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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The sketch of the system is: two strings, 1 and 2, are attached to the ceiling and to a third string, 3.The third string holds the bag of cement.
The free body diagram of the weight with the string 3, drives to the tension T3 = weihgt => T3 = 325 N
The other free body diagram is around the joint of the three strings.
In this case, you can do the horizontal forces equilibrium equation as:
T1* cos(60) - T2*cos(40) = 0
And the vertical forces equilibrium equation:
Ti sin(60) + T2 sin(40) = T3 = 325 N
Then you have two equations with two unknown variables, T1 and T2
0.5 T1 - 0.766 T2 = 0
0.866 T1 + 0.643T2 = 325
When you solve it you get, T1 = 252.8 N and T2 = 165 N
Answer: T1 = 252.8 N, T2 = 165N, and T3 = 325N
373 kelvin = 99.9 Celsius. Round makes it 100. 373 kelvin also equals 212 Fahrenheit so the correct answer is A.
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Answer:
481 m
Explanation:
To fall 235 m, the time required is
t = √(2H/g)
t= √(2
235/9.8)
t=6.92 seconds.
The supplies will travel forward
6.92
69.4 ≈ 481 m
Therefore, the goods must be dropped 481 m in advance of the recipients.