The conservation of the momentum allows to find the result of how the astronaut can return to the spacecraft is:
- Throwing the thruster away from the ship.
The momentum is defined as the product of the mass and the velocity of the body, for isolated systems the momentum is conserved. If we define the system as consisting of the astronaut and the evo propellant, this system is isolated and the internal forces become zero. Let's find the moment in two moments.
Initial instant. Astronaut and thrust together.
p₀ = 0
Final moment. The astronaut now the thruster in the opposite direction of the ship.
= m v + M v '
where m is propellant mass and M the astronaut mass.
As the moment is preserved.
0 = m v + M v ’
v ’=
We can see that the astronaut's speed is in the opposite direction to the propeller, that is, in the direction of the ship.
The magnitude of the velocity is given by the relationship between the masses.
In conclusion, using the conservation of the momentun we can find the result of how the astronaut can return to the ship is:
- Throwing the thruster away from the ship.
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Answer: Homeostasis helps to maintain body equilibrium.
Explanation:
Homeostasis is defined as the maintenance of a fairly constant internal environment in an organism. Body fluids such as blood, lymph, and tissue fluids make up the internal environment of the body, hence by the action of the nervous and endocrine system, there levels are maintained at a body equilibrium, by the help of brain, which has overall control of homeostasis while other parts such as kidney, liver, skin, hormones are also involved.
Explanation:
Given:
Solving for
:

where:

Integrating to get
with initial conditions
:

Integrating to get x with initial conditions x(0) = 0:

When t=T:


Answer
acceleration due to gravity on Jupiter's moon,g' = 1.81 m/s²
weight of water melon on earth, W = 40 N
acceleration due to gravity on earth, g = 9.8 m/s²
a) Mass on the earth surface
M = 4.08 Kg
b) Mass on the surface of Lo
Mass of an object remain same.
Hence, mass of object at the surface of Lo = 4.08 Kg.
c) Weight at the surface of Lo
W' = m g'
W' =4.08 x 1.81
W' = 7.38 N
Information that is given:
a = -5.4m/s^2
v0 = 25 m/s
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S = ?
Calculate the S(distance car traveled) with the formula for velocity of decelerated motion:
v^2 = v0^2 - 2aS
The velocity at the end of the motion equals zero (0) because the car stops, so v=0.
0 = v0^2 - 2aS
v0^2 = 2aS
S = v0^2/2a
S = (25 m/s)^2/(2×5.4 m/s^2)
S = (25 m/s)^2/(10.8 m/s^2)
S = (625 m^2/s^2)/(10.8 m/s^2)
S = 57.87 m