3. The sum of the players' momenta is equal to the momentum of the players when they're stuck together:
(75 kg) (6 m/s) + (80 kg) (-4 m/s) = (75 kg + 80 kg) v
where v is the velocity of the combined players. Solve for v :
450 kg•m/s - 320 kg•m/s = (155 kg) v
v = (130 kg•m/s) / (155 kg)
v ≈ 0.84 m/s
4. The total momentum of the bowling balls prior to collision is conserved and is the same after their collision, so that
(6 kg) (5.1 m/s) + (4 kg) (-1.3 m/s) = (6 kg) (1.5 m/s) + (4 kg) v
where v is the new velocity of the 4-kg ball. Solve for v :
30.6 kg•m/s - 5.2 kg•m/s = 9 kg•m/s + (4 kg) v
v = (16.4 kg•m/s) / (4 kg)
v = 4.1 m/s
Explanation:
Given that,
Two resistors of resistance 6 ohm and 3 ohm are connected in series and then in parallel.
For series combination,

For parallel combination,

When 6 ohm and 3 ohm are in series,

When 6 ohm and 3 ohm are in paralle,

So, the equivalent resistance in series combination is 9 ohms and in parallel combination it is 2 ohms.
Via half-life equation we have:

Where the initial amount is 50 grams, half-life is 4 minutes, and time elapsed is 12 minutes. By plugging those values in we get:

There is 6.25 grams left of Ra-229 after 12 minutes.
Answer:
F = M a
W = M g equivalent equation to express weight of object of mass M
M = W / g = 2867 N / 9.8 m/s^2 = 292.6 kg
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