Answer:
2,400kg * m/s
Explanation:
You are missing some information in the question but the rest could be found some where else.
The question gives the masses and starting velocity of each car.
Car 1: m = 600kg and sv = 4m/s
Car 2: m 400kg and sv = 0m/s
Find the momentum of both cars.
Car 1: 600 * 4 = 2400
Car 2: 400 * 0 = 0
Add both.
2400 + 0 = 2400
Best of Luck!
Feet and inches or millimeters or centimeters or meters or miles or kilometers
The force equation can easily prove this. F=ma. This states that the force on an object is equal to mass times acceleration. If the mass stays the same and the velocity of the cars increases than that means there is a larger force. This is because in both cases the cars are stopping in almost an instant and the times of the crashes are theoretically identical. Acceleration is the change in velocity over time. If the velocity is higher with the same amount of time than that means there is a higher acceleration. If you plug a higher acceleration into the force equation then you wind up with a higher force and in turn a more damaging collision.
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The Ideal Gas Law makes a few assumptions from the Kinetic-Molecular Theory. These assumptions make our work much easier but aren't true under all conditions. The assumptions are,
1) Particles of a gas have virtually no volume and are like single points.
2) Particles exhibit no attractions or repulsions between them.
3) Particles are in continuous, random motion.
4) Collisions between particles are elastic, meaning basically that when they collide, they don't lose any energy.
5) The average kinetic energy is the same for all gasses at a given temperature, regardless of the identity of the gas.
It's generally true that gasses are mostly empty space and their particles occupy very little volume. Gasses are usually far enough apart that they exhibit very little attractive or repulsive forces. When energetic, the gas particles are also in fairly continuous motion, and without other forces, the motion is basically random. Collisions absorb very little energy, and the average KE is pretty close.
Most of these assumptions are dependent on having gas particles very spread apart. When is that true? Think about the other gas laws to remember what properties are related to volume.
A gas with a low pressure and a high temperature will be spread out and therefore exhibit ideal properties.
So, in analyzing the four choices given, we look for low P and high T.
A is at absolute zero, which is pretty much impossible, and definitely does not describe a gas. We rule this out immediately.
B and D are at the same temperature (273 K, or 0 °C), but C is at 100 K, or -173 K. This is very cold, so we rule that out.
We move on to comparing the pressures of B and D. Remember, a low pressure means the particles are more spread out. B has P = 1 Pa, but D has 100 kPa. We need the same units to confirm. Based on our metric prefixes, we know that kPa is kilopascals, and is thus 1000 pascals. So, the pressure of D is five orders of magnitude greater! Thus, the answer is B.
