When they meet the 40-kg boy will have moved a distance of 6 m.
Displacement of the 40 kg boy
The displacement of the 40 kg boy is calculated from the principle of center mass.
X(40 kg) = (60 x 10 m + 40 x 0)/(40 kg + 60 kg)
X(40 kg) = (600)/(100) = 6 m
X(60 kg) = (60 x 0 + 40 x 10 m)/(40 kg + 60 kg)
X(60 kg) = (400)/(100) = 4 m
Thus, when they meet the 40-kg boy will have moved a distance of 6 m.
Learn more about center mass here: brainly.com/question/13499822
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Answer:
0.088 seconds
0.0880000273785 second
0.08800054757 seconds
Explanation:
In the question it is given each century adds 1 ms to a day due to the slowing rotation of the Earth
In 88 centuries the length of the first day of the year will be
88 × 1 = 88 ms = 0.088 seconds
1 ms = 1 century
1 century = 100 years × 365.25 days
⇒1 ms = 36525 days

Sum of the gain on the first day would be
0.088 + 2.7378×10⁻⁸ = 0.0880000273785 second
Sum of the gain on the second day would be
0.088 + 2.7378×10⁻⁸+2.7378×10⁻⁸ = 0.08800054757 seconds
Answer:
Time period, T = 1.98 seconds
Explanation:
It is given that,
Mass of the block, m = 300 g = 0.3 kg
Force constant of the spring, k = 3 N/m
Displacement in the block, x = 3 cm
Let T is the period of the motion of the block. The time period of the block is given by :

T = 1.98 seconds
So, the period of the motion of the block is 1.98 seconds. Hence, this is the required solution.