The reading of the given passage on astronomy gives;
Sentence (2) is the major supporting detailHow can the parts of the passage be identified?Sentence (1) is the main idea or background statement with regards to astronomy, with regards to the sources of concepts being used in astronomy. It conveys the subject matter of passage.
Sentence (2) gives more light to the statement of sentence (1). It gives the supporting details of the concepts astronomers inherited from antiquity, by outlining them; (angles and angular measure system used locate stellar objects), while sentence (4) gives a little information with regards to the use of angular system.
Therefore;
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are there any obvious differences between the beam observations and the cylinder observations? (select all that apply.) the maximum value of the cylinder observations is higher. the observations for beams are more variable, or spread out. the maximum value of the beams observations is higher. the observations for cylinders are more variable, or spread out.
The obvious differences between the beam observations and cylinder observations are that the maximum value of the cylinder observations is higher and the observations for cylinders are more variable or spread out.
In the above case, the material strength observations in Mega Pascals for cylinder and beams are given.
The values for cylinders are as follows
6.1, 5.8, 7.8, 7.1, 7.2, 9.2, 6.6, 8.3, 7.0, 8.3, 7.8, 8.1, 7.4, 8.5, 8.9, 9.8, 9.7 14.1, 12.6 11.2, 8.1, 7.4, 8.5, 8.9, 9.8, 9.7, 14.1, 12.6, 11.2
And the values for beams are
5.9, 7.2, 7.3, 6.3, 8.1, 6.8, 7.0, 7.6, 6.8, 6.5, 7.0, 6.3, 7.9, 9.0, 8.2 , 8.7, 7.8, 9.7, 7.4, 7.7, 9.7,7.8, 7.7, 11.6 11.3 11.8 10.7,8.2, 8.7, 7.8, 9.7, 7.4, 7.7, 9.7, 7.8, 7.7, 11.6, 11.3, 11.8, 10.7
From the above-given values, we can calculate the mean/average value of the observations.
The average value for cylinders comes out to be 8.57.
The average value for beams comes out to be 8.14.
Among the given values, the maximum value is 14.1 for cylinder observations.
The values of beam observations range from 5.9 to 11.8 while the values for cylinder observations range from 5.8 to 14.1. Therefore, the observations for the cylinder are more variable or spread out.
Thus, the maximum value of the cylinder observations is higher and the observations for cylinders are more variable or spread out.
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This question is incomplete. The data used for solving the question is given in the adjoining image.
what is nuclear energy
Answer:
Nuclear energy, also called atomic energy, energy that is released in significant amounts in processes that affect atomic nuclei, the dense cores of atoms. It is distinct from the energy of other atomic phenomena such as ordinary chemical reactions, which involve only the orbital electrons of atoms.
Explanation:
Answer: Nuclear energy, also called atomic energy, energy that is released in significant amounts in processes that affect atomic nuclei, the dense cores of atoms. It is distinct from the energy of other atomic phenomena such as ordinary chemical reactions, which involve only the orbital electrons of atoms.
Explanation: please give brainliest and thanks :)
Not really sure what it is
Answer:
water seeping through a pot of soil
Explanation:
All the others permanently change the chemical structure of the element so they are chemical reactions
Which set of words below make the following statement about energy correct?
"Energy cannot be ________ or ________. Energy ________ from one form to another."
Group of answer choices
increased, decreased, flows
added, removed, moves
created, destroyed, transforms
Answer:
created, destroyed, transforms
Explanation:
Energy cannot be created nor destroyed.
Energy is transformed from one form to another form.
This is the postulate of the law of conservation of energy.Energy is neither created nor destroyed simply implies that energy is not produce in any system. Nor is it destroyed.
What we have is energy transformation from one form to another.
What is the weight of a 225 kg space probe on the moon? The gravitational field on the moon is 1.62 N/kg. Show work.
a. 139 N
b. 364 N
c. 1.35 x 10^3 N
d. 2.21 x 10^3 N
Answer:
b. 364 [N]
Explanation:
We must remember that the gravitational force can be calculated by means of the product of mass by acceleration. In this case, the acceleration corresponds to the gravitational acceleration of the moon.
