Let say the two train cars are of masses 
and 
now if the speed of two cars are 
and 
then we can say that the momentum of two cars before they collide is given by
here two cars are moving in opposite direction so we can say that the net momentum is subtraction of two cars momentum.
Now since in these two car motion there is no external force on them while they collide
So the momentum of two cars are always conserved.
hence we can say that the final momentum of two cars will be same after collision as it is before collision
Friction force is when you rub 2 things together and they get warm. Motion, on the other hand, is if your walking along the sidewalk - you hardly get warmer -------
Unless it's a colder day outside and you're walking SO you decide to rub your hands together to get warm, but if you were just walking , its motion and only motion - no friction :):)
Answer:
10 :)
You have to divide the difference of speed and divide it by the time. So 100-20 would be 80, and if you divide that by 8 it would be 10.
Hope this helps.
Answer:
a) x = (0.0114 ± 0.0001) in
, b) the number of decks is 5
Explanation:
a) The thickness of the deck of cards (d) is measured and the thickness of a card (x) is calculated
x = d / 52
x = 0.590 / 52
x = 0.011346 in
Let's look for uncertainty
Δx = dx /dd Δd
Δx = 1/52 Δd
Δx = 1/52 0.005
Δx = 0.0001 in
The result of the calculation is
x = (0.0114 ± 0.0001) in
b) You want to reduce the error to Δx = 0.00002, the number of cards to be measured is
#_cards = n 52
The formula for thickness is
x = d / n 52
Uncertainty
Δx = 1 / n 52 Δd
n = 1/52 Δd / Δx
n = 1/52 0.005 / 0.00002
n = 4.8
Since the number of decks must be an integer the number of decks is 5
Answer:
Current = dQ/dt
or I = dQ/dt
Where I represents current.
Which is the rate of flow of charge.
Q=4 + 2t + t²
dQ/dt = 2 + 2t --- This is the relation that gives the instantaneous current.
At time t=2sec
dQ/dt = I = 2 + 2t
= 2 + 2(2)
=2 + 4
= 6A.
