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3 years ago

8

Which statements accurately describe the shape of Earth’s orbit around the Sun? Check all that apply.

2 answers:

damaskus [11]3 years ago

4 0

<span>The Earth’s orbit is a nearly circular ellipse.
</span><span>The Sun is located at one of the two focal points.</span>

The Earth moves around the Sun in an orbit that is almost but not quite circular. As Kepler proved in the seventeenth century, the orbit is actually an ellipse. A parameter called the eccentricity (e) defines the degree of departure from a circle. A value of e=0 would indicate a circle whereas a value of e=0.9 would indicate a very elongated ellipse. The eccentricity of the Earth's orbit is currently e=0.0167.

anyanavicka [17]3 years ago

4 0

2. The Earth’s orbit is a nearly circular ellipse.

5. The Sun is located at one of the two focal points.

These are the two correct answers.

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A skater with a mass of 72 kg is traveling east at 5.8 m/s when he collides with another skater of mass 45 kg heading 60° south

Akimi4 [234]

The final velocity is 5.87 m/s

<u>Explanation:</u>

Given-

mass, $m_{1}$ = 72 kg

speed, $v_{1}$ = 5.8 m/s

$Mass_{2}$ , $m_{2}$ = 45 kg

$speed_{2}$ , $v_{2}$ = 12 m/s

Θ = 60°

Final velocity, v = ?

Applying the conservation of momentum:

$m_{1}$ X $v_{1}$ + $m_{2}$ X $v_{2}$ = ( $m_{1}$ + $m_{2}$ ) v

72 X 5.8 + 45 X 12 X cos 60° = (72 + 45) v

v = 417.6 + 540 X $\frac{0.5}{117}$

v = 417.6 + $\frac{270}{117}$

v = 5.87 m/s

The final velocity is 5.87 m/s

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Which of the following describes pressure in a fluid?

PtichkaEL [24]

<span>D. Pressure increases with increasing depth.

This occurs because there is more weight above you to increase the pressure.
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aniked [119]

A heliocentric system is a sun-centered

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A rectangular barge, 5.2 m long and 2.4 m wide, floats in fresh water. Suppose that a 410-kg crate of auto parts is loaded onto

sesenic [268]

<h2>Answer:</h2><h2>The depth of barge float=3 cm</h2><h2>Explanation:</h2>

Length of rectangular barge=5.2 m

Width of rectangular barge=2.4m

Mass of crate=410 kg

Let h be the height of barge float

Volume of barge float= $l\times b\times h=5.2\times 2.4\times h=12.48h$

Density of water= $10^3kg/m^3$

Weight of water displaced by barge=Buoyant force=-Weight of horse

$Volume\;of\;water\times density\;of\;water\times g=410\times g$

$12.48h\times 1000=410$

$h=\frac{410}{12.48\times 1000}=0.03 m$

1 m=100 cm

$0.03 m=0.03\times 100=3$ cm

Hence, the depth of barge float=3 cm

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Kobotan [32]

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