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3 years ago

If a spacecraft that seems to be motionless is deep space is given some type of quick push, what will happen?

Physics

2 answers:

Sergio [31]3 years ago

7 0

According to the first Newton Law of Motion (sometimes called Law of Inertia):

An object at rest or describing a uniform straight line motion (moving at constant velocity), will remain at rest or moving unless an external force is applied to it and changes its state of rest or motion.

In other words:

An object or body will keep its state of motion until an external force changes its state

This means that objects tend to remain in its state of motion, and is the definition of the inertia, as well.

Now, going back to the spacecraft in deep space of the question, if some force acts on it, the spacecraft will change its state and move in the same direction of the force applied.

It will also keep this path until another external force stops, accelerates or changes its direction.

Therefore, the right answer is C:

The spacecraft will move and will not stop until it is stopped by an equal and opposite force.

Options A, B and D are not correct because:

A) The spacecraft cannot stop by its own; this would break Newton’s Laws of Motion

B) In deep space there is not air resistance, is vacuum.

D) The spacecraft works in space

drek231 [11]3 years ago

4 0

I think is C
Hope this helps ;)

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Solve using correct significant figures and indicating maximum absolute uncertainty.

Vera_Pavlovna [14]

We use the criterion of significant figures to find the result with reliable figures

X = 9.2 10-5

now with the propagation of errors we obtain the result with its uncertainty

X ± ΔX = (9.2 ± 0.5) 10⁻⁵

given Parameter

* expression values with their absolute errors

to find

* the result with the correct significant figures

* the absolute error of the expression

Significant figures are defined with the number of decimals that give information, the number of figures in a quantity gives information about the uncertainty of this quantity.

There are two criteria for applying significant figures:

* Add and subtract the result of going with the number of decimal places of the figure that has the least

* Product and division as a result of going with the least number of significant figures than the value that has the least.

Remember that the zero to the left do not form a pair of the significant figures

Let's apply this belief to the case presented, let's write the precaution

$x = \frac{a-b}{c}$

where in this case they are worth

a = 0.0336 ± 0.0002

b = 0.010 ± 0.001

c = 255.4 ± 0.4

We see that the significant figures of each parameterize (a, b, c) and their absolute errors are correct.

Let's apply the criteria to the operation

a-b = 0.0336 - 0.010

a- b = 0.0236

we apply the criterion of significant figures for the subtraction, the result must be with 3 decimal places

a - b = 0.024

let's do the other operation

X = $\frac{a-b}{c}$

X = 0.024 / 255.4

X = 9.24 10⁻⁵

We apply the criterion of significant figures for the division, in this case the result is left with two significant figures

X = 9.2 10⁻⁵

The uncertainty or error of the measurements is of most importance as it determines how many significative figures are reliable at a given magnitude.

If the magnitudes are measured with some type of instrument, the absolute error is given by the appreciation of the instrument, if the magnitude is calculated using some equation, the errors must be propagated using the variations of each parameter in the worst case.

the uncertainty of the calculated quantity (X) is

$\Delta X = | \frac{dX}{da}| \Delta a + | \frac{dX}{db} | \Delta b + | \frac{dX}{dc}| \Delta c$