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3 years ago

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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what was the acceleration of the cart with Low fan speed cm/s squared? what was the acceleration of the cart with medium fan spe

ArbitrLikvidat [17]

Explanation:

The attached figure shows data for the cart speed, distance and time.

For low fan speed,

Distance, d = 500 cm

Time, t = 7.4 s

Average velocity,

$v=\dfrac{d}{t}\\\\v=\dfrac{500}{7.4}\\\\v=67.56\ cm/s$

Acceleration,

$a=\dfrac{v}{t}\\\\a=\dfrac{67.56}{7.4}\\\\a=9.12\ cm/s^2$

For medium fan speed,

Distance, d = 500 cm

Time, t = 6.4 s

Average velocity,

$v=\dfrac{d}{t}\\\\v=\dfrac{500}{6.4}\\\\v=78.12\ cm/s$

Acceleration,

$a=\dfrac{v}{t}\\\\a=\dfrac{78.12}{6.4}\\\\a=12.2\ cm/s^2$

For high fan speed,

Distance, d = 500 cm

Time, t = 5.6 s

Average velocity,

$v=\dfrac{d}{t}\\\\v=\dfrac{500}{5.6}\\\\v=89.28\ cm/s$

Acceleration,

$a=\dfrac{v}{t}\\\\a=\dfrac{89.28}{5.6}\\\\a=15.94\ cm/s^2$

Hence, this is the required solution.

8 0

3 years ago

Read 2 more answers

A ball is thrown upward at a 45° angle. Inthe absence of air resistance, the ballfollows aA. tangential curve.B. sine curve.C. p

Evgesh-ka [11]

As ball is projected up in air at an angle of 45 degree without any air resistance

Let the initial speed will be v

now we will have

In x direction

$v_x = v cos45$

in y direction

$v_y = vsin45$

now displacement in x direction

$x = (vcos45)t + 0$

displacement in y direction

$y = (vsin45)t - \frac{1}{2}gt^2$

now from above two equations we have

$y = (vsin45)\frac{x}{vcos45} - \frac{1}{2}g(\frac{x}{vcos45})^2$

$y = xtan45 - \frac{1}{2v^2cos^245}gx^2$

so above equation is a quadratic equation and hence it will be a parabolic curve

so correct answer will be

C. parabolic curve.

8 0

3 years ago

The position-time equation for a cheetah chasing an antelope is:

pochemuha

Answer:

x = 1.6 + 1.7 t^2 omitting signs

a) at t = 0 x = 1.6 m

b) V = d x / d t = 3.4 t

at t = 0 V = 0

c) A = d^2 x / d t^2 = 3.4 (at t = 0 A = 3.4 m/s^2)

d) x = 1.6 + 1.7 * (4.4)^2 = 34.5 (position at 4.4 sec = 34.5 m)

4 0

3 years ago

Choose either eclipses or lunar phases and make a model of the event. Your model can take any form, providing it can explain how

babymother [125]

Answer: See the explanation below.

Explanation: For this assignment, I chose to display how eclipses are created.

My model was made utilizing a 3D displaying device program for all intents and purposes. The items utilized are three models I made for this presentation, Earth, the moon, and the sun. These three models will be utilized for the showcase.

The light that shines from the sun would create a shadow on the moon. The moon would then catch the light that should've arrived on Earth, making the shadow we call an eclipse. Earth gets a shadow of the moon and the remainder of Earth is lit up from the rest of the light, making an eclipse.

The individual I demonstrated my project to was [Someone you know], [Pronoun] said it precisely took after the occasion of an eclipse. The light from the sun being shined on to the moon rather than the Earth, creating the shadow we call an eclipse.

5 0

3 years ago

Please help.. urgent Which statement is equivalent to Newton's first law? a. 15,300 N b. 1.20*10^3 N c. 2,030 N d. 1,560 N

taurus [48]

According to Newton laws of motion,
F = m*a
Here, m = 1,560 Kg
a = 1.30 m/s²

Substitute their values,
F = 1,560 * 1.30
F = 2028 N ~ 2030 N [ Closest value ]

In short, Your Answer would be Option C

Hope this helps!

6 0

3 years ago