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: The smallest effort = 300N
Explanation:
Using one of the condition for the attainment of equilibrium:
Clockwise moment = anticlockwise moments
900 × 1 = 3 × M
Where M = the weight of the strong man
3M = 900
M = 900/3 = 300N
Therefore, 300N is the smallest effort that the strongman can use to lift the goat
The instrument would be a barometer.
Answer:
true
Explanation:
The law of conservation of charge states that whenever electrons are transferred between objects, the total charge remains the same.
Answer:
Input force of pulley system = 200 N
Explanation:
Given:
Mechanical advantage of pulley system = 5
Output force from pulley system = 1,000 N
Find;
Input force of pulley system
Computation:
Mechanical advantage = Output force / Input force
Mechanical advantage of pulley system = Output force from pulley system / Input force of pulley system
5 = 1,000 / Input force of pulley system
Input force of pulley system = 1,000 / 5
Input force of pulley system = 200 N
