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vovangra [49]
3 years ago
5

Someone help me please, I don't know how to do it :( ​

Physics
1 answer:
elena-14-01-66 [18.8K]3 years ago
3 0

Answer:

fn /2 .3 'in

Explanation:

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Use the ratio version of Kepler’s third law and the orbital information of Mars to determine Earth’s distance from the Sun. Mars
frez [133]

Answer:

<em>1.49 x </em>10^{11}<em></em>

<em></em>

Explanation:

Kepler's third law states that <em>The square of the orbital period of a planet is directly proportional to the cube of its orbit.</em>

Mathematically, this can be stated as

T^{2} ∝ R^{3}

<em>to remove the proportionality sign we introduce a constant</em>

T^{2} = kR^{3}

k = \frac{T^{2} }{R^{3} }

Where T is the orbital period,

and R is the orbit around the sun.

For mars,

T = 687 days

R = 2.279 x 10^{11}

for mars, constant k will be

k = \frac{687^{2} }{(2.279*10^{11}) ^{3} } = 3.987 x 10^{-29}

For Earth, orbital period T is 365 days, therefore

365^{2} = 3.987 x 10^{-29} x R^{3}

R^{3} = 3.34 x 10^{33}

R =<em> 1.49 x </em>10^{11}<em></em>

8 0
3 years ago
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How many atoms are in NaHCO3
Elanso [62]
There are 6 atoms total:
-one sodium atom
-one hydrogen atom
-one carbon atom
-three oxygen atoms
4 0
4 years ago
What are some sources of error in a refraction lab using pins and a glass block?
ser-zykov [4K]

DONT WORRY

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6 0
2 years ago
Electromagnetic waves can transfer energy without a(n) _____________ (medium/electric field).
KATRIN_1 [288]
Medium.
Medium is matter, and electromagnetic waves do not need matter to travel. Just like the sun, its plasma light travels across space to reach us.
7 0
3 years ago
A gun has a muzzle speed of 90 meters per second. What angle of elevation should be used to hit an object 150 meters away? Negle
Artist 52 [7]

Answer:

θ₀ = 84.78° (OR) 5.22°

Explanation:

This situation can be treated as projectile motion. The parameters of this projectile motion are:

R = Range of Projectile = 150 m

V₀ = Launch Speed of Projectile = 90 m/s

g = 9.8 m/s²

θ₀ = Launch angle (OR) Angle of Elevation = ?

The formula for range of a projectile is given as:

R = V₀² Sin 2θ₀/g

Sin 2θ₀ = Rg/V₀²

Sin 2θ₀ = (150 m)(9.8 m/s²)/(90 m/s)²

2θ₀ = Sin⁻¹ (0.18)

θ₀ = 10.45°/2

<u>θ₀ = 5.22°</u>

Also, we know that for the same launch velocity the range will be same for complementary angles. Therefore, another possible value of angle is:

θ₀ = 90° - 5.22°

<u>θ₀ = 84.78°</u>

7 0
4 years ago
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