[tex]F=m*g\\[/tex]
where:
F = force [N]
m = mass = 225 [kg]
g = moon gravity acceleration = 1.62 [m/s²]
[tex]F = 225*1.62\\F = 364.5[N][/tex]
The weight is 364 N
The calculation can be done as follows
mass= 225 kg
gravitational field= 1.62 N/Kg
Formula is
F= mg
= 225 ˣ 1.62
= 364
= 364 N
Hence the weight of the probe on the moon is 364N
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the following four waves are sent along strings with the same linear densities (x is in meters and t is in seconds). rank the waves according to (a) their wave speed and (b) the tension in the strings along which they travel, greatest first:
(a) If we consider the waves to travel in a straight line (x = 0), according to their wave speed,
Wave A is the slowest wave and it travels at a speed of 0.5 meters per secondWave B is the next slow wave and it travels at a speed of 1 meter per second Wave C is the fastest wave and it travels at a speed of 2 meters per second Wave D is the fourth and final slow wave and it travel at a speed of 1.(b) The waves are ranked according to the tension in the strings along which they travel, with the greatest first,
The waves with the lowest tension (wave A and C) travel along the strings with a lower linear density. Wave B travels along the string with a higher linear density, so it has greater tension. Wave D travels along the string with an even higher linear density, so it has even greater tension.In waves, what does linear density mean?The linear density of the string and the tension in the string both affect the wave's speed. The linear density is the mass divided by the string's length. In general, the square root of the ratio of the medium's elastic to inertial properties determines a wave's speed.
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If an airplane accelerates at a constant rate of 3.0 m/s^2 starting with a velocity of 21 m/s, then what will its displacement be after 40s?
The displacement at a constant acceleration of 3.0[tex]m/s^{2}[/tex] is 526.5m.
Distance, which is a scalar quantity, describes "how much ground an object has covered" while moving. The term "how far out of place an object is" relates to displacement, a vector number that describes the total change in position of the object.
For constant acceleration, displacement is:
[tex]d = (\frac{1}{2} ) (\frac{v_{f}^{2} - v_{i} ^{2} }{a} )[/tex]
[tex]d = \frac{60^{2}-21^{2} }{2*3}[/tex]
[tex]d = \frac{3600-441}{6}[/tex]
[tex]d = \frac{3159}{6}[/tex]
d = 526.5m
Therefore, the displacement at a constant acceleration of 3.0[tex]m/s^{2}[/tex] is 526.5m.
The complete question is: An airplane accelerates with a constant rate of 3.0m/s^2 starting at a velocity of 21m/s. If the final velocity is 60m/s what is the displacement during this period?
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what are three considerations diamond discusses as he ponders yali’s question in guns, germs, and steel
The three considerations diamond discusses as he ponders Yali's question in guns, germs and steel are:
Explaining why Europe has become more advanced than other regions tends to confirm the notion that Europe is betterFraming the question in this way contributes to a Eurocentric view of human developmentThe question assumes that the natural progression of human history is toward "civilization" and that being "civilized" is better than being "primitive."The stage of human social and cultural development and organization that is considered the most advanced is termed as civilization.
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What are the 5 basic types of reactions?
Answer:
The five basic types of chemical reactions are combination, decomposition, single-replacement, double-replacement, and combustion
Explanation:
The five basic types of chemical reactions are combination, decomposition, single-replacement, double-replacement, and combustion. Analyzing the reactants and products of a given reaction will allow you to place it into one of these categories
PLEASE HELP MEE
Ruben lifts a 125 Newton box up to the top of a shelf 2 meters high. How
much work (in Joules) did Ruben do?
Answer:
work = mgh = 125 × 2 = 250 joulesWhich characteristic of water quality provides the most information regarding the health of a water system? *
A clarity
B temperature
C concentration of heavy metals
D presence of indicator species
DO NOT LEAVE I HV ALOT MORE QUESTIONS!!!
what is the speed of a car that travels a distance of 60 meters in a time of 10 seconds
Answer:
6m/s
Explanation:
A=d÷t
A=60÷10
6
rearrange Vf=Vi+at, rearrange for t
Answer:
[tex] \sf t = \dfrac{V_f - V_i}{a} [/tex]
Explanation:
[tex] \rm V_f = V_i + at \\ \rm V_f - V_i = at \\ \rm at = V_f - V_i \\ \rm t = \dfrac{V_f - V_i}{a} [/tex]
I need help on this the first one is plant cell
Answer:
The Vacuole
Explanation:
The Vacuole
What is the gravitational potential energy of a wrecking ball that is hung 20 meters above the ground if it has a mass of 200 kg?
