A block of mass m1 = 3.5 kg moves with velocity v1 = 6.3 m/s on a frictionless surface. it collides with block of mass m2 = 1.7
1 answer:
First, let's find the speed 
of the two blocks m1 and m2 sticked together after the collision.
We can use the conservation of momentum to solve this part. Initially, block 2 is stationary, so only block 1 has momentum different from zero, and it is:
After the collision, the two blocks stick together and so now they have mass
and they are moving with speed :
For conservation of momentum
So we can write
From which we find
The two blocks enter the rough path with this velocity, then they are decelerated because of the frictional force
. The work done by the frictional force to stop the two blocks is
where d is the distance covered by the two blocks before stopping.
The initial kinetic energy of the two blocks together, just before entering the rough path, is
When the two blocks stop, all this kinetic energy is lost, because their velocity becomes zero; for the work-energy theorem, the loss in kinetic energy must be equal to the work done by the frictional force:
From which we can find the value of the coefficient of kinetic friction:
We can use the conservation of momentum to solve this part. Initially, block 2 is stationary, so only block 1 has momentum different from zero, and it is:
After the collision, the two blocks stick together and so now they have mass
For conservation of momentum
So we can write
From which we find
The two blocks enter the rough path with this velocity, then they are decelerated because of the frictional force
where d is the distance covered by the two blocks before stopping.
The initial kinetic energy of the two blocks together, just before entering the rough path, is
When the two blocks stop, all this kinetic energy is lost, because their velocity becomes zero; for the work-energy theorem, the loss in kinetic energy must be equal to the work done by the frictional force:
From which we can find the value of the coefficient of kinetic friction:
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Answer:
La entropía del vapor de mercurio cambia en 214.235 joules por Kelvin.
Explanation:
Por definición de entropía (), medida en joules por Kelvin, tenemos la siguiente expresión:
(1)
Donde:
- Ganancia de calor, en joules.
- Temperatura del sistema, en Kelvin.
Ampliamos (1) por la definición de calor latente:
(1b)
Donde:
- Masa del sistema, en kilogramos.
- Calor latente de vaporización, en joules
Puesto que no existe cambio en la temperatura durante el proceso de vaporización, transformamos la expresión diferencial en expresión de diferencia, es decir:
Como vemos, el cambio de la entropía asociada al cambio de fase del mercurio es directamente proporcional a la masa del sistema. Si tenemos que ,
and
, entonces el cambio de entropía es:
La entropía del vapor de mercurio cambia en 214.235 joules por Kelvin.
Answer:
This was my best estimation of the answers
