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tino4ka555 [31]
3 years ago
14

Existe otra variable a tener en cuenta cuando de la aplicación de un trabajo se trata: el tiempo en que se aplica. Por ejemplo,

en el deporte de carreras de caballos se evidencia un ganador, así como el último en llegar. Si consideramos la misma trayectoria competitiva, podemos afirmar que los dos hicieron el mismo trabajo al llegar a la meta, sin embargo, el caballo ganador lo hizo más rápido. La magnitud física que permite determinar en qué difieren, entre el caballo ganador y el último en llegar, es la potencia. Entonces, ¿intervendrá la potencia muscular de los caballos?, ¿qué implica la magnitud que mide la rapidez con que se realiza un trabajo?, ¿se podrá generalizar a los motores, por ejemplo, de los vehículos? A continuación, los temas desarrollados ayudarán a la comprensión de esta nueva magnitud física.Rapidez para efectuar el trabajo: Sin duda alguna habrás observado en las construcciones el trabajo de las grúas para subir bloques hacia los pisos más altos. Considerando la figura, supongamos que disponemos de dos grúas para levantar bloques de un mismo peso y a la misma altura. En el momento de iniciar el desplazamiento vertical de la carga de bloques, se registra el tiempo y se advierte que la grúa de la izquierda se demoró menos tiempo. ¿Cuál crees que realizó más rápido el trabajo? Evidentemente la grúa de la izquierda. La medida de la rapidez para realizar un trabajo se denomina POTENCIA.
Physics
1 answer:
Bogdan [553]3 years ago
5 0

Answer: what

Explanation:

what

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A 1500 kg truck travelling north at 60 km/hr collides with a 1200 kg car moving east at 15km/hr. If the two cars remain locked t
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Answer:

33.33j+6.67i km/hr

Explanation:

From the law of conservation of momentum,

Applying,

mu+m'u' = V(m+m')............... Equation 1

Where m = mass of the truck, m' = mass of the car, u = initial velocity of the truck, u' = initial velocity of the car, V = Final velocity.

Note: let j represent the north, and i  represent the east

From the question,

Given: m = 1500 kg, u = 60j, m' = 1200 kg, u' = 15i

Substitute these values into equation 1

1500*60j+1200*15i = V(1500+1200)

90000j+18000i = 2700V

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On a tiny scale, what happens to an initially neutral object’s mass when it gains a net positive charge through the exchange of
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Answer:

On a tiny scale, what happens to an initially neutral object’s mass when it gains a net positive charge through the exchange of electrons? (<em>the mass will decrease by a very small factor</em>)

(b) What happens to the mass of an initially neutral object when it gains a net negative charge through the exchange of electrons?  (<em>The mass will increase by a very small factor</em>)

Explanation:

(a) On a tiny scale, what happens to an initially neutral object’s mass when it gains a net positive charge through the exchange of electrons? (<em>the mass will decrease by a very small factor</em>)

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When an object is electrically neutral it means that it has the same number of protons and electrons. For the case of an object positively charged, the rate of protons is greater than the number of electrons. That means that atom lose electrons so the mass will decrease in a very small factor.

(b) What happens to the mass of an initially neutral object when it gains a net negative charge through the exchange of electrons?  (<em>The mass will increase by a very small factor</em>)

For the case when the object is negatively charged, it means that the atom gains electrons from another object, leading to the conclusion that the mass of the atom will increase in a very small factor.  

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A ray of light in air is incident on the mid-point of a heavy flint glass prism surface at an angle of 20º with the normal. for
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The equation for refraction is:  

(sin a1)/(sin a2) = n1/n2  

where  

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n1,n2 = index of refraction for the transmission mediums.    

For this problem, we've been given an a1 of 20Âş and an n1 of 1.60. For n2, we will use air which at STP has an index of refraction of 1.00029. So  

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1.8Âş  

(sin a1)/(sin a2) = n1/n2  

(sin 1.8)/(sin a2) = 1.6/1.00029  

0.031410759/(sin a2) = 1.599536135  

0.031410759= 1.599536135(sin a2)  

0.019637418= sin(a2)  

asin(0.019637418) = a2  

1.125213477 = a2    

68.2Âş  

(sin a1)/(sin a2) = n1/n2  

(sin 68.2)/(sin a2) = 1.6/1.00029  

0.928485827/(sin a2) = 1.599536135  

0.928485827 = 1.599536135(sin a2)  

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