Answer:
a. Liquid state
b. At point R. The physical state of water at its boiling point temperature of 100 degree Celsius will be both liquid state as well as gaseous state.
c. at 110 degrees it is in gaseous state
Answer:
80×5×10=4000J
so therefore, work done on the body is 4000J
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
Answer:
power=work done÷time taken
2×5=10
10÷10=1
ans 1J per second
A single magnetic field is shown.
