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Nastasia [14]
3 years ago
14

Which characteristics of Earth’s orbit are in agreement with Kepler’s second law? Check all that apply.

Physics
2 answers:
Nataly [62]3 years ago
6 0

Here is a picture of the correct answers.

artcher [175]3 years ago
6 0

Explanation :  

The angular velocity of the planet in the elliptical orbit will vary. Its shows that the planet  travel slower when farther from the sun, then faster when closer to the sun.

1. Yes, Earth experiences the strongest gravitational force when it is closest to the Sun.

2. Yes,  Earth moves the greatest distance in thirty days when it is closest to the Sun.

3. Yes,  Earth travels with the slowest tangential speed when it is farthest from the Sun.

4. Yes,  Earth sweeps out the same area in the same time frame anywhere in its orbit

Kepler's second law : Kepler describe the motion of the planets around the sun. the line joining the planet and the sun sweeps out  the same area in the same time.

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Was the color orange named after the fruit or was the fruit named after the color ​
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I need so much help 50 points!!!!!!<br><br>complete the graph
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6 0
3 years ago
A person has legs of length L = 1.10 m. (a) If the maximum angle between the legs during walking is ϕ = 50o , what is the person
vampirchik [111]

Answer:

a). Maximum Length L=0.929m

b). T=0.83 Hz or 1.2s

c). Longer, the effortless waling T=2.1 Hz or t=0.475s

d). t=1.2s V=0.774 \frac{m}{s}

t=0.475s V=1.95 \frac{m}{s}

Explanation:

Length legs=L=1.1m

angle=50

the step that give the person forms a triangle whose two sides are known and the angle that forms between them, then using trigonometry as the image

Divide the original triangle in two and form a right triangle so the angle is 25 and the L is hypotenuse and the opposite is the step length

a).

sin(\alpha) =\frac{op}{h}

op=h*sin(\frac{\alpha }{2})\\ op=1.1m*sin(\frac{50}{2})\\op=0.464m

Length of the step

L=0.464m*2

L=0.928m

b).

period=T

T=\frac{1}{time}=\frac{1}{t}\\ T=\frac{1}{1.2s}\\T=0.83 s^{-1}\\ T=0.83Hz

c).

T1=2\pi *\sqrt{\frac{L}{g}} \\T1=2\pi *\sqrt{\frac{1.1m}{9.8\frac{m}{s^{2}}}}\\ T1=2.1 Hz

The period is the inverse of the time of the motion so, the T1 is faster that the T because

t=\frac{1}{T1}=t= \frac{1}{2.1}=0.47s

d).

The speed is the relation between the distance with time so:

Vt=\frac{0.928m}{1.2s} \\Vt=0.773 \frac{m}{s} \\Vt=\frac{0.928m}{0.475s} \\Vt=1.953 \frac{m}{s}

3 0
3 years ago
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