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:
, assuming that the gravitational field strength is .
Explanation:
Notice that both the speed and the direction of motion of this block are constant. In other words, the velocity of this block is constant.
By Newton's Second Law, the net force on this block would be . External forces on this block should be balanced. Thus, the magnitude of the (downward) weight of this block should be equal to the magnitude of the (upward) force that the boy applies on this block.
Let denote the mass of this block. It is given that . The weight of this block would be:
.
Hence, the force that the boy applies on this block would be upward with a magnitude of .
The mechanical work that a force did is equal to the product of:
- the magnitude of the force, and
- the displacement of the object in the direction of the force.
The displacement of this block (upward by ) is in the same direction as the (upward) force that this boy had applied. Thus, the work that this boy had done would be the product of:
- the magnitude of the force that this boy exerted, , and
- the displacement of this block in the direction, .
.
Answer:
5.3 cm
Explanation:
This question is an illustration of real and apparent distance.
From the question, we have the following given parameters
Real Distance, R = 8.0cm
Refractive Index, μ = 1.5
Required
Determine the apparent distance (A)
The relationship between R, A and μ is:
μ = R/A
i.e.
Refractive Index = Real Distance ÷ Apparent Distance
Substitute values in the above formula
1.5 = 8/A
Multiply both sides by A
1.5 * A = A * 8/A
1.5A = 8
Divide both side by 1.5
1.5A/1.5 = 8/1.5
A = 8/1.5
A = 5.3cm
Hence, the letters would appear at a distance of 5.3cm