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son4ous [18]
4 years ago
9

Malika is writing an essay about the Sun. Below is the first paragraph in the essay. The Sun is an average-size star that is loc

ated at the center of the solar system. This hot, glowing ball of gas has a diameter of 1,393,000 km. This means 109 Earths would fit side by side across the diameter of the Sun. At the surface, the Sun’s temperature is about 550 degrees Celsius. Earth is located just under 150 million kilometers from the Sun. Malika quickly realizes this paragraph contains an error. Which statement identifies the error? The Sun is a large star, not an average-size star. The Sun is not a glowing ball of gas—it is solid. The Sun’s surface temperature is about 5500 degrees Celsius. Earth is located over 300 million kilometers from the Sun.
Physics
2 answers:
joja [24]4 years ago
5 0

The answer would be:


<u>The Sun's surface temperature is about 5,500 degrees Celsius. </u>


The surface of the Sun is about 5,778 Kelvin. If you want to be exact the sun is 5,505 degrees Celsius and that's just the surface. The corona gets even hotter reaching about 10 million degrees Celsius. The other information about the Sun is correct.

harina [27]4 years ago
4 0

Answer: Option (c) is the correct answer.

Explanation:

Sun is a large star and as it is given that it's diameter is 1,393,000 km. Since there are nuclear reactions that take place inside the Sun, therefore, there is emission of lot of radiations.

As a result, lot of heat energy is there. So, the statement at the surface Sun’s temperature is about 550 degrees Celsius will be an error.

Thus, we can conclude that this error is identified by the statement the Sun’s surface temperature is about 5500 degrees Celsius.

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You are studying a population of flowering plants for several years. When you present your research findings you make the statem
Anit [1.1K]

Answer:

Option A

Explanation:

The graph for this problem must depict the following ""Increased allocation of resources to reproduction relative to growth diminished future fecundity."

Hence, the survivor ship must be on the Y axis and the resources on the X axis.

Here the resources include the number of seeds produced.

hence, the higher is the number of seeds (resource), the lower is the survivorship (future fecundity)

Hence, option A is correct

6 0
3 years ago
Earth rotates on its axis once every 24 hours, so that objects on its surface execute uniform circular motion about the axis wit
cestrela7 [59]

Answer:

v=1667.9km/h

a_{cp}=436.6km/h^2

Explanation:

The speed is the distance traveled divided by the time taken. The distance traveled in 24hs while standing on the equator is the circumference of the Earth C=2\pi R, where R=6371km is the radius of the Earth.

We have then:

v=\frac{C}{t}=\frac{2\pi R}{t}=\frac{2\pi (6371km)}{(24h)}=1667.9km/h

And then we use the centripetal acceleration formula:

a_{cp}=\frac{v^2}{R}=\frac{(1667.9km/h)^2}{(6371km)}=436.6km/h^2

6 0
3 years ago
Which experiment proved light was a wave?
pshichka [43]

Answer:

The answer is B) Double slit

8 0
3 years ago
Estimate the mass of the Great Pyramid of Giza, in tons. You make may use of the following information: the Great Pyramid is in
postnew [5]

Answer:

6005803.83105 short tons

Explanation:

The definition of density is \rho = \frac{m}{V}, and the volume of a pyramid is (confusingly written on the proposal) V=\frac{1}{3} Ah, so we can write:

m=\rho V=\rho V \frac{1}{3} Ah=\rho V \frac{1}{3} s^2h

Where s is the side of the base, being s^2 the area of that square.

We will write everything in S.I., and the best way to convert units is using conversion factors, for example, since 1m=100cm, we know that \frac{1m}{100cm}=1, and we can use this factor to convert anything written in cm to anything written in m. Example:

500cm=500cm\frac{1m}{100cm}=5m

Here we just multiplied 500cm by something that is equal to 1 (as every conversion factor must), so <em>it's not doing anything but changing the units</em>.

We can use this tool like this:

2.1\frac{g}{cm^3}=2.1\frac{g}{cm^3}(\frac{1Kg}{1000g})(\frac{100cm}{1m})^3=2100Kg/m^3

Where we have used the fact that 1^3=1 (<u>we can elevate any conversion factor to any number and they still will be 1</u>) and where we have placed strategically what is the numerator and what in the denominator so the units we don't want cancel out and the units we want appear.

Substituting then our values:

m=\rho V \frac{1}{3} s^2h=(2100Kg/m^3)\frac{1}{3} (230.34m)^2(146.7m)=5448373586.96Kg

And now we will convert to short tons using two conversion factors at the same time:

m=5448373586.96\ Kg(\frac{1\ lb}{0.45359237\ Kg})(\frac{1\ short\ ton}{2000\ lb} )=6005803.83105\ short \ tons

Remember, their value is 1, and we place the units to cancel the ones we don't want and keep the ones we want, here Kg cancel out, and lb cancel out, leaving the short tones.

8 0
3 years ago
Atoms contain empty space true or false​
olga nikolaevna [1]
The answer would be false
5 0
3 years ago
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