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Lady_Fox [76]
4 years ago
8

When the balloon hits the ground, it rebounds slightly. What is the source of the energy for this rebound? A. When the balloon h

its the ground, the water warms; this thermal energy is converted to gravitational potential energy of the rebounding balloon. B. As the balloon hits the ground, the edges rise up; this gravitational potential energy is shared with the rest of the balloon. C. When the balloon hits the ground, the rubber envelope stretches, storing elastic potential energy; this elastic potential energy is converted to the gravitational potentiaL
Physics
1 answer:
nevsk [136]4 years ago
7 0

Answer:

The correct answer is c.    When the balloon hits the ground, the rubber envelope stretches, storing elastic potential energy; this elastic potential energy is converted to the gravitational potentiaL

Explanation:

Let's analyze the situation of the globe

When it touches the ground, the part that is in contact decreases its velocity to zero, but the upper part of the ball continues to move, which creates that the molecules approach slightly, if we approximate the spring links, a repulsive force is created that after all the particles reach zero speed. The force of the springs moves the ball up until the force decreases to zero.

We can relate this force of Hooke with an elastic energy

This energy can be stored in the deformation of the system, as elastic potential energy, which is subsequently transformed into gravitational potential energy when the balloon is lifted.

The correct answer is c

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 True or False? Anything that moves up and down or back and forth in a rhythmic way is vibrating.
ivanzaharov [21]
True, for example, sound waves are known for vibrating and they move up and down in a particular pattern depending on the pitch and volume. :) Hope this helps x
5 0
3 years ago
What is the angle θ between vectors A⃗ and B⃗ if A⃗ =4ı^−4ȷ^ and B⃗ =−5ı^+7ȷ^?
ololo11 [35]

The characteristics of the scalar product allows to find the angle between the two vectors is:

  • The angle θ = 170º

The scalar product is the product between two vectors whose result is a scalar.

            A . B = |A|  |B| cos θ

Where A and B are the vectors, |A| and |B| are the modules of the vectors and θ at the angle between them.

The vector is given in Cartesian coordinates and the unit vectors in these coordinates are perpendicular.

            i.i = j.j = 1

            i.j = 0

            A . B = (4 i - 4j). * -5 i + 7j)

            A . B = - 4 5 - 4 7

            A. B = -48

We look for the modulus of each vector.

           |A| = \sqrt{x^2 +y^2 }

           |A| = \sqrt{4^2 + 4^2}  

           |A| = 4 √2

          |B| = \sqrt{5^2 +7^2}

          |B| = 8.60

We substitute.

            -48 = 4√2  8.60  cos θ

            -48 = 48.66 cos θ

            θ = cos⁻¹   \frac{-48}{48.664}  

            θ = 170º

In conclusion using the dot product we can find the angle between the two vectors is:

  • the angle θ = 170º

Learn more about the scalar product here:  brainly.com/question/1550649

8 0
3 years ago
A ball is thrown vertically upward with a speed of 1.86
Inessa05 [86]

Answer:

t = 1.09 s.

Explanation:

This is a one-dimensional kinematics question, so the equations of kinematics will be sufficient to solve the question.

y-y_0 = v_0t + \frac{1}{2}at^2\\0 - 3.82 = 1.86t +\frac{1}{2}(-9.8)t^2\\-3.82 = 1.86t - \frac{1}{2}9.8t^2\\4.9t^2 - 1.86t - 3.82 = 0

This quadratic equation can be solved using determinant.

\Delta = b^2 - 4ac\\t_{1,2} = \frac{-b \pm \sqrt{\Delta} }{2a}\\t_1 = 1.09~s\\t_2 = -0.71~s

Of course, we will choose the positive time.

4 0
3 years ago
The closest stars are 4 light years away from us. How far away must you be from a 854 kHz radio station with power 50.0 kW for t
Lubov Fominskaja [6]

Answer:

The distance from the radio station is 0.28 light years away.

Solution:

As per the question:

Distance, d = 4 ly

Frequency of the radio station, f = 854 kHz = 854\times 10^{3}\ Hz

Power, P = 50 kW = 50\times 10^{3}\ W

I_{p} = 1\ photon/s/m^{2}

Now,

From the relation:

P = nhf

where

n = no. of photons/second

h = Planck's constant

f = frequency

Now,

n = \frac{P}{hf} = \frac{50\times 10^{3}}{6.626\times 10^{- 34}\times 854\times 10^{3}} = 8.836\times 10^{31}\ photons/s

Area of the sphere, A = 4\pi r^{2}

Now,

Suppose the distance from the radio station be 'r' from where the intensity of the photon is 1\ photon/s/m^{2}

I_{p} = \frac{n}{A} = \frac{n}{4\pi r^{2}}

1 = \frac{8.836\times 10^{31}}{4\pi r^{2}}

r = \sqrt{\frac{8.836\times 10^{31}}{4\pi}} = 2.65\times 10^{15}\ m

Now,

We know that:

1 ly = 9.4607\times 10^{15}\ m

Thus

r = \frac{2.65\times 10^{15}}{9.4607\times 10^{15}} = 0.28\ ly

5 0
3 years ago
A swimming pool is an example of an open system. The pool loses 10,500 J
zaharov [31]

Answer: A a decrease of 8000

Explanation: 10,500 - 2,500= 8000

8 0
4 years ago
Read 2 more answers
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