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3 years ago

Wenen an astronaut travels to different planets, which of the following planets will the astronaut's weight stay constant to

Physics

1 answer:

devlian [24]3 years ago

3 0

Answer:

None.

Explanation:

The weight of the astronaut is equal to the force of gravity the astronaut feels that depends on the mass of the planet.

Because the mass of the Earth is different than that of Jupiter, Saturn and Mars, the astronaut's weight will change when he/she travels to either planets.

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You whirl a stone in a horizontal circle in such a way that the stone is in uniform circular motion. Which of the following is t

horsena [70]

a. The direction of the stone's velocity changes as it moves around the circle.

b. The magnitude of the stone's velocity does not change.

d. The change in direction of the stone's motion is due to the centripetal force acting on the stone.

Above given are true for the given situation.

<u>Answer:</u> Option A, B and D

<u>Explanation:</u>

Circular motion may be characterized as the moving of an objects along the diameter of the circle or any circular direction. It may be standardized and non-uniform based on whether or not the rate of rotation is unchanged.

The velocity, a vector quantity is constant in a uniform circle motion speed is constant as its direction continues to change. Centripetal force works inward toward the core to counterbalance the centrifugal force from the center moving outward.

4 0

3 years ago

Pls help A car starts from rest and gains a velocity of 20m/s in 10 seconds calculate its acceleration and the distance covered

Soloha48 [4]

Answer:

$\boxed{\sf Acceleration \ (a) = 2 \ m/s^{2}}$

$\boxed{\sf Distance \ covered \ (s) = 100 \ m}$

Given:

Initial velocity (u) = 0 m/s

Final velocity (v) = 20 m/s

Time taken (t) = 10 sec

To Find:

(i) Acceleration (a)

(ii) Distance covered (s)

Explanation:

$\sf (i) \ From \ 1^{st} \ equation \ of \ motion:$

$\sf \implies v = u + at$

$\sf \implies 20 = 0 + a(10)$

$\sf \implies 10a = 20$

$\sf \implies \frac{10a}{10} = \frac{20}{10}$

$\sf \implies a = 2 \: m/ {s}^{2}$

$\sf (ii) \ From \ 2^{nd} \ equation \ of \ motion:$

$\sf \implies s = ut + \frac{1}{2} a {t}^{2}$

$\sf \implies s = (0)(10) + \frac{1}{2} \times 2 \times {(10)}^{2}$

$\sf \implies s = \frac{1}{ \cancel{2}} \times \cancel{2} \times {(10)}^{2}$

$\sf \implies s = {10}^{2}$

$\sf \implies s = 100 \: m$

6 0

3 years ago

A 40 g ball rolls around a 30 cm -diameter L-shaped track, shown in the figure, (Figure 1)at 60 rpm . What is the magnitude of t

levacccp [35]

Answer:

0.47 N

Explanation:

Here we have a ball in motion along a circular track.

For an object in circular motion, there is a force that "pulls" the object towards the centre of the circle, and this force is responsible for keeping the object in circular motion.

This force is called centripetal force, and its magnitude is given by:

$F=m\omega^2 r$