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Anni [7]
3 years ago
14

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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It’s North of the equator
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Arrange an 8-, 12-, and 16-Ω resistor in a combination that has a total resistance of 8.89 Ω pls with de work
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20 ohms in parallel with 16 ohm= 8.89

20x16/20+16. Product over sum


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4 years ago
Physicists at CERN study the conditions present during the big bang by using machines to do what?
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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

(C) Height will be 0.92\times 10^6m

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

8 0
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:

v_f=v_o+a.t

\displaystyle x=v_o.t+\frac{a.t^2}{2}

Solving the first one for t

\displaystyle t=\frac{v_f-v_o}{a}

We have

v_o=-6.5,\ v_f=-1.5,\ a=2.7

Using these values

\displaystyle t=\frac{-1.5+6.5}{2.7}=1.852\ s

We now compute x

\displaystyle x=(-6.5)(1.852)+\frac{(2.7)(1.852)^2}{2}

x=-7,41\ m

The motorboat ends up 7.41 meters to the west of the initial position

5 0
3 years ago
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