Answer:
The recoil speed of Astronaut A is 0.26 m/s.
Explanation:
Given that,
Mass of astronaut A, 
Mass of astronaut B, 
Astronaut A pushes B away, with B attaining a final speed of 0.4, 
We need to find the recoil speed of astronaut A. The momentum remains conserved here. Using the law of conservation of linear momentum as :
So, the recoil speed of Astronaut A is 0.26 m/s.
Answer:
7) 5 m/s
8) 1.5 m/s
9) -9 m/s^2
10) 2.2 m/s
11) 5 s
Explanation:
These problems make use of the relations:
a = ∆v/∆t
d = 1/2at^2 . . . . acceleration to/from rest
v^2 = 2ad . . . . . acceleration to/from rest
In each case, choose the formula appropriate to the question, fill in the given values, and solve for what's missing.
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7) v^2 = 2ad
v = √(2(9.8 m/s^2)(1.5 m)) = √(29.4 m^2/s^2) ≈ 5 m/s
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8) d = 1/2at^2
a = 2d/t^2 = 2(75 m)/(10 s)^2 = 1.5 m/s^2
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9) a = ∆v/∆t
a = (-45 m/s)/(5 s) = -9 m/s^2
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10) a = ∆v/∆t
∆v = a·∆t = (0.09 m/s^2)(10 s) = 0.9 m/s
Vivian's final speed is the initial speed plus the change in speed:
1.3 m/s + 0.9 m/s = 2.2 m/s
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11) a = ∆v/∆t
∆t = ∆v/a = (0.50 cm/s -0.75 cm/s)/(-0.05 cm/s^2) = -.25/-.05 s = 5 s
Answer:
J=0.211kg.m/s
Explanation:
The impulse-momentum theorem states:
The velocity before the impact is given by:
For the velocity after the impact:
so:
The combined amount of kinetic and potential energy of its molecules
