Answer:
20.4m/s²
Explanation:
Given parameters:
Initial velocity = 0m/s
Distance = 53m
Time = 5.2s
Unknown:
Acceleration = ?
Solution:
This is a linear motion and we use the right motion equation;
S = ut + at²
S is the distance
u is the initial velocity
a is the acceleration
t is the time
Insert the parameters and solve;
53 = (0x 5.2) + x a x 5.2
53 = 2.6a
a = = 20.4m/s²
The change in the speed of the space capsule will be -0.189 m/s.
The average force exerted by each on the other will be 567 N.
The kinetic energy of each after the push for the astronaut and the capsule are 459.27 J and 32.14 J.
<h3>Given:</h3>Mass of the astronaut, = 126 kg
Speed he acquires, = 2.70 m/s
Mass of the space capsule, = 1800kg
The initial momentum of the astronaut-capsule system is zero due to rest.
= 0
m/s
Therefore,
According, to the impulse-momentum theorem;
FΔt = ΔP
ΔP = m Δv
ΔP = 126×2.70
= 340.2 kgm/sec
t is time interval = 0.600s
F = ΔP/Δt
F = 340.2/0.600
= 567 N
Therefore, the average force exerted by each on the other will be 567 N.
The Kinetic Energy of the astronaut;
K.E =
× 126 ×
= 459.27 J
The Kinetic Energy of the capsule;
K.E =
= ×1800×
= 32.14 J
Therefore, the kinetic energy of each after the push for the astronaut and the capsule are 459.27 J and 32.14 J.
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