_{oy}² - 2 g (y –yo)

The origin of the coordinate system is on the floor and the ball is thrown from a height

           y-yo = v_{oy}² /2 g&#10;            y- 0 = 10.0²/2 9.8&#10;            y - 0 = 5.10 m&#10;            &#10;The height from the ground is the height that rises from the reference system plus the depth of the ground from the reference system&#10;             y₂ = 5.1 + 44&#10;             y₂ = 49.1 m&#10;Let's use the other equation to find the time&#10;              [tex]v_{y} = v_{oy} - g t

              t = v_{oy} / g

              t = 10 / 9.8

              t = 1.02 s

b) the maximum height

            y- 44.0 = v_{y}² / 2 g

            y - 44.0 = 5.1

            y = 5.1 +44.0

            y = 49.1 m

The time is the same because it does not depend on the initial height

              t = 1.02 s

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