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 .
Answer:
<u>Searching in google I found the total mass and the radius of the ball (m = 1.5 kg and r = 10 cm) which are needed to solve the problem!</u>
The ball rotates 6.78 revolutions.
Explanation:
<u>Searching in google I found the total mass and the radius of the ball (m = 1.5 kg and r = 10 cm) which are needed to solve the problem!</u>
At the bottom the ball has the following angular speed:
Now, we need to find the distance traveled by the ball (L) by using θ=28° and h(height) = 2 m:
To find the revolutions we need the time, which can be found using the following equation:
(1)
So first, we need to find the acceleration:
(2)
By entering equation (2) into (1) we have:
Since it starts from rest (v₀ = 0):
Finally, we can find the revolutions:
Therefore, the ball rotates 6.78 revolutions.
I hope it helps you!
Explanation:
A) Use Hooke's law to find the spring constant.
F = kx
40 N = k (0.4 m)
k = 100 N/m
B) Period of a spring-mass system is:
T = 2π √(m / k)
T = 2π √(2.6 kg / 100 N/m)
T = 1 s
Frequency is the inverse of period.
f = 1 / T
f = 1 Hz