Answer:
u₂ = 3.7 m/s
Explanation:
Here, we use the law of conservation of momentum, as follows:
where,
m₁ = mass of the car = 1250 kg
m₂ = mass of the truck = 2020 kg
u₁ = initial speed of the car before collision = 17.4 m/s
u₂ = initial speed of the tuck before collision = ?
v₁ = final speed of the car after collision = 6.7 m/s
v₂ = final speed of the truck after collision = 10.3 m/s
Therefore,
<u>u₂ = 3.7 m/s</u>
Complete Question: A charge q1 = 2.2 uC is at a distance d= 1.63m from a second charge q2= -5.67 uC. (b) Find a point between the two charges on the horizontal line where the electric potential is zero. (Enter your answer as measured from q1.)
Answer:
d= 0.46 m
Explanation:
The electric potential is defined as the work needed, per unit charge, to bring a positive test charge from infinity to the point of interest.
For a point charge, the electric potential, at a distance r from it, according to Coulomb´s Law and the definition of potential, can be expressed as follows:
V =
We have two charges, q₁ and q₂, and we need to find a point between them, where the electric potential due to them, be zero.
If we call x to the distance from q₁, the distance from q₂, will be the distance between both charges, minus x.
So, we can find the value of x, adding the potentials due to q₁ and q₂, in such a way that both add to zero:
V = +
= 0
⇒:
Replacing by the values of q1, q2, and k, and solving for x, we get:
⇒ x = (2.22 μC* 1.63 m) / 7.89 μC = 0.46 m from q1.