Match what is seen through a telescope in Column A with the information that is learned from it in Column B. Column A 1. the visible light spectrum emitted by a star : the visible light spectrum emitted by a star 2. the brightness of a star : the brightness of a star 3. the color of a star : the color of a star Column B a. the distance between the star and Earth b. the chemical composition of the star c. the temperature of the star
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1. the visible light spectrum emitted by a star (1) ⇔ the chemical composition of the star (b)
2. the brightness of a star (2) ⇔ the distance between the star and Earth (a)
3. the color of a star (3) ⇔ the temperature of the star (c)
Explanation:
The brighter the star is, the closer it will be from the Earth. Hence, the brightness of a star depends upon the distance between the star and the Earth. The color of a star tells you how cool or hot it is. Blue colored stars are cool while red colored stars are hot (like sun). Hence, the color of star tells us its temperature. Stars emit different kinds of radiations (which we see in the radiation spectrum). That radiation spectrum tells us the chemical composition of the star.
NOTE: Do not confuse a star with a planet! (most students make this mistake)
Answer:
the visible light spectrum emitted by a star → the chemical composition of the star
the brightness of a star → the distance between the star and Earth
the color of a star → the temperature of the star
Explanation: plato :3
Related Questions
Franklin’s hometown is flooding due to a severe storm. He is evacuating the area in his vehicle. What should he do if he comes across a flooded area or flowing water?
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Franklin must not drive through a flood for there may no road at all under the water, unless he is familiar with the road and the flowing water is below one foot.
Moreover, if negotiating a flooded section of road, he must drive in the middle where the water will be at its shallowest and he must not drive through water against approaching vehicle to consider other drivers.
how does applying force to an object affect its motion
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calculate the speed of a sound wave produced from a guitar if it creates 38 compressions each second with 0.57 meter wavelength
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what is the mass of a 4900 N bobsled
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4900 / 9.8 = 500
Answer:
500 Kg.
Explanation:
The problem is giving us the weight of the object the weight is different from the mass of an object
Weight is the mass of any object multiplied by gravity (the value of gravity is approximately [tex]9.8\frac{m}{s^2}[/tex] ).
The unit of weight are newtons that equals
[tex]N= Kg.\frac{m}{s^2}[/tex]
[tex]W= m.g[/tex]
we cleared the mass of the formula
[tex]W=m.g\\ m=\frac{W}{g} \\ m= \frac{4900N}{9.8 \frac{m}{s^2} }= 500 Kg[/tex]
Batteries should only be removed from the premises by a licensed battery hauler for recycling. A) True B) False
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True
Explanation:Batteries carry acids and electrolytes which if comes in contact with human skin can cause severe effects and sometimes it can become fatal. Therefore a great care is needed while replacing batteries. To avoid any danger and harmful circumstances, batteries should always be replaced by a licenced battery haulers for recycling.
whats the basic particle that makes up matter
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the half-life of iodine-131 is 8 days. if a sample originally contains 0.96 g of the isotope, how much will remain after 16 days
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In this problem, the half-life of iodine-131 is 8 days. This means that after 8 days, only half of the initial sample will be left. After another 8 days (16 days), the sample will be reduced by another factor 1/2, so only 1/4 of the initial sample will be left.
Since the initial sample was 0.96 g, after 16 days the amount of iodine left is
[tex]m= \frac{1}{4}(0.96 g) =0.24 g[/tex]
After 16 days, there will be approximately 0.24 grams of iodine-131 remaining in the sample.
How the half-life of iodineThe half-life of iodine-131 is 8 days, which means that every 8 days, the amount of iodine-131 in a sample will be reduced by half.
After 16 days, which is two half-lives, the amount of iodine-131 remaining can be calculated as follows:
First half-life: 0.96 g / 2 = 0.48 g remaining
Second half-life: 0.48 g / 2 = 0.24 g remaining
Therefore, after 16 days, there will be approximately 0.24 grams of iodine-131 remaining in the sample.
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The earth takes 1 year to revolve around the sun at 1 A.U. distance (an astronomical unit = 93,000,000 miles). If a planet were 4 A.U. from the sun, how many years would it take to make 1 orbit?
