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lana66690 [7]
4 years ago
14

You drop a ball from a height of 2.0 m, and it bounces back to a height of 1.5 m. a) What fraction of its initial energy is lost

? b) What is the ball's speed after the bounce? c) Where did the energy go?
Physics
1 answer:
tangare [24]4 years ago
4 0
The fraction of energy that is lost is 25%, it depends how fast the ball was going until it lost 25% of its energy, the gravitational energy was transferred into the kinetic energy that helped the ball bounce back
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coasting due west on your bicycle at 8 m/s, you encounter a sandy patch of road 7.2m across. when you leave the sandy patch your
LuckyWell [14K]

Its acceleration in the sandy patch is 1.5 m/s^2

Explanation:

The average speed

V = (v1 + v2) / 2 = (8 + 6.5) / 2 = 7.25 m/s  \\

Time to pass patch

t = 7.2 / 7.25 = .993 sec                      \\

The acceleration

a = (v2 - v1) / t = (6.5 - 8) / .993 = -1.51 m/s^2     or 1.5 m/s^2

Hence,  1.5 m/s^2 that the acceleration in the sandy patch.

<h3>What do you understand by the acceleration? </h3>

Any procedure where the velocity varies is referred to as acceleration. There are only two ways to accelerate: changing your speed or changing your direction, or changing both. This is because velocity is both a speed and a direction.

The meter per second per second (m/s2) unit of acceleration. Definition. The newton is the unit of force that, when applied to a mass of one kilogram, results in a one-meter-per-second acceleration.

Changing the speed at which an object is travelling is what acceleration is all about. A substance is not accelerating if its velocity is not changing.

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5 0
2 years ago
1 question 20points<br> How is frequency related to the sound we here
pishuonlain [190]
Frequency is the vibration of noise and the vibration determines the pitch, which we depend on to be a pitch or frequency we can hear. If it's too high or too low our ears can't hear it 
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3 years ago
Help!!! I need the order abcde
goldenfox [79]

Answer:

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8 0
3 years ago
1. Una carga eléctrica de 5*10-6 C genera un campo eléctrico a su alrededor. Calcula la intensidad de este campo a una distancia
Jlenok [28]

Answer:

La intensidad del campo eléctrico es 70312.5 \frac{N}{C}.

Explanation:

La perturbación que crea en torno a ella una carga eléctrica se representa mediante un vector denominado campo eléctrico.

Se dice que un campo eléctrico es uniforme en una región del espacio cuando la intensidad de dicho campo eléctrico es el mismo en todos los puntos de dicha región.

El campo eléctrico E creado por la carga puntual q en un punto cualquiera P se define como:

E=k*\frac{q}{r^{2} }

donde q es la carga creadora del campo, k es la constante electrostática y r es la distancia desde la carga fuente al punto P.

En este caso, los datos son:

  • k= 9*10⁹ \frac{N*m^{2} }{C^{2} }
  • q= 5*10⁻⁶ C
  • r= 0.8 m

Reemplazando:

E=9*10^{9}\frac{N*m^{2} }{C^{2} }  *\frac{5*10^{-6} C}{(0.8 m)^{2} }

Resolviendo:

E= 70312.5 \frac{N}{C}

<em><u>La intensidad del campo eléctrico es 70312.5 </u></em>\frac{N}{C}<em><u>.</u></em>

5 0
3 years ago
In this graph, what is the displacement of the particle in the last two seconds?
nikklg [1K]
In this graph, what is the displacement of the particle in the last two seconds?of the particle in the last two seconds? 

<span>0.2 meters
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In this graph, the displacement of the particle in the last two seconds is 2 meters.
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3 years ago
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