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sesenic [268]
3 years ago
9

A child’s toy that is made to shoot ping pong balls consists of a tube, a spring (k = 18 N/m) and a catch for the spring that ca

n be released to shoot the balls. When a ball is loaded into the tube, it compresses the spring 9.5 cm. If you shoot a ping pong ball straight up out of this toy, how high will it go?
Physics
1 answer:
UkoKoshka [18]3 years ago
6 0

Answer:

The height is 3.1m

Explanation:

Here we have a conservation of energy problem, we have a conversion form eslastic potencial  energy to gravitational potencial energy, so:

E_e=\frac{1}{2}K*x^2\\E_e=\frac{1}{2}18N/m*(9.5*10^{-2}m)^2\\E_e=0.081J

then we have only gravitational potencial energy when the ball is at its maximun height.

E_g=m*g*h

because all the energy was transformed Eg=Ee

h=\frac{0.081J}{9.8m/s^2*m}

searching the web, the mass of a ping pong ball is 2.7 gr in average. so:

h=\frac{0.081J}{9.8m/s^2*(2.7*10^{-3}kg)}\\h=3.1m

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Responder:

<h2>64 Julios </h2>

Explicación:

La energía cinética se expresa mediante la fórmula KE = 1 / 2mv² donde;

m es la masa del cuerpo

v es la velocidad del objeto

Dado que el cuerpo se mueve horizontalmente con v = 4 m / sy después de un período de tiempo se mueve con v = 20 m / s, entonces la variación en la velocidad será de 20 m / s - 4 m / s = 16 m / s.

Parámetros dados

masa del objeto m = 0,5 kg

Variación de velocidad = 16 m / s

Variación de la energía cinética = 1/2 * 0,5 * 16²

Variación de la energía cinética = 1/2 * 0,5 * 256

Variación de la energía cinética = 0,5 * 128

<em>Variación de la energía cinética = 64 Julios</em>

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3 years ago
Which of the following statements accurately describes work being done.
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En un experimento de calorimetría, 0.50 kg de un metal a 100°C se añaden a 0.50 kg de agua a 20°C en un vaso de calorímetro de a
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Answer:

c=0.14J/gC

Explanation:

A.

2) The specific heat will be the same because it is a property of the substance and does not depend on the medium.

B.

We can use the expression for heat transmission

Q=mc(T_2-T_1)

In this case the heat given by the metal (which is at a higher temperature) is equal to that gained by the water, that is to say

Q_1=-Q_2

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c = 4.18J / g ° C

replacing we have

c_{metal}*(500g)(100\°C-25\°C)=-(250g)(4.18\frac{J}{g\°C})(20\°C-25\°C)\\c_{metal}=0.14\frac{J}{g\°C}

I hope this is useful for you

A.

2) El calor específico será igual porque es una propiedad de la sustancia y no depende del medio.

B.

Podemos usar la expresión para la transmisión de calor

Q=mc(T_2-T_1)

En este caso el calor cedido por el metal (que está a mayor temperatura) es igual al ganado por el agua, es decir

Q_1=-Q_2

para el agua tenemos que

c=4.18J/g°C

reemplazando tenemos

c_{metal}*(500g)(100\°C-25\°C)=-(250g)(4.18\frac{J}{g\°C})(20\°C-25\°C)\\c_{metal}=0.14\frac{J}{g\°C}

7 0
3 years ago
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