Answer:
The minimum speed must the car must be 13.13 m/s.
Explanation:
The radius of the loop is 17.6 m. We need to find the minimum speed must the car traverse the loop so that the rider does not fall out while upside down at the top.
We know that, mg be the weight of car and rider, which is equal to the centripetal force.
So, the minimum speed must the car must be 13.13 m/s.
When they meet the 40-kg boy would have moved a distance of 6 m.
<h3>Distance moved by the 40 kg boy</h3>
Apply the principle of center mass;
Take the 40 kg mass as the reference point;
X(40 kg) = (40kg x 0 + 60kg x 10 m)/(40 kg + 60 kg)
X(40 kg) = (600)/(100)
X(40 kg) = 6 m
Thus, when they meet the 40-kg boy would have moved a distance of 6 m.
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Answer:
A.) 27000 kgm/s
18000 kgm/s
B.) Va = 22 m/s
C.) 19800 kgm/s
25200 kgm/s
Explanation: Given that the velocity of A and B are 30 m/s and 20 m/s. And of the same mass M = 9 × 10^5g
M = 9×10^5/1000 = 900 kg
A.) Initial momentum of A
Mu = 900 × 30 = 27000 kgm/s
Initial momentum of B
Mu = 900 × 20 = 18000 kgm/s
B.) if they have an accident and then the velocity of the B is 28 m/s, find out velocity of A.
Momentum before impact = momentum after impact
Given that Vb = 28 m/s
27000 + 18000 = 900Va + 900 × 28
45000 = 900Va + 25200
900Va = 45000 - 25200
900Va = 19800
Va = 19800/900
Va = 22 m/s
C.) Momentum of A after impact
MV = 900 × 22 = 19800 kgm/s
Momentum of B after impact
MV = 900 × 28 = 25200 kgm/s
Answer:
D
Explanation:
D. A swing moving back and forth
Answer:
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