Biografia

Hoja de respuestas (Preguntas tipo Icfes)

1. C: 42.5Hz

λ = V/F
8m = 340 / F
340/8 = 42,5 Hz

2. A: 216

Fo /Co ± VF = Ff /C ± Vf 90 x 900 = 9000/3600 = 25m/seg
Fo /340 = 200 /340 – 25m/seg
Fo= (340)(200) / 315 = 215.8 = 216 

3. B: I/2

Distancia: r Intensidad: I r’ = 2r I=?

I= P/4πr2 I´= P/4π2r2
I´= ½ x P/4πr2
I´= I/2 

4. B: La longitud de onda del sonido disminuyemodificando el tono del sonido escuchado

5. A: 275

Frecuencia: 55Hz Tiempo: 5seg
55Hz x 5seg = 275

14. An object with “high heat content” does not have to have a high temperature. If a given amount of heat energy is transferred into equal-mass samples of two substances initially at the same
temperature, the substance with the lower specific heat will have the higher final temperature. Butboth substances would have the same “heat content” relative to their original state. So an object with
“high heat content” might be made of material with a very high specific heat, and therefore not necessarily be at a high temperature.

16. A ceiling fan makes more of a “breeze” when it is set to blow the air down (usually called the “forward” direction by fan manufacturers). This is thesetting for the summer, when the breeze will feel cooling since it accelerates evaporation from the skin. In the winter, the fan should be set to pull air up. This forces the warmer air at the top of the room to move out towards the walls and down. The relocation of warmer air keeps the room feeling warmer, and there is less “breeze” effect on the occupants of the room.

18. For all mechanisms ofcooling, the rate of heat transfer from the hot object to the cold one is dependent on surface area. The heat sink with fins provides much more surface area than just a solid
piece of metal, and so there is more cooling of the microprocessor chip. A major mechanism for cooling the heat sink is that of convection. More air is in contact with the finned heat sink than
would be in contact with a solidpiece of metal. There is often a cooling fan circulating air around that heat sink as well, so that heated air can continually be replaced with cool air to promote more
cooling

20. We assume that the temperature in the house is higher than that under the house. Thus heat will flow through the floor out of the house. If the house sits directly on the ground or on concrete, the heat flow willwarm the ground or concrete. Dirt and concrete are relatively poor conductors of heat, and so the thermal energy that goes into them will stay for a relatively long time, allowing their temperature to rise and thus reducing the heat loss through the floor. If the floor is over a crawlspace, then the thermal energy from the floor will be heating air instead of dirt or concrete. If that warmed airgets moved away by wind currents or by convection and replaced with colder air, then the temperature difference between the inside and outside will stay large, and more energy will leave through the floor, making the inside of the house cooler.
14 . Un objeto con " alto contenido de calor " no tiene que tener una temperatura alta . Si una cantidad dada de energía calorífica se transfiere a muestrasiguales de masa de dos sustancias inicialmente a la misma
la temperatura, la sustancia con el calor específico más bajo tendrá la temperatura final más alta. Pero ambas sustancias tendrían el mismo " contenido de calor " en relación a su estado original . Así que un objeto con
" Alto contenido de calor " puede ser de un material con un alto calor específico , y por lo tanto no sernecesariamente a una temperatura alta .

16 . Un ventilador de techo hace más de una " brisa " cuando se pone a soplar el aire hacia abajo ( normalmente llamada dirección "hacia adelante " por los fabricantes de ventiladores ) . Este es el escenario para el verano , cuando la brisa se sentirá de enfriamiento , ya que acelera la evaporación de la piel. En el invierno , el ventilador se debe establecer en...