Answer:
The particle comes to rest before reaching the position x=4m.
Explanation:
When the object is at x=0m, all of its energy is kinetic energy. Using the equation for kinetic energy yields KE=1/2mv^2=(12)(2)(3)^2=9J. Using the given equation for potential energy when the object is at x=4m yields U=4x^2=4(4)^2=64J. Since the system only has 9J of energy, the object comes to rest before reaching x=4m.
Answer:
x' = 1.01 m
Explanation:
given,
mass suspended on the spring, m = 0.40 Kg
stretches to distance, x = 10 cm = 0. 1 m
now,
we know
m g = k x
where k is spring constant
0.4 x 9.8 = k x 0.1
k = 39.2 N/m
now, when second mass is attached to the spring work is equal to 20 J
work done by the spring is equal to


x'² = 1.0204
x' = 1.01 m
hence, the spring is stretched to 1.01 m from the second mass.
Answer : The equilibrium concentration of T(g) is 0.5 M
Solution :
Let us assume that the equilibrium reaction be:
The given equilibrium reaction is,

The expression of
will be,
![K_c=\frac{[Z][X]^2}{[R][T]^2}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BZ%5D%5BX%5D%5E2%7D%7B%5BR%5D%5BT%5D%5E2%7D)
where,
= equilibrium constant = 16
[Z] = concentration of Z at equilibrium = 2.0 M
[R] = concentration of R at equilibrium = 2.0 M
[X] = concentration of X at equilibrium = 2.0 M
[T] = concentration of T at equilibrium = ?
Now put all the given values in the above expression, we get:
![16=\frac{(2.0)\times (2.0)^2}{(2.0)\times [T]^2}](https://tex.z-dn.net/?f=16%3D%5Cfrac%7B%282.0%29%5Ctimes%20%282.0%29%5E2%7D%7B%282.0%29%5Ctimes%20%5BT%5D%5E2%7D)
![[T]=0.5M](https://tex.z-dn.net/?f=%5BT%5D%3D0.5M)
Therefore, the equilibrium concentration of T(g) is 0.5 M