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Phoenix [80]
3 years ago
8

A 500 g model train car traveling at 0.8 m/s collides with a 300 g stationary car. The cars hook up and move off down the track

together. How fast are they going? ( also how do I do it )
Mathematics
2 answers:
ss7ja [257]3 years ago
8 0

Answer:

0.5 m/s

Step-by-step explanation:

A 500 g model train car traveling at 0.8 m/s collides with a 300 g stationary car.

Initial velocity of train, v_T=0.8\ m/s

Initial velocity of car, v_C=0\ m/s

Mass of train, m_T=500\ g

Mass of car, m_C=300\ g

The cars hook up and move off down the track together.

Let the final velocity of car and train, v_T=v_C=v

Using conservation of momentum,

Momentum before collision = Momentum after collision

m_T\times v_T+m_C\times v_C=(m_T+m_C)\times v

500\times 0.8+300\times 0=(500+300)\times v

v=\dfrac{400}{800}

v=0.5\ m/s

Hence, After collision they will going with 0.5 m/s

sdas [7]3 years ago
3 0

Answer:

0.5ms^{-1}

Step-by-step explanation:

Let v be the speed after the collision of both the cars.

Now, momentum is given by: mass × velocity, then by equating the total momentum before and after the collision of the two cars, we get

500{\times}0.8+300{\times}0=(500+300)v

⇒400+0=800v

⇒400=800v

⇒v=0.5ms^{-1}

Thus, they are moving at the velocity of 0.5ms^{-1}

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timama [110]

The solution of the equation is  x=\frac{11}{7}

Step-by-step explanation:

To simplify an equation of x

  • Simplify each side of the equation
  • Collect x in side and the numerical terms in the other side
  • Find the value of x

∵ The equation is 2x + \frac{1-x}{4}=3

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∵ The equation is 4(2x) +4 (\frac{1-x}{4})=4(3)

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- Divide both sides by 7

∴ x=\frac{11}{7}

The solution of the equation is  x=\frac{11}{7}

Learn more:

You can learn more about the solution of an equation in brainly.com/question/11306893

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3 0
3 years ago
Solve each equation. Check your solution.<br> 3(n - 7) = -30
uranmaximum [27]
3(n-7)=-30
3n-21= -30
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5 0
3 years ago
Read 2 more answers
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