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Dmitrij [34]
4 years ago
9

Suppose astronomers discover a new planet farther away from the Sun than Earth. How would the day and year of this planet compar

e to Earth's?
A. The planet's day would be longer, and its year would be shorter.
B. The planet's day would be shorter, and its year would be longer.
C. The planet's year would be longer, but it is impossible to predict how its day would compare to Earth's.
D. The planet's year would be shorter, but it is impossible to predict how its day would compare to Earth's. PLEASE HELP ❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️
Physics
2 answers:
Volgvan4 years ago
8 0
The year would definitely be longer, due to Keller’s 3rd law, but I know of several objects orbiting farther from the sun than earth with both much longer and much shorter day lengths.
posledela4 years ago
6 0
I think the answer would be C. because the orbit around the sun compared to the Earth's (Identifies the seasons change and the year) would be longer, but the day is unknown because that is where it would come down to the total mass of the planet and other important factors about the planets properties.
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Answer:

Explanation:

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TRUE OR FALSE. if an object does not change its position at a given time interval, then it is at rest or its speed is zero or no
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true

Explanation:

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3 years ago
The diagram below shows a wave with its wavelength indicated in red.
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4 years ago
Read 2 more answers
To tugboats pull a disabled supertanker. Each tug exerts a constant force of 1.50•10^6 N, one at an angle 19.0° west of north, a
irina1246 [14]

Answer:

Work done by a tug boat, W = 1.735 x 10⁸ J

Explanation:

Given,

The of each tugboat, F = 1.5 x 10⁶ N

The angle of each tugboat forms with the resultant force, θ = 19°

The displacement of the supertanker, s = 710 m

The individual tugboat will be responsible for the displacement, d = 710/2

                                                                                                               = 355 m

The displacement component in each tugboat direction = 355 · sin θ meter

Therefore, the work done by each tugboat is

                                           W = F x S    joules

Substituting the values in the above equation

                                            W = 1.5 x 10⁶  x  355 · sin θ

                                                = 1.735 x 10⁸ J

Hence, the work done by each tugboat is, W = 1.735 x 10⁸ J                        

6 0
3 years ago
A cement block accidentally falls from rest from the ledge of a 53.0-m-high building. When the block is 14.0 m above the ground,
katrin [286]

Answer:

0.405 seconds

Explanation:

Consider the amount of time it takes the block to fall from 53 m up to 14 m above the ground; then consider the amount of time it takes the block to fall from 53 m up to 2 m above  the ground.

First,     d = (1/2) gt^2    or     t=   ( 2 d / g)^1/2

= ( 2 × 39 / 9.8)^1/2 = 2.8212 seconds

Then, to fall from 53 down to 2 meters...

 d = (1/2) gt^2    or     t=   ( 2 d / g)^1/2

= ( 2 * 51/ 9.8 )^1/2 = 3.2262 seconds

So the amount of time it takes for the block to fall from 14 m upto 2 m above the ground

3.2262 - 2.8212 = 0.405 seconds      

this is how much time there is from when the man sees the block until it hits him. Not much time...

5 0
3 years ago
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