Suppose a 0.04-kg car traveling at 2.00 m/s can barely break an egg. what is the minimum sped a 0.08-kg car would need to have in order to break the egg?

V= 1m/s

Step by step Explanation:

1.  The car needs a certain impulse to break the egg.  The impulse is given by:

                                          I=m. ΔV

where I  is the impulse, m is the mass of the car and ΔV is the change of velocity of the car.   The impulse excerted by the car of 0.04kg is:

I=    (0.04kg)( 2m/s ) =0.08 kg.m/s

take into account that the change of velocity is taken as 2 m/s because the instruction says the car can "barely" break the egg, the word "barely" suggest us that after breaking it, the car stops, i.e., its final velocity is 0 m/s.

2. With this information we know that, to break an egg, we need an impulse of 0.08 kg.m/s at least.

Another car of 0.08kg could break the egg if this achieve an impulse of 0.08 kg.m/s.

Replacing in the equation of impulse, we can find the minimum speed:

0.08 kg.m/s = (0.08kg)(V)

V= 1m/s

answer: at 24 february, 2014 the population gets twice of its initial population.

explanation:

since we have given that

the population for alpha city, t years after january 1, 2004 is given as

first we find out the initial population, i.e. t=0,

so, our quadratic equation becomes,

according to question, we have said that if the population has twice its initial population, then it becomes

so, time taken to reach this above mentioned population is given by

now, we know that time can't be negative so, we take t=9.149 years.

hence, at 24 february, 2014 the population gets twice of its initial population.

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