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3 years ago

The particles in a gaseous substance can move easily because the____

Physics

2 answers:

oksano4ka [1.4K]3 years ago

8 0

Answer:

particles are free to move

AVprozaik [17]3 years ago

3 0

Answer:

a option okkkkkkkkkkkkkk

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4 years ago

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3 years ago

A satellite of mass 5600 kg orbits the Earth and has a period of 6200 s.

Troyanec [42]

Answer:

(a) Radius of orbit will be $=7.32\times10^6m$

(b) Earth gravitational force will be $=4.18\times 10^4N$

Explanation:

We have given

Mass of the earth, $M=6\times 10^{24}kg$

Mass of the satellite, m = 5600 kg

Radius of earth, $R=6.4\times 10^6m$

Time period T = 6200 sec

We know that $\omega =\frac{2\pi }{T}=\frac{2\times 3.14}{6200}=0.00101rad/sec$

Now

(a) We know that $\omega ^2=\frac{GM}{R^3}$

$R^3=\frac{GM}{\omega ^2}$

$R^3=\frac{6.67\times 10^{-11}\times 6\times 10^{24}}{0.00101 ^2}$

$R^3=3.92\times 10^{20}$

Radius of the orbit $R=7.32\times 10^6m$

(b)

Force $F=\frac{GMm}{R^2}=\frac{6.67\times 10^{-11}\times 6\times 10^{24}\times 5600}{(7.32\times 10^6)^2}=4.18\times 10^4N$

(c)

Altitude $h=radius\ of\ orbit-radius\ of\ earth=7.32\times 10^6-6.4\times 10^6=0.92\times 10^6m$

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3 years ago

A motorboat accelerates uniformly from a velocity of 6.5 m/s west to a velocity of 1.5 m/s west. If its acceleration was 2.7 m/s

expeople1 [14]

Answer:

<em>The motorboat ends up 7.41 meters to the west of the initial position </em>

Explanation:

<u>Accelerated Motion </u>

The accelerated motion describes a situation where an object changes its velocity over time. If the acceleration is constant, then these formulas apply:

$\vec v_f=\vec v_o+\vec a.t$

$\displaystyle \vec r=\vec v_o.t+\frac{\vec a.t^2}{2}$

The problem provides the conditions of the motorboat's motion. The initial velocity is 6.5 m/s west. The final velocity is 1.5 m/s west, and the acceleration is $2.7 m/s^2$ to the east. Since all the movement takes place in one dimension, we can ignore the vectorial notation and work with the signs of the variables, according to a defined positive direction. We'll follow the rule that all the directional magnitudes are positive to the east and negative to the west. Rewriting the formulas: