Question 4 is true, question 5 is B.
Respuesta:
0,0560 cal / gºC.
Explicación:
Cantidad de calor; (Q)
Q = mcΔt; Δt = t2 - t1
m = masa, c = capacidad calorífica específica; Δt = cambio de temperatura
c de agua = 1 cal / gºC
c de aluminio = 0,22 cal / gºC
QTotal = Q de agua + Q de aluminio
Q de agua = 450 * 1 * (26 - 23) = 1350 cal
Q de aluminio = 60 * 0.22 * (26 - 23) = 39.6 cal
QTotal = 1350 + 39,6 = 1389,6 cal
Calor perdido = calor ganado
QTotal = calor perdido
- 1389,6 = 335,2 * c * (26 - 100)
-1389,6 = −24804,8 * c
c = 1389,6 / 24804,8
c = 0,056021 cal / gºC.
Capacidad calorífica específica de la plata = 0,0560 cal / gºC.
Answer:
A_resulting = 0.2 m
Explanation:
Let's analyze the impact of the pulse with the pole, this is a fixed obstacle that does not move therefore by the law of action and reluctant, the force that the pole applies on the rope is of equal magnitude to the force of the rope on the pole (pulse), but opposite directional, so the reflected pulse reverses its direction and sense.
With this information we analyze a point on the string where the incident pulse is and each reflected with an amplitude A = 0.1 m, the resulting is
A_res = 2A
A_resultant = 2 .01
A_resulting = 0.2 m
Answer: a. This would be exciting, but not surprising. Heat from Martian volcanoes may well be enough to melt water under the Mars' surface.
Explanation: It was recently observed by a team of geological researchers that there exist some activity at the crust of the planet mars. This activity are volcanic in nature and estimated to be about 10kilometers large. Also this volcanic eruptions in the planet mars core are described as among the largest in our solar system. Therefore it won't be a surprise that Heat from Martian volcanoes may well be enough to melt water under the Mars' surface.
Answer : The temperature when the water and pan reach thermal equilibrium short time later is, 
Explanation :
In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.
where,
= specific heat of aluminium = 
= specific heat of water = 
= mass of aluminum = 0.500 kg = 500 g
= mass of water = 0.250 kg = 250 g
= final temperature of mixture = ?
= initial temperature of aluminum = 
= initial temperature of water = 
Now put all the given values in the above formula, we get:
Therefore, the temperature when the water and pan reach thermal equilibrium short time later is,
