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SSSSS [86.1K]
4 years ago
6

The table below shows the average distances of Venus and Earth from the sun. -Name of Planet -Distance from the sun (in AU) Venu

s 0.72 Earth 1.0 What is the difference between the orbital periods of Venus and Earth? a. 1.72 years b. 1.62 years c. 0.62 years d. 0.28 years
Physics
2 answers:
Serhud [2]4 years ago
6 0
<span>Venus goes around faster than Earth, so its period must be less than Earths 1.0 years. That eliminates the first two answers. My recollection is that T^2 = k r^3, relating period of orbit to radius, so T should be (0.72/1.00)^3/2 = 0.61 of the Earths period, so 0.62 is ans. </span>
Paul [167]4 years ago
6 0

Answer: 0.39 years (Venus orbits the Sun in 0.62 Earth years)

Explanation:

The orbital period is related to the distance of the planet from the star as:

T² = R³

where T is in years and R is in AU.

<u>Orbital period of Venus:</u>

T² = (0.72)³

⇒ T = 0.62 years

<u>Orbital period of Earth </u>

T² = (1.0)³

⇒T = 1 year

The difference between the orbital periods of Venus and Earth is: (1 -0.62) = 0.38 years. Venus orbits the Sun in 0.62 Earth years.

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A 50-n force is applied to a 10-kg chair, but the chair does not change its motion. What could explain this?.
Zepler [3.9K]

Explanation:

it will not change it's motion because it's force is less than the force applied

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2 years ago
A jet airliner moving initially at 504 mph (with respect to the ground) to the east moves into a region where the wind is blowin
adelina 88 [10]

Answer

given,

intial velocity = 504 mph

wind speed = 219 mph

at an angle of 29◦

from the data

 The resultant velocity =

V = V_x \hat{i} + V_y\hat{j}

 V = (504 + 219 cos 29^0 ) \hat{i} +(219 sin 29^0 )\hat{j}

 V = 695.54\hat{i} + 106.17 \hat{j}

the magnitude of velocity

V = \sqrt{695.54^2+106.17^2}

V = 703.59 m/s

direction

tan θ =  \dfrac{106.17}{695.54}

θ = 8.676°

8 0
4 years ago
Water drops fall from the edge of a roof at a steady rate. a fifth drop starts to fall just as the first drop hits the ground. a
Alchen [17]

The height of the roof is <u>3.57m</u>

Let the drops fall at a rate of 1 drop per t seconds. The first drop takes 5t seconds to reach the ground. The second drop takes 4t seconds to reach the bottom of the 1.00 m window, while the 3rd drop takes 3t s to reach the top of the window.

Calculate the distances traveled by the second and the third drops s₂ and s₃, which start from rest from the roof of the building.

s_2=\frac{1}{2} g(4t)^2=8gt^2\\  s_3=\frac{1}{2} g(3t)^2=(4.5)gt^2

The length of the window s is given by,

s=s_2-s_3\\ (1.00 m)=8gt^2-4.5gt^2=3.5gt^2\\ t^2=\frac{1.00 m}{(3.5)(9.8m/s^2)} =0.02915s^2

The first drop is at the bottom and it takes 5t seconds to reach down.

The height of the roof h is the distance traveled by the first drop and is given by,

h=\frac{1}{2} g(5t)^2=\frac{25t^2}{2g} =\frac{25(0.02915s^2)}{2(9.8m/s^2)} =3.57 m

the height of the roof is 3.57 m



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4 years ago
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3 years ago
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Phobos's Orbit. Phobos orbits Mars at a distance of 9,380 km from the center of the planet and has a period of 0.3189 days. Assu
Elenna [48]

Answer:

Explanation:

The relation between time period of moon in the orbit around a planet can be given by the following relation .

T² = 4 π² R³ / GM

G is gravitational constant , M is mass of the planet , R is radius of the orbit and T is time period of the moon .

Substituting the values in the equation

(.3189 x 24 x 60 x 60 s)²  = 4 x 3.14² x ( 9380 x 10³)³ / (6.67 x 10⁻¹¹ x M)

759.167 x 10⁶ = 8.25 x 10²⁰ x 39.43 / (6.67 x 10⁻¹¹ x M )

M = .06424  x 10²⁵

= 6.4 x 10²³ kg .

4 0
3 years ago
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