Alan

1. Energy is a scalar, not a vector quantity, so you would add not subtract the energies.

2. If it were an elastic collision they wouldn't stick together.

3. It would be unusual not to lose a lot of energy (dissipated through frictional heat loss, for example) in a head-on car crash! 

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Let N = number enrolled

N- N/12 present or 11N/12 present

11N/(12*5) on field trip, so 4*11*N/(12*5) in school or 11N/15 in school

11N/15 = 704

N = 15*704/11

N = 960

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No problem SevenUP!

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Yes,  if the rod was positively charged the ball in step a will be charged positively.

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The acceleration is given by ω²r = 0.35²*55 = 6.7375 m/s²

The number of 'g's is 6.7375/9.81 ≈ 0.7g

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Yes, you could do, as there is a "c" to wrap up the constants (though you would change the cos term to a sin term as per your original question!)

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$$\frac{1}{2}\int (1+\cos{2(x+1)})dx=\frac{1}{2}\int dx+\frac{1}{2}\int \cos{2(x+1)}dx$$

so the single x comes from the first integral on the right-hand side. However, you could add whatever constant you like, because, as it's an indefinite integral, any other constants are accounted for in the "c" at the end.

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Momentum is a vector quantity.  In the absence of any losses, the momentum of the system after the collision is the same as that before the collision. So, taking north as positive the total momentum of the system is:

1300*30 - 900*15 =  25500 kg.m/s

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After the collision the velocity of the two cars, assuming they are stuck together, is

v = 25500/(1300 + 900)

or v = 11.59 m/s  (positive, so going north)

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Using an obvious notation we have:

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sin^-1x is the same as arcsinx or asinx.  You can get it (asin) on this calculator by pressing the 2nd button first

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