Which graph BEST represents the motion of a car that runs out of gas on while driving along a flat highway?
2 answers:
Hey there!
So, when reviewing this question, we would want to find a graph that would start fresh and end back down.
This would mean that, the car would have to first start the car, accelerate, and then go on the highway, and then (sadly), he run's out of gas.
For example, if I were to run a race, I would have to start, which mean's that I would start on the bottom of the graph, and then when I keep running, the line would get higher and higher, but then (when I run out of breath) and will then slow down, and eventually stop.
So, it's the same in this case, the car was driving at a constant speed, and then ran out of gas, and then the car would have to turn off. This mean's once again that the graph line would have to start low, and end low.
From your option's listed above, the graph that I see that does this would be (graph A). It start's low, and end's low. B would <em>NOT </em> be the answer because the line stay's straight going up the whole time. C would <em>NOT </em> be the answer because he started high, and ended low.
This would be tricky. You can not start high and end low. When I did that race, I didn't start going 5 (mph) fast. I slowly got up there. And (for the last option) D, D would <em>NOT </em>be your answer because he did not start like that.
Your correct answer would be . . .
Hope this helps.
~Jurgen
So, when reviewing this question, we would want to find a graph that would start fresh and end back down.
This would mean that, the car would have to first start the car, accelerate, and then go on the highway, and then (sadly), he run's out of gas.
For example, if I were to run a race, I would have to start, which mean's that I would start on the bottom of the graph, and then when I keep running, the line would get higher and higher, but then (when I run out of breath) and will then slow down, and eventually stop.
So, it's the same in this case, the car was driving at a constant speed, and then ran out of gas, and then the car would have to turn off. This mean's once again that the graph line would have to start low, and end low.
From your option's listed above, the graph that I see that does this would be (graph A). It start's low, and end's low. B would <em>NOT </em> be the answer because the line stay's straight going up the whole time. C would <em>NOT </em> be the answer because he started high, and ended low.
This would be tricky. You can not start high and end low. When I did that race, I didn't start going 5 (mph) fast. I slowly got up there. And (for the last option) D, D would <em>NOT </em>be your answer because he did not start like that.
Your correct answer would be . . .
Hope this helps.
~Jurgen
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Hi there! :)
Reference the diagram below for clarification.
1.
We must begin by knowing the following rules for resistors in series and parallel.
In series:
In parallel:
We can begin solving for the equivalent resistance of the two resistors in parallel using the parallel rules.
Now that we have reduced the parallel resistors to a 'single' resistor, we can add their equivalent resistance with the other resistor in parallel (15 Ohm) using series rules:
2.
We can use Ohm's law to solve for the current in the circuit.
3.
For resistors in series, both resistors receive the SAME current.
Therefore, the 15Ω resistor receives 6A, and the parallel COMBO (not each individual resistor, but the 5Ω equivalent when combined) receives 6A.
In this instance, since both of the resistors in parallel are equal, the current is SPLIT EQUALLY between the two. (Current in parallel ADDS UP). Therefore, an even split between 2 resistors of 6 A is <u>3A for each 10Ω resistor</u>.
4.
Since the 15.0 Ω resistor receives 6A, we can use Ohm's Law to solve for voltage.
