Answer:
W = 71J
Explanation:
Given force F = (5i+5j−1k)N
d = Δr
r1 = (−5,−3,−4)m
r2 = (2,5,0)m
Δr = r2 – r1 = (2-(-5), 5-(-3), 0-(-4))
Δr = (2+5, 5+3, 0+4) = (7i+ 8j +4k)m
W = F•d = (5i+5j−1k)•(7i+ 8j +4k)
W = 5×7 + 5×8 +-1×4 = 35 + 40 - 4
W = 71J
Given the values to proceed to solve the exercise, we resort to the solution of the exercise through differential equations.
The problem can be modeled through a linear equation, in the form:

With the initial conditions as,


Where Q(t) is the charge.
<em>The general solution of a linear equation is given as:</em>
<em>
</em>
Applying this definiton in our differential equation we have that

To find b and a we use the first equation and find the roots:


Then we have

To find the values of the Constant we apply the initial conditions, then

And for the derivate:



We have a system of 2x2:


Solving we have:


The we can replace at the equation and we have that the Charge at any moment is given by,

If we obtain the derivate we find also the Current, then

Answer:
Car B has a mass of 800 kg.
General Formulas and Concepts:
<u>Momentum</u>
Law of Conservation of Momentum: 
Explanation:
<u>Step 1: Define</u>
<em>Identify variables</em>
[Given] m₁ = 1200 kg
[Given] v₁i = 10 m/s
[Solve] m₂
[Given] v₂i = 0 m/s
[Given] vf = 6 m/s
<u>Step 2: Solve for m₂</u>
- Substitute in variables [Law of Conservation of Momentum]: (1200 kg)(10 m/s) + m₂(0 m/s) = (1200 kg + m₂)(6 m/s)
- Multiply: 12000 kg · m/s = (1200 kg + m₂)(6 m/s)
- Isolate m₂ term: 2000 kg = 1200 kg + m₂
- Isolate m₂: 800 kg = m₂