Answer : The value of the constant for a second order reaction is, 
Explanation :
The expression used for second order kinetics is:
![kt=\frac{1}{[A_t]}-\frac{1}{[A_o]}](https://tex.z-dn.net/?f=kt%3D%5Cfrac%7B1%7D%7B%5BA_t%5D%7D-%5Cfrac%7B1%7D%7B%5BA_o%5D%7D)
where,
k = rate constant = ?
t = time = 17s
= final concentration = 0.0981 M
= initial concentration = 0.657 M
Now put all the given values in the above expression, we get:


Therefore, the value of the constant for a second order reaction is, 
Answer:
a) v_average = 11 m / s, b) t = 0.0627 s
, c) F = 7.37 10⁵ N
, d) F / W = 35.8
Explanation:
a) truck speed can be found with kinematics
v² = v₀² - 2 a x
The fine speed zeroes them
a = v₀² / 2x
a = 22²/2 0.69
a = 350.72 m / s²
The average speed is
v_average = (v + v₀) / 2
v_average = (22 + 0) / 2
v_average = 11 m / s
b) The average time
v = v₀ - a t
t = v₀ / a
t = 22 / 350.72
t = 0.0627 s
c) The force can be found with Newton's second law
F = m a
F = 2100 350.72
F = 7.37 10⁵ N
.d) the ratio of this force to weight
F / W = 7.37 10⁵ / (2100 9.8)
F / W = 35.8
.e) Several approaches will be made:
- the resistance of air and tires is neglected
- It is despised that the force is not constant in time
- Depreciation of materials deformation during the crash
If the wagon travels 18.75 m, then the work done on the wagon is
(18.75 m) x (the steady force applied to the wagon all the way, in Newtons) .
The unit is Joules .
Answer:
F = 520 N
Explanation:
For this exercise the rotational equilibrium equation should be used
Σ τ = 0
Let's set a reference system with the origin at the back of the refrigerator and the counterclockwise rotation as positive. On the x-axis it is horizontal directed outward, eg the horizontal y-axis directed to the side and the z-axis vertical
Torque is
τ = F x r
the bold indicate vectors, we analyze each force
the applied force is horizontal along the -x axis, the arm (perpendicular distance) is directed in the z axis,
The weight of the body is the vertical direction of the z-axis, so the arm is on the x-axis
-F z + W x = 0
F z = W x
F =
W
The exercise indicates the point of application of the force z = 1.5 m and the weight is placed in the center of mass of the body x = 0.6 m, we are assuming that the force is applied in the wide center of the refrigerator
let's calculate
F = 1300 0.6 / 1.5
F = 520 N