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Korvikt [17]
4 years ago
10

Two cars, C and D, travel in the same direction on a long, straight section of highway. During a particular time interval Ato, c

ar D is ahead of car C and is speeding up while car C is slowing down. During the interval At, it is observed that car C gains on car D (i.e., the distance between the cars decreases). Explain how this is possible, and give a specific example of such a case.
Physics
1 answer:
a_sh-v [17]4 years ago
6 0

Answer:

possibly because the car is running out of gas

Explanation:

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W=225 J<br> P=25 W<br> t=?<br> what's T
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Explanation:

p=w/t

t=w/p

t=225/25=9seconds

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3) A tolley of mass 4kg, moving with a velocity
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Answer:

3.57 m/s

Explanation:

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q3=m3*v3

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3 years ago
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2200 kg semi truck driving down the highway has lost control. The truck rolls across the median and into oncoming traffic. The t
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Answer:

The semi truck travels at an initial speed of 69.545 meters per second downwards.

Explanation:

In this exercise we see a case of an entirely inellastic collision between the semi truck and the car, which can be described by the following equation derived from Principle of Linear Momentum Conservation: (We assume that velocity oriented northwards is positive)

m_{S}\cdot v_{S}+m_{C}\cdot v_{C} = (m_{S}+m_{C})\cdot v (1)

Where:

m_{S}, m_{C} - Masses of the semi truck and the car, measured in kilograms.

v_{S}, v_{C} - Initial velocities of the semi truck and the car, measured in meters per second.

v - Final speed of the system after collision, measured in meters per second.

If we know that m_{S} = 2200\,kg, m_{C} = 2000\,kg, v_{C} = 45\,\frac{m}{s} and v = -15\,\frac{m}{s}, then the initial velocity of the semi truck is:

m_{S}\cdot v_{S} = (m_{S}+m_{C})\cdot v -m_{C}\cdot v_{C}

v_{S} = \frac{(m_{S}+m_{C})\cdot v - m_{C}\cdot v_{C}}{m_{S}}

v_{S} = \left(1+\frac{m_{C}}{m_{S}} \right)\cdot v - \frac{m_{C}}{m_{S}} \cdot v_{C}

v_{S} = v +\frac{m_{C}}{m_{S}}\cdot (v-v_{C})

v_{S} = -15\,\frac{m}{s}+\left(\frac{2000\,kg}{2200\,kg} \right) \cdot \left(-15\,\frac{m}{s}-45\,\frac{m}{s}  \right)

v_{S} = -69.545\,\frac{m}{s}  

The semi truck travels at an initial speed of 69.545 meters per second downwards.

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