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3 years ago

What is kinetic and potential energy?

Physics

2 answers:

BaLLatris [955]3 years ago

8 0

Answer:

the energy produced in a body due to its motion is kinetic energy.

energy produced in a body due to its position is potential energy.

Afina-wow [57]3 years ago

7 0

Explanation:

kinetic energy is the energy possessed by a body in motion

Potential energy is the energy possessed by a body by virtue of its position with respect to a reference level

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A uniform solid sphere has mass m= 7 kg and radius r= 0. 4 m. What is its moment of inertia about an axis tangent to its surface

lilavasa [31]

The moment of inertia of a uniform solid sphere is equal to 0.448 $kgm^2$ .

<u>Given the following data:</u>

Mass of sphere = 7 kg.

Radius of sphere = 0.4 meter.

<h3>How to calculate moment of inertia.</h3>

Mathematically, the moment of inertia of a solid sphere is given by this formula:

$I=\frac{2}{5} mr^2$

<u>Where:</u>

I is the moment of inertia.

m is the mass.

r is the radius.

Substituting the given parameters into the formula, we have;

$I=\frac{2}{5} \times 7 \times 0.4^2\\\\I=2.8 \times 0.16$

I = 0.448 $kgm^2$ .

Read more on inertia here: brainly.com/question/3406242

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2 years ago

Io experiences tidal heating primarily because __________. hints io experiences tidal heating primarily because __________. io i

maxonik [38]

Lo experiences tidal heating primarily because lo’s elliptical orbit causes the tidal force on lo to vary as it orbits the Jupiter. Thus, lo’s elliptical orbit is essential to its tidal heating. This elliptical orbit, in turn, is an end result of the orbital resonance among lo, Europa and ganymade. This orbital resonance origin lo to have a more elliptical orbit than it would because lo intermittently passes Europa and ganymade in the same orbital position. We cannot perceive tidal forces of tidal heating in lo but rather we foresee that they must occur based on the orbital characteristic of the moons and active volcanoes on lo is the observational evidence that tidal heating is significant in lo.

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3 years ago

What energy transfer will a stretched rubber band have when let go

GarryVolchara [31]

Answer:

when the rubber band is realeased the potential energy is quickly converted to kinetic energy this is equal to one mass of the the rubber band multiplied by its velocity( in meters per second)

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3 years ago

Debbie places two shopping carts in a cart Corral. she pushes the first cart, which then pushes a second cart. what force is bei

Bingel [31]

When Debbie pushes the first cart she is using an applied force. An applied force is created when someone or something pushes another thing using, of course, an applied force. Now, when the second cart is being pushed by the first cart, this is also an applied force. You can tell because the first cart is being pushed using forced and this causes the second cart to be pushed using some of the force that is being transmitted to the first cart.

Debbie exerts applied force on the first cart. The first cart exert applied force on the second cart.

- Marlon Nunez

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3 years ago

Una placa de cobre a 20°C tiene unas dimensiones de 65cm x 78 cm. Encuentra el área de la placa a 400°C; Coeficiente de dilataci

ValentinkaMS [17]

Answer:

El área de la placa es aproximadamente 5102.752 centímetros cuadrados.

Explanation:

Asumamos que el cambio dimensional como consecuencia de la temperatura es pequeña, entonces podemos estimar el área de la placa de cobre en función de la temperatura mediante la siguiente aproximación:

$A_{f} = w\cdot l \cdot [1 + 2\cdot \alpha\cdot (T_{f}-T_{o})]$ (1)

Donde:

$w$ - Ancho de la placa, en centímetros.

$l$ - Longitud de la placa, en centímetros.

$\alpha$ - Coeficiente de dilatación, en $\frac{1}{^{\circ}C}$ .

$T_{o}$ - Temperatura inicial, en grados Celsius.

$T_{f}$ - Temperatura final, en grados Celsius.

Si sabemos que $w = 65\,cm$ , $l = 78\,cm$ , $\alpha = 17\times 10^{-6}\,\frac{1}{^{\circ}C}$ , $T_{o} = 20\,^{\circ}C$ and $T_{f} = 400\,^{\circ}C$ , entonces el área de la placa a la temperatura final:

$A_{f} = (65\,cm)\cdot (78\,cm)\cdot \left[1+\left(17\times 10^{-6}\,\frac{1}{^{\circ}C} \right)\cdot (400\,^{\circ}C-20\,^{\circ}C)\right]$

$A_{f} = 5102.752\,cm^{2}$

El área de la placa es aproximadamente 5102.752 centímetros cuadrados.

4 0

3 years ago