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olga55 [171]
3 years ago
14

Two 110 kg bumper cars are moving toward each other in opposite directions. Car A is moving at 8 m/s and Car Z at –10 m/s when t

hey collide head–on. If the resulting velocity of Car A after the collision is –10 m/s, what is the velocity of Car Z after the collision?
Physics
1 answer:
salantis [7]3 years ago
5 0

Explanation:

Mass of bumper cars, m_1=m_2=110\ kg

Initial speed of car A, u_1=8\ m/s

Initial speed of car Z, u_2=-10\ m/s

Final speed of car A after the collision, v_1=-10\ m/s

We need to find the velocity of car Z after the collision. Let it is equal to v_2. Using the conservation of momentum as :

m_1u_1+m_2u_2=m_1v_1+m_2v_2

110\times 8+110\times (-10)=110\times (-10)+110v_2

v_2=\dfrac{-1320}{110}\ m/s

v_2=-12\ m/s

So, the velocity of car Z after the collision is (-12 m/s). Hence, this is the required solution.

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A moving sidewalk in an airport terminal moves at 1 m/s and is 35 m long. If a person steps on at one end of the sidewalk and wa
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4 0
3 years ago
The lengths of the mercury column of a mercury thermometer are 1.26cm and 20.86cm respectively at the standard fixed points. Wha
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The temperature of the body is equal to 39.48°C

Why?

To find the temperature of the body that produces 9.0cm of this mercury column, we need to use the formula for linear interpolation. Also, the temperatures to use are based on the Celsius degrees scale.

For the fixed tempetatures, we have:

1.26cm=0\°C(ColdWaterPoint)\\20.86cm=100\°C(BoilingWater)Point\\

We need to use the following formula for linear interpolation:

y=\frac{x-x_{1}}{x_{2}-x_{1}}*(y_{2}-y_{1})+y_{1}

Then, we have:

1.26cm=0\°C\\20.86cm=100\°C\\9.0cm=???\\\\x_{1}=1.26cm,y_{1}=0\°C\\\\x_{2}=20.86cm,y_{2}=100\°C\\\\x=9.0cm,y=???

So, interpolating we have:

y=\frac{(9cm-1.26cm)}{(20.86cm-1.26cm)}*(100\°C -0\°C)+0\°C\\\\y=0.39*100\°C=39.48100\°C

Hence, the temperature of the body is equal to 39.48°C

Have a nice day!

8 0
3 years ago
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