6CO2 + 6H20 + (energy)
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Answer:
Let's define t = 0s (the initial time) as the moment when Car A starts moving.
Let's find the movement equations of each car.
A:
We know that Car A accelerations with a constant acceleration of 5m/s^2
Then the acceleration equation is:
To get the velocity, we integrate over time:
Where V₀ is the initial velocity of Car A, we know that it starts at rest, so V₀ = 0m/s, the velocity equation is then:
To get the position equation we integrate again over time:
Where P₀ is the initial position of the Car A, we can define P₀ = 0m, then the position equation is:
Now let's find the equations for car B.
We know that Car B does not accelerate, then it has a constant velocity given by:
To get the position equation, we can integrate:
This time P₀ is the initial position of Car B, we know that it starts 100m ahead from car A, then P₀ = 100m, the position equation is:
Now we can answer this:
1) The two cars will meet when their position equations are equal, so we must have:
We can solve this for t.
This is a quadratic equation, the solutions are given by the Bhaskara's formula:
We only care for the positive solution, which is:
Car A reaches Car B after 11.48 seconds.
2) How far does car A travel before the two cars meet?
Here we only need to evaluate the position equation for Car A in t = 11.48s:
3) What is the velocity of car B when the two cars meet?
Car B is not accelerating, so its velocity does not change, then the velocity of Car B when the two cars meet is 20m/s
4) What is the velocity of car A when the two cars meet?
Here we need to evaluate the velocity equation for Car A at t = 11.48s
Answer:
Explanation:
As we know that due to a thin sheet electric field at a point near its surface is given as
now we have two sheets of opposite charges so the net electric field is double that of field due to one sheet
so we have
now we have