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3 years ago

Answer please just 1-7 it should be easy

Physics

2 answers:

Dmitrij [34]3 years ago

6 0

1. As of right now, our solar system only includes asteroids and they are found in Kuiper Belt next to Saturn.

On May 6, 2020, scientists have discovered a mysterious blanket of darkness approximately 1,000 light years away from our solar system. However, in this case, black holes would not be a correct answer.

Even though constellations may seem close, they are actually far away since the size of the stars are immense and can be visible from light years away. Constellations are not in our solar system.

Galaxies are groups of solar systems so therefore galaxies is not an answer.

2. Ptolemy was the man who proposed the early Earth-centered model of the universe and his model was called the Ptolemaic model also known as the geocentric model.

3. The planet that is most similar to Earth in mass and diameter is Venus.

4. The inner planets in our solar system are formed from dense elements and that is why they are able to withstand the intense of the sun. Meanwhile the outer planets were formed from gaseous materials but Pluto is so cold that its elements are hardened and dense.

5. The answer would be D since they orbit in the same direction and on the same plane.

6. The speed of rotation determines the length and duration of day on a planet.

7. The answer would be D since they have to have their own gravity, have a path clear of smaller bodies , and orbit a star,

Margaret [11]3 years ago

4 0

1. Asteroids

2. Ptolemly

3. Mars

4. They are formed from dense elements

5. In the same direction

6. Speed of rotation

7. All of thee above

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3 years ago

Why does New and Full moons produce stronger tides.

serious [3.7K]

Because at the time of New and Full moons, the Earth, Sun,
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4 0

4 years ago

A bicycle travels 6.10 km due east in 0.210 h, then 11.30 km at 15.0° east of north in 0.560 h, and finally another 6.10 km due

Virty [35]

Answer:

Explanation:

Given

First bicycle travels 6.10 km due to east in 0.21 h

Suppose its position vector is $r_1$

$r_1=6.10\hat{i}$

After that it travels 11.30 km at $15^{\circ}$ east of north in 0.560 h

suppose its position vector is $r_2$

$r_{2}=11.30\left ( cos15\hat{j}+sin15\hat{i}\right )$

after that he finally travel 6.10 km due to east in 0.21 h

suppose its position vector is $r_3$

$r_{3}=6.10\hat{i}$

so position of final position is given by

$r=r_1+r_{2}+r_{3}$

$\vec{r}=15.12\hat{i}+10.91\hat{j}$

$\vec{v_{avg}}=\frac{\vec{r}}{t}$

t=0.21+0.56+0.21=0.98 h

$\vec{v_{avg}}=15.42\hat{i}+11.13\hat{j}$

$|v_{avg}|=\sqrt{361.71}=19.01 km/hr$

For direction

$tan\theta =\frac{11.13}{15.42}=0.721$

$\theta =35.791^{\circ}$ w.r.t to x axis

5 0

4 years ago

In simple harmonic motion, the acceleration is proportional to In simple harmonic motion, the acceleration is proportional to:__

erik [133]

Answer:

e. all of the above

Explanation:

In simple harmonic motion, the acceleration is given as;

a = -ω²x = -(2πf)²x

$a_{max} = -\omega ^2A$

where;

ω is the angular velocity

f is the frequency

x is the displacement

A is the amplitude

Thus, In simple harmonic motion, the acceleration is proportional to the amplitude, velocity, frequency, and displacement.

The correct option is E.

"all of the above"

5 0

3 years ago

You place a 3.0-m-long board symmetrically across a 0.5-m-wide chair to seat three physics students at a party at your house. If

wlad13 [49]

Answer:

between locations that are 14 cm outboard of the chair edges
the weightless board is centered and end sitters are 25 cm from the ends

Explanation:

We can assume the .5 m-wide chair means that it is comfortable for each student to sit 0.25 m from the end of the board. If the board is centered on the chair, then each student is 1 m from the edge of the chair.

When Dan and Tahreen are seated on the board, their center of mass is ...

(50 kg×2.5 m)/(50 kg +67 kt) = 1.068 m

to the right of the position where Dan is seated. Since this location is over the chair, the board is stable.

Komila can sit as much as x distance from the chair toward Dan, where ...

67(1) +54(x) = 50(1.5)

x = 8/54 ≈ 0.148 . . . . meters

Or, Komila can sit as much as x distance from the chair toward Tahreen, where ...

67(1.5) = 54(x) +50(1)

x = 50.5/54 ≈ 0.935 . . . . meters

Scenario 1

Assuming the (weightless) board is centered on the chair, Komila can sit anywhere between 14.8 cm left of the chair and 93.5 cm right of the chair and the board will remain stable. Sitting on the board centered on the chair is a suitable location. The two students sitting on the ends must become (and stay) seated at the same time. They both must be seated 0.25 m from the end of the board for the other dimensions to remain valid.

Scenario 2

Assuming the (weightless) board is located so its left end is 1.068 m from the chair, and Dan and Tahreen are seated 0.25 m from the ends of the board, Komila can sit anywhere within (117/54×.25 m) = 0.54 m of the chair and the board will remain stable. Again, sitting centered on the chair is a suitable location.

There does not appear to be any location where Komila can sit and have the board remain stable with only Dan or Tahreen seated on one end (assuming a width of 0.5 m for each sitter).

_____

Comment on the question

For the board to remain stable, the sum of moments about either edge of the chair must tend to rotate the board toward the chair. This sum will depend on the locations of the sitters relative to each edge of the chair, so there is significant freedom in choosing locations. To make the problem tractable, we have made some specific assumptions about where the board is and what the locations of the sitters might be. YMMV

3 0

4 years ago