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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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A wildlife researcher is tracking a flock of geese. The geese fly 4.0 km due west, then turn toward the north by 40º and fly ano
Kobotan [32]

Wildlife researcher starts from a and then reaches b, he turns towards north 40 degree to move towards c.

Total displacement is ac

Total horizontal displacement = 4+4 cos40 =7.06 km

Total vertical displacement = 4 sin40 =2.57 km

Total displacement = \sqrt{7.06^2+2.57^2} = 7.51 km

7 0
3 years ago
Suppose that the space shuttle Columbia accelerates at 14.0 m/s2 for 8.50 minutes after takeoff.
givi [52]

Answer:

A. speed = 7.14 Km/s

B. distance = 1820.7 Km

Explanation:

Given that: a = 14.0 m/s^{2}, t = 8.50 minutes.

But,

t = 8.50 = 8.50 x 60

  = 510 seconds

A. By applying the first equation of motion, the speed of the shuttle at the end of 8.50 minutes can be determined by;

v = u + at

where: v is the final velocity, u is the initial velocity, a is the acceleration and t is the time.

u = 0

So that,

v = 14 x 510

 = 7140 m/s

The speed of the shuttle at the end of 8.50 minute is 7.14 Km/s.

B. the distance traveled can be determined by applying second equation of motion.

s = ut + \frac{1}{2}at^{2}

where: s is the distance, u is the initial velocity, a is the acceleration and t is the time.

u = 0

s = \frac{1}{2}at^{2}

  = \frac{1}{2} x 14 x (510)^{2}

 = 7 x 260100

 = 1820700 m

The distance that the shuttle has traveled during the given time is  1820.7 Km.

5 0
3 years ago
If air resistance acts on a falling object will all of its potential energy be converted into kinetic energy?
Mademuasel [1]

Answer:

No, some energy will be dissipated energy due to work of air resistance.

6 0
2 years ago
How do you calculate distance using deceleration and time?​
Effectus [21]

Answer:

U = initial velocity, t = time taken, s = distance covered. Deceleration Formula is used to calculate the deceleration of the given body in motion.

6 0
3 years ago
What is required for the maximum high tide to occur?.
daser333 [38]

Answer:

Tides are very long waves that move across the oceans. They are caused by the gravitational forces exerted on the earth by the moon, and to a lesser extent, the sun. ... Because the gravitational pull of the moon is weaker on the far side of the Earth, inertia wins, the ocean bulges out and high tide occurs.

Explanation:

4 0
3 years ago
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