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Mariulka [41]
3 years ago
8

True or false earth orbit is nearly circular

Physics
2 answers:
Simora [160]3 years ago
4 0
The answer would be true
Naddik [55]3 years ago
4 0
I guess that depends on what you mean by "nearly".
"Nearly" is one of those rubber words that means different things
to different people, and may even mean different things to the same
person at different times and in different situations.

I can tell you this much:

-- The Earth's orbit is an ellipse with eccentricity of  0.0167... .

-- At closest approach to the sun (perihelion, around January 3), the Earth
is about  3.3% closer to the sun than at its farthest (aphelion, around July 5).

-- When the true shape of the orbit is drawn on paper, its departure from
a perfect circle is not noticeable.

So to me, Earth's orbit is "nearly circular".  Your interpretation of the same
words may be different.
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What is the gravitational potential energy of a 15.0kg object that is 5.00m above the ground relative to a point 8.00m above the
Licemer1 [7]

Explanation:

an object's gravitational potential energy Eg is m×g×h where:

m=mass

g=9.8m/s²

h=height relative to the closest object below it (because it cannot potentially fall through it

so Eg = 15×9.8×5=735J

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2 years ago
What study did james e hansen focus on
Brut [27]
James E. Hansen studied climate change
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2 years ago
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An object is placed in front of a diverging lens, such that the object-to-image distance is 71 cm.
Pachacha [2.7K]

Explanation:

Given that,

Object-to-image distance d= 71 cm

Image distance = 26 cm

We need to calculate the object distance

u -v= d

u=71+26=97\ cm

We need to calculate the focal length

Using formula of lens

\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}

put the value into the formula

\dfrac{1}{f}=\dfrac{1}{-26}+\dfrac{1}{97}

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f=-35.52\ cm

The focal length of the lens is 35.52.

(B). Given that,

Object distance = 95 cm

Focal length = 29 cm

We need to calculate the distance of the image

Using formula of lens

\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}

Put the value in to the formula

\dfrac{1}{-29}=\dfrac{1}{v}-\dfrac{1}{95}

\dfrac{1}{v}=\dfrac{1}{-29}-\dfrac{1}{95}

\dfrac{1}{v}=-\dfrac{124}{2755}

v=-22.21\ cm

We need to calculate the magnification

Using formula of magnification

m=\dfrac{v}{u}

m=\dfrac{22.21}{95}

m=0.233

The magnification is 0.233.

The image is virtual.

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4 0
3 years ago
2.5 gram sample of a radioactive element was formed in a 1960 explosion of an atomic bomb at Johnson Island in the Pacific Test
kow [346]

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0.15625 grams

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Half life: It is related to the decay of radioactive material. The duration in which  half of the material will be degraded/decayed. That means after half life 50% of the radioactive material will be left. Here the half life is 28 years.

Initial quantity of the sample: 2.5 grams.

After 28 years, the leftover quantity = 1.25 grams

After 56 years, the leftover quantity = 0.625 grams

After 84 Years, the leftover quantity = 0.3125 grams

After 112 years, the leftover quantity = 0.15625 grams

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3 years ago
How long will it take to travel 200.000 m [N] traveling 10 m/s [N]?
gtnhenbr [62]

Answer: here are 1,000m in a km, so 200km is 200,000m

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Explanation:

8 0
2 years ago
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