6 m/s
This type of collision is a <u>Inelastic collision </u>
Explanation:
We begin by finding the momentum of the initial train car;
Remember that momentum is given by the formulae mass (kg) * velocity (m/s)
Therefore;
Momentum = 6000 * 10
= 60,000 kg⋅m/s
The other car has zero momentum because;
4000 * 0
= 0
When the two train cars collide, the total momentum will be;
60,000 + 0 = 60,000
Therefore to find the velocity, well use the same formulae;
p = mv whereby;
p – momentum
m – mass
v – velocity
60,000 = (6000 + 4000) * v
v = 60,000 / 10,000
v = 6
= 6 m/s
This is an elastic collision because we are assuming that no energy is lost in the collision. Most collisions, however, are not elastic but rather inelastic. In inelastic collision some of the kinetic energy is lost to the environment in some other form of energy such as heat energy.
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Answer:
Also called Darwinian fitness, means the ability to survive to reproductive age find a mate, and produce offspring. Basically the more offspring an organism produces during its lifetime the greater its biological fitness.
Explanation:
Answer:
B) venous blood and aveolar
Explanation:
Partial pressure of carbon dioxide in the aveolar is the lowest, it is 40mmHg (millimetre mecury ) while the partial of carbon dioxide in venous blood is the highest at 45 to 50 mmHg (millimetre mecury)
Plotting a graph of the mass or volume of the product created against time allows you to determine the reaction's pace. This is depicted for two reactions on the graph. The rate of reaction is inversely proportional to the gradient of the line; that is, the steeper the line, the higher the rate of reaction
<h3>What is Rate of reaction ?</h3>
The result is a straight line with a positive gradient on a graph of reaction rate against concentration (a graph showing proportionality).
- The half-life is constant in a concentration-time graph of first order. As a result, the period of time it takes for the concentration to decrease to 50% of its initial value is constant.
- It would be considered first order if you obtained a straight line with a negative slope. If you graph the inverse of the concentration for second order
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