Given :
Two astronauts, one of mass 64 kg and the other 86 kg , are initially at rest together in outer space.
To Find :
How far apart are they when the lighter astronaut has moved 10 mm.
Solution :
Since, there is no force in space.
So, applying conservation of momentum :
We know, distance travelled by smaller astronaut is :
For bigger astronaut distance travelled :
Therefore, they are ( 10 + 7.44 ) = 17.44 mm apart from each others.
Hence, this is the required solution.
Answer:
27.82m
Explanation:
The distance of the cyclist can be calculated using the formula:
S = ut + 1/2 (at^2)
Where; S = distance
u = initial velocity
a = acceleration
t = time
However, to find acceleration, we use:
a = v - u/t, where v is the final velocity
According to this question, u = 4.2m/s, v = 6.3m/s, t = 5.3s
a = 6.3 - 4.2 / 5.3
a = 2.1/5.3
a = 0.396m/s^2
Hence, using S = ut + 1/2 (at^2)
S = (4.2 × 5.3) + 1/2 (0.396 × 5.3 × 5.3)
S = 22.26 + 5.56
S = 27.82
Therefore, the cyclist travel in 27.82m
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