If an object suspended by a scale shows a weight of 3 N in air, and 2 N when submerged in water, the buoyant force on the submer
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The given question is incomplete. The complete question is as follows.
The block has a weight of 75 lb and rests on the floor for which = 0.4. The motor draws in the cable at a constant rate of 6 ft/sft/s. Neglect the mass of the cable and pulleys.
Determine the output of the motor at the instant .
Explanation:
We will consider that equilibrium condition in vertical direction is as follows.
N - W = 0
N = W
or, N = 75 lb
Again, equilibrium condition in the vertical direction is as follows.
= 0
=
= 30 lb
Now, the equilibrium equation in the horizontal direction is as follows.
or, T =
=
=
= 17.32 lb
Now, we will calculate the output power of the motor as follows.
P = Tv
=
=
= 0.189 hp
or, = 0.2 hp
Thus, we can conclude that output of the given motor is 0.2 hp.
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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