-- If the system is 'closed', then nothing ... including energy ... can get in or out, and the total energy inside has to be constant.
If half of the energy in the system starts out as potential energy and the rest starts out as kinetic, and then the potential energy increases, there's only one place the increase could have come from ... it could only have been converted from kinetic energy. So the <em>kinetic energy</em> in the system <em>must</em> <em>decrease</em>.
In fact, this isn't even a "result". The kinetic energy has to decrease <em><u>before</u></em> the potential energy can increase, because that's where the increase has to come from.
If the system is 'open', then energy can come in and go out. If the potential energy inside suddenly increases, we don't know where it came from, so we can't say anything about what happens to the system.
Nuclear reaction you are literally splitting an atom and in a chemical reaction you are not
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:
x = 0.176 m
Explanation:
For this exercise we will take the condition of rotational equilibrium, where the reference system is located on the far left and the wire on the far right. We assume that counterclockwise turns are positive.
Let's use trigonometry to decompose the tension
sin 60 =
/ T
T_{y} = T sin 60
cos 60 = Tₓ / T
Tₓ = T cos 60
we apply the equation
∑ τ = 0
-W L / 2 - w x + T_{y} L = 0
the length of the bar is L = 6m
-Mg 6/2 - m g x + T sin 60 6 = 0
x = (6 T sin 60 - 3 M g) / mg
let's calculate
let's use the maximum tension that resists the cable T = 900 N
x = (6 900 sin 60 - 3 200 9.8) / (700 9.8)
x = (4676 - 5880) / 6860
x = - 0.176 m
Therefore the block can be up to 0.176m to keep the system in balance.