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Anarel [89]
3 years ago
14

Why would the planets move in a straight path if there was no gravitational energy from the sun​

Physics
2 answers:
omeli [17]3 years ago
6 0

Explanation:

The sun's gravitational force is very strong. If it were not, a planet would move in a straight line out into space. The sun's gravity pulls the planet toward the sun, which changes the straight line of direction into a curve. This keeps the planet moving in an orbit around the sun

Lemur [1.5K]3 years ago
5 0

Answer and Explanation:

The planets would move in a straight path since there would be no source that will be acting upon them (if the sun wasn't there).

Gravitational pull is the amount of force that an object emits. Due to the Sun having a larger mass and a larger gravitational pull, the planets orbit around it.  

If, for example, a house was bigger than everything, including planets and stuff, those things would orbit around the house.

A real life example is the moon orbiting Earth. The Earth has a greater magnitude and mass, so it is able to pull the Moon.

Right now the Sun is the external force, as its gravitational pull is moving the planets around. IF the sun was to magically disappear, the planets won't follow an orbit anymore, since there isn't a gravitational pull on them. This leads them to going in a straight line for possibly a couple years.

Eventually, the planets will stop going in a straight line and start to orbit each other, leading to the planets eventually crashing.

External Force = Sun's gravitational pull because of bigger mass

The planets themselves were created after the sun (though some minerals were created before the sun, but that's minerals and not the planet entirely). The sun gives the planet's their motion. The planets, however (though I'm not sure about mercury and Venus) are going super fast, fast enough to where they aren't being pulled into the sun, but not fast enough to go out of orbit.

If the sun goes out magically, the planets, actually yes will continue orbit due to them being in that motion at such a fast rate, but eventually when they are supposed to turn again, (like, to another part of the sun), the would go straight. Eventually, the will get into the orbit of something else and start going around that object's orbit, most likely outcome being that they crash.

The excess force from no gravitational pull will make the planets go in a straight path. Since nothing is acting upon it, the planets would move in a straight line. Imagine if you were on those spinning circle things, and then let go. You would go around it for like a second and then go immediately straight in one direction.

<em><u>#teamtrees #PAW (Plant And Water)</u></em>

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1. Which of the following statements best describes matter?
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Matter is anything that has mass

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Calculate the pressure exerted by a 4000N camel on the sand. The camel’s feet have a
Harrizon [31]

Answer:

The pressure exerted by camel feet is <u>2000 N/m²</u>.

Step-by-step explanation:

<h3><u>Solution</u> :</h3>

Here, we have given that ;

  • Force applied on camel feet = 4000 N
  • Total area of camel feet = 2 m²

We need to find the pressure exerted by camel feet.

As we know that :

{\longrightarrow{\pmb{\sf{Pressure= \dfrac{Area}{Force}}}}}

Substituting all the given values in the formula to find the pressure exerted by camel feet.

\begin{gathered} \begin{array}{l} {\longrightarrow{\sf{Pressure= \dfrac{Area}{Force}}}} \\  \\ {\longrightarrow{\sf{Pressure= \dfrac{4000}{2}}}}  \\  \\ {\longrightarrow{\sf{Pressure= \cancel{\dfrac{4000}{2}}}}} \\  \\ {\longrightarrow{\sf{Pressure= 2000 \: N/{m}^{2}}}} \\  \\\star \:  \small\underline{\boxed{\sf{\purple{Pressure= 2000 \: N/{m}^{2}}}}} \end{array}\end{gathered}

Hence, the pressure exerted by camel feet is 2000 N/m².

\rule{300}{2.5}

3 0
2 years ago
With detailed explaniation
belka [17]
  • Ø=37°
  • Initial velocity=u=20m/s
  • g=10m/s²

#A

\\ \rm\Rrightarrow H_{max}=\dfrac{u^2sin^2\theta}{2g}

\\ \rm\Rrightarrow H_{max}=\dfrac{20^2(sin37)^2}{2(10)}

\\ \rm\Rrightarrow H_{max}=\dfrac{400sin^237}{20}

\\ \rm\Rrightarrow H_{max}=20sin^237

\\ \rm\Rrightarrow H_{max}=7.2m

#B

\\ \rm\Rrightarrow R=\dfrac{u^2sin2\theta}{g}

\\ \rm\Rrightarrow R=\dfrac{20^2sin74}{10}

\\ \rm\Rrightarrow R=40sin74

\\ \rm\Rrightarrow R=38.5m

#C

\\ \rm\Rrightarrow T=\dfrac{2usin\theta}{g}

\\ \rm\Rrightarrow T=\dfrac{2(20)sin37}{10}

\\ \rm\Rrightarrow T=4sin37

\\ \rm\Rrightarrow T=2.4s

Now

\\ \rm\Rrightarrow v=u-gt

\\ \rm\Rrightarrow v=20-10(2.4)

\\ \rm\Rrightarrow v=20-24

\\ \rm\Rrightarrow v=-4m/s

4 0
2 years ago
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