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P^2 / (4 AU)^3 = 1P^2 = 64P = 8 years
Unsaved Earth is located in a region known as __________.
options:
a.the asteroid belt
b.the Kuiper Belt
c.the Oort Cloud
d.the Goldilocks Zone
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The Goldilocks zone refers to a zone around the star that is habitable. This means that it is not too hot and not too cold.
The Kuiper belt is located beyond Neptune. The Oort cloud is even further out in the solar system, past Pluto. The asteroid belt is located in between the orbit of Mars and Jupiter.
When a book is resting on a level table and you are not pushing on it , is there a frictional force between the book and table ?
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Makori drew the diagram below to show how a light microscope works.
Which best describes how Makori can correct the error in his diagram?
A) Change the label “Virtual image from objective lens” to “Real image from objective lens.”
B) Switch the locations of the eyepiece lens and the objective lens.
C) Move the light so it is shining downward from the eyepiece lens toward the objective lens.
D) Replace “Convex” with “Concave” on the label for the eyepiece lens.
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Makori can correct the error in his diagram by changing the label “Virtual image from objective lens” to “Real image from objective lens.”
What are the two types of lenses?Concave lensConvex lensWhich lens is used in light microscope?Convex lens is used in light microscope.
What is the difference between concave lens and convex lens?A concave lens is thinner in the middle and thicker at the edges. A convex lens is thicker in the middle and thinner at the edges.What is the objective of convex lens?They focus light for a clear picture.It is used in front of the eye to bend the incoming light.Focal point shortens and the light focuses properly on the retina.Thus , the real and inverted image is produced by objective lens of light microscope, which goes through convex lens of eyepiece.Therefore, Makori can correct the error in his diagram by changing the label “Virtual image from objective lens” to “Real image from objective lens.”
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A 12 volt parallel circuit is drawing 2 amps what is the total amount of power consumed?
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Please mark as Brainliest, thanks.
PLZ HELP!!!
*best answer gets brainliest*
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Have you ever seen the two different sides of a battery? One side is often labelled with a positive sign, and the other with a negative sign. Inside the battery are four cells where two processes take place. One process uses electrons, and the other creates electrons. When the electrons flow, it creates electricity.
Drums convert mechanical energy into sound energy.
When a musician bangs his/her drumstick on the drum, the motion makes a sound. This movement turned motion into sound.
n the picture of a planet in orbit around its star area a is double that of area b. What is true of the time the planet takes to travel A1A2 as compared to B1B2
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1. The orbit of a planet is an ellipse with the Sun (the sun is a star!) at one of the two focus.
2. A line segment joining a planet and the Sun sweeps out equal areas during equal intervals of time.
3. The square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit.
So, let's use second law. The Sun sweeps out equal areas during equal intervals of time means that if A = B, the time the planet takes to travel A1A2 is equal to the time the planet takes to travel B1B2, but given that A = 2B, then takes twice the time to travel A1A2 compared to B1B2.
Answer:
twice as much
Two footballs (A & B) are thrown with the same amount of force. Football A has a greater mass. Describe the acceleration for each football.
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[tex]F=ma[/tex]
where F is the force, m the mass and a the acceleration.
A force F is applied to football A, whose mass is [tex]m_A[/tex], and so the acceleration of this football will be given by (re-arranging the previous equation)
[tex]a_A = \frac{F}{m_A} [/tex]
Similarly, the acceleration of football B will be
[tex]a_B= \frac{F}{m_B} [/tex]
where [tex]m_B[/tex] is the mass of football B, and where the force F applied to the two footballs is the same.
Since football A has greater mass than football B, [tex]m_A \ \textgreater \ m_B[/tex], if we compare the two previous formula we see that the acceleration of football B is greater than the acceleration of football A:
[tex]a_B \ \textgreater \ a_A[/tex]
Therefore, if the same force is applied to the two footballs, football B will accelerate more than football A.
n the picture of a planet in orbit around its star area a is double that of area b. What is true of the time the planet takes to travel A1A2 as compared to B1B2
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1) the planets move in ellipses (with one focus on the sun)
2) A line joining the sun and the planet traverses equal areas in equal times
3) The square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit.
Applying the second law, we get that the time to transverse A1A2 is double the time needed to transverse B1B2 (even though B1B2 is longer, because the planet is closer to the sun, it goes faster). The second law simply states that time is proportional to area in such graphs.
A parent isotope has a half life of 1 million years. If a rock contained 20 mg of the parent isotope when it formed, how much parent isotope would be left after 2 million years?
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What si the name of the state change water is going through as it boils?
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~~~hope this helps~~~
~~have a beautiful day~~
~davatar~
what two forces are there when you skydive
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2). The force of reason and self preservation, telling you that this is crazy, and you must NOT jump out of a perfectly good airplane that's not on the ground.
a ball is thrown horizontally from the roof of a building 45m tall and lands 24m from the base what is the initial speed?
Answers
To find the initial speed of a ball thrown horizontally from a 45m tall building, we first calculate the time of flight using the vertical motion equation and then determine the initial horizontal speed using the range and time of flight.
Finding the Initial Speed of a Horizontally Thrown Ball
To calculate the initial speed of a ball thrown horizontally from the roof of a building that is 45 meters tall and lands 24 meters from the base, we can use the kinematic equations of motion for projectile motion.
Step 1 - Time of Flight
To find the time it takes for the ball to hit the ground, we use the vertical motion equation:
s = ut + 0.5*at^2
Since the ball is thrown horizontally, the initial vertical velocity (u) is 0 m/s. s is the height of the building (45 m) and a is the acceleration due to gravity (9.8 m/s2).
By substituting the known values, we can solve for t.
Step 2 - initial Horizontal Speed
Now, knowing the time of flight, we can find the initial horizontal speed(vx) using the horizontal motion equation:
Range = vxt
The range is given as 24 m. Therefore, vx = Range / t.
Final Answer
After calculating, we get the initial horizontal speed of the ball, which is necessary to cover the horizontal distance within the derived time of flight.
How much heat is absorbed by a 66g iron skillet when its temperature rises from 7oC to 20oC?
Answers
Q = mCΔT
Q = heat absorbed
m = mass = 66 g
C = specific heat of iron = 0.449 J/g.°C
ΔT = temperature change = 20 - 7 =13°C
Therefore,
Q = 66*0.449*13 = 385.242 J
a blank is an electric circuit that produces a magnetic field.
whats the blank?
Answers
Answer:
Explanation: A coil is an electric circuit that produces a magnetic field
Match the adaptations with their benefits.Match the adaptations with their benefits.
the wasp’s stinger
the peacock’s bright plumage
the owl’s large eyes
the bear’s fur
help to search for food
arrowBoth
protection against predators
arrowBoth
attraction to obtain a mate
arrowBoth
maintain homeostasis
arrowBoth
Answers
Answer: the wasp’s stinger - protection against predators
the peacock’s bright plumage - attraction to obtain a mate
the owl’s large eyes - help to search for food
the bear’s fur - maintain homeostasis
Explanation:
An adaptation can be define as any change that occurs in the morphology, physiology and genetic make up of the organism which makes it suitable for survival in it's natural environment even in adverse conditions and reproduce.
The following are the adaptations with their benefits.
the wasp’s stinger - protection against predators
The wasp stinger generally exhibit toxic substance or allergy causing agent which the wasp use when it finds a predator around it.
the peacock’s bright plumage - attraction to obtain a mate
The male peacock exhibit bright and showy feathers which is a phenotypic advantage for attraction of female mates for copulation. This phenomena is called as sexual selection.
the owl’s large eyes - help to search for food
The owl's large eyes allows the night vision. There are many rods and cones present in the eyes of the owl so as to allow better vision in night and allow it to observe and capture prey.
the bear’s fur - maintain homeostasis
The bear's fur are insulating in nature that means it does not allow the entry of external air. Hence, helps in retaining the heat in the body in otherwise cold environment. Therefore, helps in maintaining the homeostasis (the tendency of the body retain it's constant internal environment as compared to changing external environment).
What are non examples of Asexual reproduction
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Many single-celled organisms rely on binary fission to reproduce themselves. All bacteria reproduce by binary fission.
Budding and Hydras-
Hydras are a type of freshwater cnidarian related to jellyfish, sea anemones and corals. Hydras reproduce asexually by developing small, genetically identical polyps that protrude from the parent. These polyps break off from the parent to form a new organism in a process called "budding."
Parthenogenesis and Copperheads- Though most of the organisms that use asexual reproduction are invertebrates, several species of vertebrate animals do make use of it. Some, like copperhead snakes, can reproduce asexually as an alternative to their typical sexual reproduction. Through a process called parthenogenesis -- literally "virgin birth" -- copperhead females can give birth to live young without having been inseminated by a male, even when males are available, according to National Geographic. In this form of parthenogenesis, the polar body, or a cell produced alongside the female's egg, acts as a sperm cell and essentially "fertilizes" the egg. Parthenogenesis has been observed in several species of reptiles, birds and sharks, and is especially common in insects.
We have that Non examples of Asexual reproduction will be the reproduction that requires two cells to reproduce.
Examples
BuddingHumansBirdsinsects
From the question we are told
What are non examples of Asexual reproduction
Generally
Asexual reproduction this are the type of reproduction in which a single sex cell is able to reproduce The Traditional union of sex cells in this case is not needed This can be found in plants and microorganismsTherefore
non examples of Asexual reproduction will be the reproduction that requires two cells to reproduce.
Examples
BuddingHumansBirdsinsectsFor more information on this visit
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Food borne pathogens are harmful to the human body in any amount
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It does vary between different pathogens & between the condition of the host (immunocompromised vs. healthy), but if the minimum infectious dose is not met- then the body is typically able to snuff out the intruders before they can do any real damage.
Hope that helps!
A wire delivers 12.0 C of charge in 4.0 s.
What is the current in the wire?
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C=As
12=A(4)
A=3
The current of the wire is 3 Amperes.
A remote control car travels at a uniform speed of 3.5 m and reaches its destination in five minutes. How far did the remote control car travel
Answers
The elapsed time is 5 minutes.
You can use the formula
speed = distance / time.
Clearing the distance, which is what we want to find, we have:
distance = speed*time.
However, the speed is in units of seconds and the time in units of minutes. To perform the operation we need to have everything in the same units.
So.
5minutes x 60seconds / 1minute = 300 seconds
Finally:
distance = 3.5m / s * 300s = 1050 meters.
The remote control car traveled 1050 meters
How does the current in a circuit change if the resistance is doubled?
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What is the electric potential of a 2.2 µC charge at a distance of 6.3 m from the charge? Recall that Coulomb’s constant is k = 8.99 × 109 N • ___ V
What is the electric potential at a distance of 99 m from the charge?___ V
Answers
3100 V
and
200v
The electric potential of a 2.2 µC charge at a distance of 6.3 m is 3.14 × 10^6 V. At a distance of 99 m, the electric potential is 2.02 × 10^4 V.
Explanation:To calculate the electric potential, we can use the equation V = kQ/r, where V is the electric potential, k is Coulomb's constant (8.99 × 10^9 N • V^-2 • m^2 • C^-2), Q is the charge, and r is the distance from the charge.
For a charge of 2.2 µC at a distance of 6.3 m, we can plug in the values to find:
V = (8.99 × 10^9 N • V^-2 • m^2 • C^-2)(2.2 × 10^-6 C)/(6.3 m) = 3.14 × 10^6 V
Similarly, for a distance of 99 m, the calculation would be:
V = (8.99 × 10^9 N • V^-2 • m^2 • C^-2)(2.2 × 10^-6 C)/(99 m) = 2.02 × 10^4 V
The difference in the composition of the terrestrial planets and the gas giants can be explained by the different condensation ____________________ in the regions near and far from the sun during the formation of the solar system.
Answers
According to the Solar Nebula theory, temperatures high enough to condense metals were closer to the Sun, while towards the outside of the System, temperatures where lower.
This is the reason why terrestrial rocky planets are closer to the Sun, while gaseous planets are further from the Sun.
Hence, the difference in the composition of the terrestrial planets and the gas giants can be explained by the different condensation temperature in the regions near and far from the Sun during the formation of the Solar System.