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Sveta_85 [38]
3 years ago
15

Which graph BEST represents the motion of a car that runs out of gas on while driving along a flat highway?

Physics
2 answers:
Lady_Fox [76]3 years ago
5 0
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  . . .

\boxed{\boxed{A, because \ it \ started \ low \ and \ ended \ low \ also}}

Hope this helps.
~Jurgen
cluponka [151]3 years ago
3 0
Hey there!
The answer would be Graph A. The driver picks up speed but over time, the speed decreases because he ran out of gas.
hope this helps! :)
~ 
erudite
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Answer:

A. 0.199 J

B. 0.0663 C

C = 0.0221 F

D. 12.68 ohms

Explanation:

From the question:

time duration, t = 0.28 seconds

Average power, P = 0.71 W

Average voltage, V = 3 V

A) Energy is given as:

E = P * t

=> E = 0.71 * 0.28 = 0.199 J

B) Electrical energy is also given as:

E = qV

where q = charge

=> q = E / V

∴ q = 0.199 / 3 = 0.0663 C

C) Capacitance is given as charge over voltage:

C = q / V

=> C = 0.0663 / 3 = 0.0221 F

D) Electrical power, P, can also be given as:

P = V^2 / R

where R = resistance

=> R = V^2 / P

R = 3^2 / 0.71 = 9 / 0.71 = 12.68 ohms

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An air-track glider with a mass of 239 g is moving at 0.81 m/s on a 2.4 m long air track. It collides elastically with a 513 g g
HACTEHA [7]

Answer:

Glider it stops just when it reaches the end of the runway

Explanation:

This is a shock between two bodies, so we must use the equations of conservation of the amount of movement, in the instant before the crash and the subsequent instant, with this we calculate the second glider speed, as the shock that elastic is also keep it kinetic energy

        Po = pf

        Ko = Kf

 Before crash

       Po = m1 Vo1 + 0

       Ko = ½ m1 Vo1²

 

After the crash

       Pf = m1 Vif + Vvf

       Kf = ½ m1 V1f² + ½ m2 V2f²

 

      m1 V1o = m1 V1f + m2 V2f           (1)

      m1 V1o² = m1 V1f² + m2 V2f²      (2)

We see that we have two equations with two unknowns, so the system is solvable,  we substitute in 1 and 2

   

     m1 (V1o -V1f) = m2 V2f      (3)

      m1 (V1o² - V1f²) = m2 V2f²

Let's use the relationship      (a + b) (a-b) = a² -b²

     m1 (V1o + V1f) (V1o -V1f) = m2 V2f²

We divide  with 3 and simplify

      (V1o + V1f) = V2f      (4)

Substitute in 3, group and clear

         m1 (V1o - V1f) = m2 (V1o + V1f)

         m1 V1o - m2 V1o = m2 V1f + m1 V1f

         V1f (m1 -m2) = V1o (m1 + m2)

         V1f = V1o  (m1-m2 / m1+m2)

We substitute in (4) and group

         V2f = V1o + (m1-m2 / m1 + m2) V1o

         V2f = V1o [1+ + (m1-m2 / m1 + m2)]

         V2f = V1o (2m1 / (m1+m2)

We calculate with the given values

         V1f = 0.81 (239-513 / 239 + 513)

         V1f = 0.81 (-274/752)

         V1f = - 0.295 m/s

The negative sign indicates that the planned one moves in the opposite direction to the initial one

         V2f = 0.81 [2 239 / (239 + 513)]

        V2f = 0.81 [0.636]

        V2f = 0.515 m / s

Now we analyze in the second glider movement only, we calculate the energy and since there is no friction,

         Eo = Ef

Where Eo is the mechanical energy at the lowest point and Ef is the mechanical energy at the highest point

         Eo = K = ½ m2 vf2²

         Ef = U = m2 g Y

   

         ½ m2 v2f² = m2 g Y

         Y = V2f² / 2g

         Y = 0.515²/2 9.8

         Y = 0.0147 m

At this height the planned stops, let's use trigonometry to find the height at the end of the track of the track

         tan θ = Y / x

         Y = x tan θ

The crash occurs in the middle of the track whereby x = 1.2 m

        Y = 1.2 tan 0.7

        Y = 0.147 m

As the two quantities are equal in glider it stops just when it reaches the end of the runway

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