1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
GrogVix [38]
3 years ago
7

Consider again the set of observations from Part A. This time, classify each observation according to whether it is consistent w

ith only the Earth-centered model, only the Sun-centered model, both models, or neither model. (Note that an observation is "consistent" with a model if that model offers a simple explanation for the observation.)
(1) Mercury goes through a full cycle of phases.
(2) Moon rises in east, sets in west.
(3) stars circle daily around north or south celestial pole.
(4) positions of nearby stars shift slightly back & forth each year.
(5) a distant galaxy rises in east, sets in west each day.
(6) a planet beyond Saturn rises in west, sets in east.
(7) we sometimes see a crescent Jupiter.
Physics
1 answer:
DaniilM [7]3 years ago
8 0

Answer:

1. Sun-centered.

2.Both models.

3. Both models.

4. Sun-centered.

5.Both models.

6. Earth-centered only

7. Neither models.

Explanation:

Mercury goes through a full cycle phase. This is not sun-centered because the earth rotes from west to east,therefore objects in the sky moves east to west.

We sometimes see a crescent in Jupiter . These models cannot classify the observation because for an object to be seen as a crescent,the object must come between the earth and the sun.

Observation 3 and 5 can be classified with both models because the rise and set of all objects in the sky depend only on the rotation of the earth.

You might be interested in
Red light is bent the least of all colors as it passes through a prism. What does this tell you about red light? It has a short
Alik [6]

Answer:

Longest wavelength, lowest intensity

Explanation:

7 0
4 years ago
Two charged particles, q1 and q2, are located on the x-axis, with q1 at the origin and q2 initially atx1 = 14.7 mm.In this confi
Sladkaya [172]

Answer:

The force magnitude is 1.75 μN acting outward from the origin towards x₂

Explanation:

The given parameters, are;

The location of q₁ = The origin

The location of q₂ = 14.7 mm from q₁

The repulsive force exerted on q₂ by q₁ = 2.62 μN

The location the particle q₂ is located to =  18.0 mm from q₁

By Coulomb's law, we have;

F = k\dfrac{q_1 \times q_2 }{r^2}

Where;

k = Coulomb constant ≈ 8.99 × 10⁹ kg·m³/(s²·C²)

r = The distance between the particles

F = The force acting between the particles

When r = 14.7 mm F = 2.62 μN

∴ q₂ × q₁ = r² × F/k = (14.7×10^(-3))²×2.62×10^(-6)/(8.9875517923^9) ≈ 1.48 × 10⁻¹⁸

q₂ × q₁ = 1.48 × 10⁻¹⁸ C²

When the distance is increased to 18.0 mm, we have;

F = (8.9875517923^9) × 1.4796647 × 10^(-18)/((18×10^(-3))²) = 1.75 × 10⁻⁶ N

∴ F = 1.75 μN.

Therefore;

The force magnitude is 1.75 μN outward from the origin towards  x₂.

8 0
4 years ago
A cancer or cancerous tumour on the skin<br> as a result of extensive exposure to UV rays.
nekit [7.7K]

Answer:

Almost all skin cancers (approximately 99% of non-melanoma skin cancers and 95% of melanoma) are caused by too much UV radiation from the sun or other sources such as solaria (solarium, sunbeds, and sun lamps).  

Explanation:

Skin cancer develops in the cells in the epidermis – the top or outer layer of the skin.  UV radiation is made up of UVA and UVB rays which are able to penetrate the skin and cause permanent damage to the cells below:

UVA penetrates deeply into the skin (the dermis) causing genetic damage to cells, photo-ageing (wrinkling, blotchiness etc) and immune-suppression.

UVB penetrates into the epidermis (top layer of the skin) causing damage to the cells. UVB is responsible for sunburn – a significant risk factor for skin cancer, especially melanoma.

If the body is unable to repair this damage the cell can begin to divide and grow in an uncontrolled way. This growth can eventually form a tumor.

UVA and UVB both contribute to sunburn, skin ageing, eye damage and melanoma and other skin cancers.

3 0
4 years ago
PLEASE HELPPPPPPPPPPPPPP
MrRa [10]
Your answer is c hope I helped
7 0
3 years ago
A tugboat tows a ship with a constant force of magnitude F1. The increase in the ship's speed during a 10 s interval is 3.0 km/h
Yuki888 [10]

Answer:

The magnitude of F₁ is 3.7 times of F₂

Explanation:

Given that,

Time = 10 sec

Speed = 3.0 km/h

Speed of second tugboat = 11 km/h

We need to calculate the speed

v_{1}=\dfrac{3.0\times10^{3}}{3600}

v_{1}=0.833\ m/s

The force F₁is constant acceleration is also a constant.

F_{1}=ma_{1}

We need to calculate the acceleration

Using formula of acceleration

a_{1}=\dfrac{v}{t}

a_{1}=\dfrac{0.833}{10}

a_{1}=0.083\ m/s^2

Similarly,

F_{2}=ma_{2}

For total force,

F_{3}=F_{2}+F_{1}

ma_{3}=ma_{2}+ma_{1}

The speed of second tugboat is

v=\dfrac{11\times10^{3}}{3600}

v=3.05\ m/s

We need to calculate total acceleration

a_{3}=\dfrac{v}{t}

a_{3}=\dfrac{3.05}{10}

a_{3}=0.305\ m/s^2

We need to calculate the acceleration a₂

0.305=a_{2}+0.083

a_{2}=0.305-0.083

a_{2}=0.222\ m/s^2

We need to calculate the factor of F₁ and F₂

Dividing force F₁ by F₂

\dfrac{F_{1}}{F_{2}}=\dfrac{m\times0.83}{m\times0.22}

\dfrac{F_{1}}{F_{2}}=3.7

F_{1}=3.7F_{2}

Hence, The magnitude of F₁ is 3.7 times of F₂

3 0
3 years ago
Other questions:
  • The mixture you separated was mixture oof iron filings, sand, and salt. Based on your understanding of matter, is this mixture a
    12·2 answers
  • WHY did NASA scientist Madhulika Guhathakurta compare the solar eclipse to the 1969 moon landing?
    11·1 answer
  • A neutron is confined in a one-demensional potential box of width 5.0 x 10^-15 m. Calculate the minimum kinetic energy of the ne
    6·1 answer
  • A grating has 460 rulings/mm. What is the longest wavelength for which there is a 6.0th-order diffraction line
    10·1 answer
  • 7. A woman steps in front of a child to keep him from running off. Which term best describes this example? (2 po
    8·2 answers
  • What is the vertical displacement below any line?
    13·1 answer
  • Hawk maintains a consistent internal temperature in both the heat of the day in the cool of the night. This is called
    13·1 answer
  • What part of the Milky Way allows for determination of its rotational direction? A. Celestial body B. Galactic body C. Galactic
    13·2 answers
  • Please help I literally don’t understand
    15·1 answer
  • The average threshold of dark-adapted (scotopic) vision is 4.00 10-11 W/m2 at a central wavelength of 500 nm. If light with this
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!