A spring has a spring constant of 330 n/m. how far is the spring compressed when 150 newtons of force are used?
1 answer:
You might be interested in
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.
Answer : The maximum concentration of silver ion is
Solution : Given,
for AgBr =
Concentration of NaBr solution = 0.1 m
The equilibrium reaction for NaBr solution is,
The concentration of NaBr solution is 0.1 m that means,
The equilibrium reaction for AgBr is,
At equilibrium s s
The expression for solubility product constant for AgBr is,
The concentration of = s
The concentration of = 0.1 + s
Now put all the given values in expression, we get
By rearranging the terms, we get the value of 's'
Therefore, the maximum concentration of silver ion is .