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3 years ago

What advantages do space travelers have using Newton’s first law of motion?

Physics

1 answer:

Lina20 [59]3 years ago

5 0

They can continue moving without the need of an external force

Explanation:

Newton's first law of motion states that:

"An object in motion at constant velocity (or at rest) will continue moving at constant velocity (or will stay at rest) unless acted upon an external force"

In normal daily life, we normally experience something different: when we throw an object on a table, for instance, we notice that the object will eventually stop, instead of keep moving. This is because there is another force acting on the object, the force of friction, opposite to the motion, that "resists" the motion of the object eventually bringing it to a stop.

In space, however, there is no force of friction or air resistance. Therefore, Newton's first law can be applied: and so, if space travellers are in a ship, and this ship is moving at a certain constant velocity, it will keep moving at this velocity without slowing down, and no external force would be needed in order to keep the ship in motion.

Learn more about Newton laws of motion:

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A professor sits on a rotating stool that is spinning at 10.0 rpm while she holds a heavy weight in each of her hands. Her outst

MAVERICK [17]

Answer:

0.5864 m

Explanation:

$r_1$ = 0.705 m

$\omega_1$ = 10 rpm

$\omega_2$ = 20.5 rpm

$\left(m r_{1}^{2}\right) \omega_{1} &=\left(m r_{2}^{2}\right) \omega_{2}$

$r_{2}^{2} &=\frac{\left(m r_{1}^{2}\right) \omega_{1}}{\left(\omega_{2}\right)}$

$=\frac{\left(r_{1}^{2}\right) \omega_{1}}{\left(\omega_{2}\right)}$

$=\frac{\left((0.705 \mathrm{m})^{2}\right)(10 \mathrm{rpm})}{(20.5 \mathrm{rpm})}$

$=\sqrt{\frac{\left((0.705 \mathrm{m})^{2}\right)(10 \mathrm{rpm})}{(20.5 \mathrm{rpm})}}$

$r_{2} =0.5864\ m$

The weights are 0.5864 m apart.

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4 years ago

Which example provides a complete scientific description of an object in motion? the vibration of the jackhammer broke through t

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The hiker followed the north trail a distance of two kilometers in thirty minutes is an example that provides a complete scientific description of an object in motion.

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3 years ago

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5. George walks to a friend's house. He walks 750 meters North, then realizes he walked too far.

dedylja [7]

Answer:

Average speed: approximately $76.9\; {\rm m\cdot s^{-1}}$ .

Average velocity: approximately $38.5\; {\rm m \cdot s^{-1}}$ (to the north.)

Explanation:

Consider an object that travelled along a certain path. Distance travelled would be equal to the length of the entire path.

In contrast, the magnitude of displacement is equal to distance between where the object started and where it stopped.

In this question, the path George took required him to travel $750\; {\rm m} + 250\; {\rm m} = 1000\; {\rm m}$ in total. Hence, the distance George travelled would be $1000\; {\rm m}$ . However, since George stopped at a point $(750\; {\rm m} - 250\; {\rm m}) = 500\; {\rm m}$ to the north of where he started, his displacement would be only $500\; {\rm m}$ to the north.

Divide total distance by total time to find the average speed.

Divide total displacement by total time to find average velocity.

The total time of travel in this question is $13\; {\rm s}$ .. Therefore:

$\begin{aligned}\text{average speed} &= \frac{\text{total distance}}{\text{total time}} \\ &= \frac{1000\; {\rm m}}{13\; {\rm s}} \\ &\approx 76.9\; {\rm m\cdot s^{-1}}\end{aligned}$ .

$\begin{aligned}\text{average velocity} &= \frac{\text{total displacement}}{\text{total time}} \\ &= \frac{500\; {\rm m}}{13\; {\rm s}} && \genfrac{}{}{0px}{}{(\text{to the north})}{}\\ &\approx 38.5\; {\rm m\cdot s^{-1}} && (\text{to the north})\end{aligned}$ .

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2 years ago