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Katarina [22]
2 years ago
13

Which of the following is NOT an argument for showing that the Earth must be round: a. during an eclipse of the Moon, the shadow

of the Earth is always seen to be round b. when ships travel a large distance away, we see their hulls disappear first and their masts disappear last c. the height of the North Star changes as we travel to different latitudes d. photographs of the Earth from space always show a round body e. the Sun is seen blocking different constellations in the course of a year Submit Your Answer
Physics
1 answer:
Hitman42 [59]2 years ago
4 0

Answer: The correct option is option E (the Sun is seen blocking different constellations in the course of a year.

Explanation:

The earth, which is one of the planets of the solar system that supports life, is shperical in shape. The spherical ( round) shape of the earth is marked by the intervening highlands and oceans on its surface.

Evidence to show that the earth is shperical are:

--> The Lunar eclipse: During an eclipse of the Moon, the shadow of the Earth is always seen to be round.

--> Ships Visibility: When ships travel a large distance away, we see their hulls disappear first and their masts disappear last.

-->Altitude of Polaris (North Star): The height of the North Star changes as we travel to different latitudes. That is ,increases as you move toward the North pole, or decreases as you move toward the equator.

--> Aerial photographs: Photographs of the Earth from space always show a round body.

The statement that doesn't prove that the earth is spherical in shape is (the Sun is seen blocking different constellations in the course of a year). The sun is seen in front of stars blocking different constellation in a year because the earth orbits round the sun in a year and not that it is shperical in shape.

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Sunspots appear dark because they are cooler than the surrounding solar surface. True or false
DiKsa [7]

Answer:

False, Sunspots appear dark (in visible light) due to their low temperature(cooler) than rest of the sun

Explanation:

Sunspots appear dark because they are much cooler( have low temperature than the rest of the surface contained by Sun. As they appear dark, but still they have very temperature that's why so hot. Sunspots have temperatures ranges  3,500 Celsius (3773 kelvin) and the surrounding surface of the sun has a temperature much higher of about 5,500 Celsius(5773 Kelvin). Even if we see a sunspot alone in space,  it will glow so brightly.

Learn more about sunspots :

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6 0
1 year ago
What are the benefits of solar cookers?
Elanso [62]

Answer:

Using clean, renewable, and readily available solar energy as fuel.

Preserving natural resources by not requiring the use of wood or other biomass fuels to cook.

Not producing dangerous emissions which pollute local environments and contribute to climate change.

Explanation:

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2 years ago
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On a nice winter day at the South Pole, the temperature rises to −54°F. What is the approximate temperature in degrees Celsius?
Evgesh-ka [11]

Answer:

Its going to be about -47.78°C

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3 years ago
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Help! How do you find the Eccentricity??
denis23 [38]

Explanation:

The formula to determine the eccentricity of an ellipse is the distance between foci divided by the length of the major axis

8 0
2 years ago
You are working in cooperation with the Public Health department to design an electrostatic trap for particles from auto emissio
Mademuasel [1]

Answer:

Explanation:

Given that:

Charge (q) on the particle = 3 × 10⁻⁸ C

mass (m) of the particle = 6 × 10⁻⁹ kg

at a distance x = 15 cm , the velocity in the plate = 900 m/s²

For the square plate, the surface charged density σ = -8 × 10⁻⁶ C/m²

To start with calculating the electric field as a result of the square plate; we use the formula;

E = \dfrac{\sigma }{2 \varepsilon_o}

E = \dfrac{8 \times 10^{-6} }{2 \times  8.85 \times 10^{-12}}

E = 4.51977 \times 10^5 \ V/m

On the square plate; The electric force F = Eq

F = (4.51977 \times 10^5 \ V/m )(3\times 10^{-8} \ C)

F = 1.3559 \times 10^{-2} \ N

The acceleration a =\dfrac{ F}{m}{

a = \dfrac{1.3559\times 10^{-2} \ N}{6 \times 10^{-9} \ Kg}

a = 2.25988 \times 10^6 \ m/s^2

For the particle, the velocity at distance x = 7 m can be calculated by using the formula:

(\dfrac{1}{2}) mv^2 = \Delta Vq

v^2 = \dfrac{2 Eq}{dm}

v^2 = \dfrac{2 * 4.51977 \times 10^5 \times 3 \times 10^{-8} }{0.07 \times 6\times 10^{-9} }

v^2 = 64568142.86  \ m/s

v =\sqrt{ 64568142.86  \ m/s}

\mathbf{v = 8.035 \times 10^3 \ m/s}

From the calculation, we realize that the charge acting between the particle and the plate is said to be "opposite".

Hence, the force is an attractive force.

Similarly, there is a gradual increase exhibited by the velocity of the particle.

Therefore, the particles get to the detector, but the detector failed to get detect due to the velocity which is greater than 1000 m/s.

6 0
3 years ago
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