Answer:
The gravitational potential energy of the ball is 39,200 J
Explanation:
Gravitational Potential Energy
It is the energy stored in an object because of its vertical position or height in a gravitational field.
It can be calculated with the equation:
U=m.g.h
Where m is the mass of the object, h is the height with respect to a fixed reference, and g is the acceleration of gravity or [tex]9.8 m/s^2[/tex].
The wrecking ball is hung at a height of h=20 m and has a mass of m=200 Kg. Substituting in the formula:
U=200*9.8*20
U = 39,200 J
The gravitational potential energy of the ball is 39,200 J
12.
A rock on the surface of Mars is projected vertically upwards with an intial speed of 9.4ms The
rock rises to a height of 12 m above the surface.
Assume there is no atmosphere on Mars
What is the acceleration of free fall near the surface of Mars?
Answer:
3.68 m/s².
Explanation:
From the question given above, the following data were obtained:
Initial velocity (u) = 9.4 ms¯¹
Maximum height (h) reached = 12 m
Final velocity (v) = 0 (at maximum height)
Acceleration of free fall (g) =?
v² = u² – 2gh (since the rock is going against gravity)
0 = 9.4² – 2 × g × 12
0 = 88.36 – 24g
Collect like terms
0 – 88.36 = –24g
–88.36 = –24g
Divide both side by –24
g = –88.36 / –24
g = 3.68 m/s²
Thus, the acceleration of free fall near the surface of Mar is 3.68 m/s².
The acceleration of free fall near the surface of Mars is [tex]3.68 \;\rm m/s^{2}[/tex].
Given data:
The magnitude of initial velocity of rock is, u = 9.4 m/s.
The height raised up by rock is, h = 12 m.
Here, we need to find the acceleration of free fall, which is nothing but the gravitational acceleration. So we will use the Third kinematic equation of motion, which is given as,
[tex]v^{2}=u^{2}+2gh[/tex]
Solving as,
[tex]0^{2}=9.4^{2}+(2 \times -g \times 12)\\\\g = 3.68 \;\rm m/s^{2}[/tex]
Thus, we can conclude that the acceleration of free fall near the surface of Mars is [tex]3.68 \;\rm m/s^{2}[/tex].
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This hurts. Help me.
Answer:
a. T₂ = 60 N
b. W = 25.98 N
Explanation:
a.
Taking right side as positive x direction, the equation for sum of forces in x-direction can be written as follows:
[tex]T_{2}Sin\ 30^o - T_{1} = 0\\T_{2}Sin\ 30^o = T_{1}\\T_{1} = (T_{2})(Sin\ 30^o)\\where,\\T_{2} = 30\ N\\Therefore,\\T_{1} = (30\ N)(Sin\ 30^o)\\[/tex]
T₁ = 15 N
b.
Taking upward direction as positive y direction, the equation for sum of forces in y-direction can be written as follows:
[tex]T_{2}Cos\ 30^o - W = 0\\T_{2}Cos\ 30^o = W\\W = (30\ N)Cos\ 30^o\\[/tex]
W = 25.98 N
The si unit of mass is the Newton. True or false.
a student is rotating an object on a row that is 4.5 M long if we increase the length of the Rope so that the student rotates and uniform circular motion with the mass rotating 9.0 M away from the center of the rotation 2x farther what will be the effect of the tangential velocity for the athlete if we keep the time for one Revolution the same
Answer:
is there any options
Explanation:
A 8 kg ball is moving with a speed of 10 m/s directly toward a 4 kg ball
which is at rest. The two balls collide and stick together. What is their
velocity immediately after the collision?
Answer:
v₃ = 6.67 [m/s]
Explanation:
To solve this problem we must use the linear momentum conservation theorem, which tells us that momentum is preserved before and after the collision.
Let's take the ball movement of 8 [kg] as positive.
