Answer:
a) m=20000Kg
b) v=0.214m/s
Explanation:
We will separate the problem in 3 parts, part A when there were no coals on the car, part B when there is 1 coal on the car and part C when there are 2 coals on the car. Inertia is the mass in this case.
For each part, and since the coals are thrown vertically, the horizontal linear momentum p=mv must be conserved, that is, 
, were each velocity refers to the one of the car (with the eventual coals on it) for each part, and each mass the mass of the car (with the eventual coals on it) also for each part. We will write the mass of the hopper car as 
, and the mass of the first and second coals as 
and 
respectively
We start with the transition between parts A and B, so we have:
Which means
And since we want the mass of the first coal thrown () we do:
Substituting values we obtain
For the transition between parts B and C, we can write:
Which means
Since we want the new final speed of the car (
) we do:
Substituting values we obtain
If you mark off a beginning time and ending time on the graph,
then the area under the part of the graph between those limits
is the distance covered during that period of time.
Answer:
Incomplete question
This is the completed question
If the resistor in the circuit had a larger resistance then the current would be then have to be proportionally smaller. Because the batteries each give off 1.5 volts then the current would have to be the variable that would change. What affect would using a 12V car battery have on the operation of your circuit? (Do not try this.) What would happen to the current? What would happen to the resistor?
Explanation:
Using ohms law as our basis
Ohms law state that, the voltage in an ohmic conductor is directly proportional to the current
V∝I
Resistance is the constant of proportionality
Then
V=iR
Since we want a relationship between current and resistance.
then, I=V/R
So, current is inversely proportional to Resistance
as the current increase the resistance reduce and as the current reduces the resistance increases.
a. So, increasing the voltage from 1.5V to 12V increases the current In the circuit because voltage Is directly proportional to I.
From ohms law
V=iR
When v =1.5V
I=1.5/R
When V increase to 12V
I=12/R
I.e, it increases by a factor of 8. Eight times it's initial value
b. Now, the resistance in the circuit is the constant of proportionality and it doesn't change in a given circuit expect when using a variable resistoa r like rheostat.
Answer:
yea that is very true ◠﹏◠✿
Explanation:
