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
The first one is electrical energy
Answer:
A) B = 5.4 10⁻⁵ T, B) the positive side of the bar is to the West
Explanation:
A) For this exercise we must use the expression of Faraday's law for a moving body
fem = 
fem =
- d (B l y) / dt = - B lv
B = 
we calculate
B = - 7.9 10⁻⁴ /(0.73 20)
B = 5.4 10⁻⁵ T
B) to determine which side of the bar is positive, we must use the right hand rule
the thumb points in the direction of the rod movement to the south, the magnetic field points in the horizontal direction and the rod is in the east-west direction.
Therefore the force points in the direction perpendicular to the velocity and the magnetic field is in the east direction; therefore the positive side of the bar is to the West
D. 5.0A because this is right and will lead to the right answer okay you got this girl letssssss goooo googoggo Gogol