Therefore we can built the following equation:
[tex](m_{1}*v_{1})+(m_{2}*v_{2})=(m_{1}+m_{2})*v_{3}[/tex]
where:
m₁ = mass of the 8 [kg] ball
m₂ = mass of the 4 [kg ] ball
v₁ = velocity of the 8 [kg} ball before the colllision = 10 [m/s]
v₂ = velocity of the 4 [kg] ball before the colllision = 0 [m/s] (at rest)
v₃ = velocity of the two balls after the collision [m/s]
[tex](8*10)+(4*0)=(8+4)*v_{3}\\80 = 12*v_{3}\\v_{3}=6.67 [m/s][/tex]
(b) What If? A particle with charge q is located a very small distance from the center of a very large square on the line perpendicular to the square and going through its center. Determine the approximate electric flux through the square due to the charged particle.
A particle with charge q is a very small distance from the centre of a very large square on the line perpendicular to the square and going through its centre.
Since the distance between the charge and the centre of the square is very small, the square contains the field lines emerging from the charge.
[tex]\Phi_{E,square} = \frac{q}{2_{\in_{0} } }[/tex]
A particle is a small localized object which can be ascribed to several physical or chemical properties, such as volume, density, or mass.[1][2] They vary greatly in size or quantity, from subatomic particles like the electron to microscopic particles like atoms and molecules, to macroscopic particles like powders and other granular materials. Particles can also be used to create scientific models of even larger objects depending on their density, such as humans moving in a crowd or celestial bodies in motion.
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you read in an astronomy book that our milky way galaxy consists of about 200 billion stars. how do you suppose this number was determined?
By utilising a galaxy model that accurately explains all of the data that has been collected thus far.
Is each star in a different galaxy a sun?In other words, while every star is a star, not every star is a Sun. Since the Sun is bigger than most stars, it is also more brighter. In our galaxy alone, there are billions of Suns, and as was already established, many of the stars we can see are Suns. However, a lot of what you see while gazing up into space is not a star.
The sizes of stars range from neutron stars, which can be about 12 miles (20 kilometres) across, to supergiants, which have a diameter that is 1,000 times larger than the sun. The brightness of a star is influenced by its size. Particularly, brightness and radius are related.
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A player hits a ball with a bat. The action is the force of the bat against the ball.
What is the reaction to this force?
Answer: Newton's third law of motion states that for every action, there is an equal and opposite reaction. When a player hits a ball with a bat, the force the player exerted on the ball through the bat gave the ball a push force in the forward direction.
Explanation:
A car weighs 3600 kg is traveling at 21. 0 m/s. The driver doesn?t notice a red light and rear-ends another car at rest. Both cars stick together and move with a speed of 9. 0 m/s. What is the mass of the car that was at rest?.
Answer:
[tex]4800\; {\rm kg}[/tex].
Explanation:
When an object of mass [tex]m[/tex] travels at a velocity of [tex]v[/tex], the momentum [tex]p[/tex] of that object would be [tex]p = m\, v[/tex].
Momentum is conserved immediately after the collision. Hence, the momentum that the [tex]3600\; {\rm kg}[/tex] vehicle lost in the collision would be equal to the momentum that the other vehicle has gained.
Momentum that the [tex]3600\; {\rm kg}[/tex] vehicle lost:
[tex]\begin{aligned}(\text{momentum change}) &= (\text{new momentum}) - (\text{original momentum}) \\ &= m\, v(\text{new}) - m\, v(\text{old}) \\ &= m\, (v(\text{new}) - v(\text{old})) \\ &= 3600\; {\rm kg} \, (9.0\; {\rm m\cdot s^{-1}} - 21.0\; {\rm m\cdot s^{-1}}) \\ &= (-43200)\; {\rm kg \cdot m \cdot s^{-1}}\end{aligned}[/tex].
The vehicle that was originally at rest would have gained that [tex]43200\; {\rm kg \cdot m\cdot s^{-1}}[/tex] of momentum. The momentum of that vehicle was [tex]0\; {\rm kg \cdot m \cdot s^{-1}}[/tex] before the collision since that vehicle was initially not moving. After gaining the [tex]43200\; {\rm kg \cdot m\cdot s^{-1}}\![/tex] of momentum, the new momentum of that vehicle would be [tex]p = 43200\; {\rm kg \cdot m\cdot s^{-1}}\!\![/tex].
Rearrange the equation [tex]p = m\, v[/tex] to obtain an expression for mass: [tex]m = (p / v)[/tex]. It is given that the velocity of this vehicle is [tex]v = 9.0\; {\rm m\cdot s^{-1}}[/tex] after the collision. Substitute both momentum [tex]p[/tex] and velocity [tex]v[/tex] into the equation [tex]m = (p / v)\![/tex] to find mass [tex]m\![/tex]:
[tex]\begin{aligned} p &= \frac{m}{v} \\ &= \frac{43200\; {\rm kg \cdot m \cdot s^{-1}}}{9.0\; {\rm m\cdot s^{-1}}} \\ &= 4800\; {\rm kg} \end{aligned}[/tex].
In other words, the mass of the vehicle that was originally not moving would be [tex]4800\; {\rm kg}[/tex].
if the pumpkin is thrown from a height of 14 meters above the ground. the velocity of the pumpkin when it is thrown is 23 m/s at 40 degrees above the horizontal. what is the value of the horizontal velocity when the pumpkin is thrown
Answer:
The value of the horizontal velocity when the pumpkin is thrown is approximately 17.619 m/s
Explanation:
The height from which the pumpkin is thrown = 14 meters
The velocity with which the pumpkin is thrown = 23 m/s
The direction in which the pumpkin is thrown = 40° above the horizontal
The horizontal component of the velocity = 23 × cos(40) ≈ 17.619
Therefore, the value of the horizontal velocity when the pumpkin is thrown ≈ 17.619 m/s
Please find the average of the following numbers
9,10,100,50,20
Answer:
Average: 22.142857142857
Average =
Sum
Count
=
155
7
= 22.142857142857
Sum 155
Count 7
Median 23
Geometric Mean 16.412764443111
Largest 38
Smallest 2
Range 36
Answer:
15.4
Explanation:
4) What amount of energy is radiated by the sun per second
Answer:
"The sun releases energy at a mass–energy conversion rate of 4.26 million metric tons per second, which produces the equivalent of 384.6 septillion watts (3.846×1026 W)."
Explanation:
This might help you, also this is searched.
sam is an astronomer on planet hua, which orbits the distant star barnard. it has recently been accepted that hua is spherical in shape, although its exact size is unknown. while studying in the library, in the city of joy, sam learns that during equinox, barnard is directly overhead in the city of bar, located 1100.0 km north of his location. on the equinox, sam goes outside and measures the altitude of barnard at 78 degrees. what is the radius of hua in km?
There are 90 degrees directly overhead. Barnard is at 74 degrees at Sam's position, 1300.0 kilometers south. The part of the 360 degree circle that corresponds to Hua's circumference is the difference between 90 and 74, or 16 degrees. 1300.0 km is therefore 16/360 of the circle, or 0.044 times the circumference. 29.250 km is the circumference (1300 divided by 0.044). Radius divided by two gives the circumference equation. Therefore, 2r=29250 and the radius of Hua=4655.3 km
What do you mean by Astronomy?Astronomy is a scientific science that examines celestial objects and phenomena (from the Ancient Greek "o" (astronoma), "science that investigates the laws of the stars"). In order to describe their creation and evolution, it makes use of mathematics, physics, and chemistry. Planets, moons, stars, nebulae, galaxies, and comets are among the interesting celestial bodies. Supernova explosions, gamma ray bursts, quasar, blazar, pulsar, and cosmic microwave background radiation are examples of relevant phenomena. Astronomy is the study of everything that comes from outside the atmosphere of the Earth. Astronomical cosmology is the study of the cosmos as a whole.
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what is the benefit of using neutral arguments to express oneself while in a conflict.
Answer:
Neutral arguments avoid using "you" statements and, therefore, can assist in avoiding being perceived as offensive or indicating that you believe your opponent is unreasonable. It is best to use "I" statements rather than "you" statements, because it helps keep the tone of statements neutral, which avoids escalating the conflict.
Explanation:
Answer:
the person above is correct ^^^^^^^^^^^
Explanation:
26 POINTS!! will give brailest! can anyone help me? if you can't well, Merry Christmas!
Have an amazing day you are an amazing person:)