The statement that correctly differentiates between astronomy and cosmology is as follows: Astronomy is a much broader study of the universe as a whole, while cosmology is a more focused study on specific objects or celestial bodies contained within the universe (option A).
What is astronomy?Astronomy is the study of the physical universe beyond the Earth's atmosphere, including the process of mapping locations and properties of the matter and radiation in the universe.
On the other hand, cosmology is the study of the physical universe, its structure, dynamics, origin and evolution, and fate.
Astronomy studies objects and phenomena beyond Earth, whereas cosmology is a branch of astronomy that studies the origin of the universe and how it has evolved.
Therefore, it can be said that astronomy studies the universe from a much broader sense while cosmology is much more specific study.
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A parking lot is going to be 60 m wide and 240 m long. which dimensions could be used for a scale model of the lot?A. 90 in x 360 cmB. 1 m x 3 mC. 20 cm x 80 cmD. 120 cm x 480 m
Answer:
C. 20 cm x 80 cm
Explanation:
The dimensions of the scale model have the same ratio of the original dimensions. The ratio of the original dimenstions is
240/60 = 4
Then, the ratio for each option is
A. 360/90 = 4
B. 3/1 = 3
C. 80/20 = 4
D. 480/120 = 4
Therefore, B is not a correct option.
Additionally, we don't use A because the units for length and width are different
And we don't use D because the measures are too long for a model.
Therefore, the answer is
C. 20 cm x 80 cm
Gravitation always does negative work. Is this true or false?
The given statement 'Gravitation always does negative work' is false.
The work done by the gravitation depends upon the reference level of the system. Therefore, according to reference level it is taken as positive or negative.
A pencil is viewed through a concave lens. Which of the following best describes the image?A.RealB.InvertedC.VirtualD.Enlarged
A concave lens produces a virtual image like the one seen in the diagram (we know is virtual since it is on the same side to the original object), we also see that the image is smaller, and not inverted.
Therefore, the only option that describes this image is C which is virtual.
The force diagram below is for a 10 kg object. Determine its acceleration.
As the buoyant force balances the weight of the object.
Thus, there is no motion of the object in the vertical direction.
Whereas, the force acting on the object towards the right is more than the drag force.
Thus, the net force acting on the object along the horizontal direction is,
[tex]F_{net}=F_T-F_{drag}[/tex]Substituting the known values,
[tex]\begin{gathered} F_{\text{net}}=35-10 \\ F_{\text{net}}=25N_{} \end{gathered}[/tex]According to Newton's second law, the net force acting on the object in terms of the acceleration of the object is,
[tex]F_{\text{net}}=ma[/tex]where m is the mass, and a is the acceleration,
Substituting the known values,
[tex]\begin{gathered} 25=10\times a \\ a=\frac{25}{10} \\ a=2.5ms^{-2} \end{gathered}[/tex]Thus, the acceleration of the object is 2.5 meters per second.
H
Calculate the area of the plates of a 1 pF parallel plate capacitor in a vacuum if the separation of the plates is 0.1 mm. [ε0=8.85×10-12C2N-1m-2 ]
For parallel plate capacitors the capacitance C=Aϵ0/d
dAϵ 0
Area, A Cd/ ϵ0 = 1 * 1×[tex]10^{-3}[/tex]/8.85× [tex]10^{-12}[/tex]
= 1.13×10 8 m 2.
The capacitance depending on geometry is given by the formula C=ϵ⋅Ad C = ϵAd. where C is the capacitance value A is the area of each plate, and d is the distance between the plates. plates, and ϵ is the dielectric constant of the material between the plates of the parallel capacitor.
The governing equations for capacitor design are C = εA/d where C is the capacitance. ε is the permittivity, a term that describes how well a dielectric holds an electric field. A is the area of the parallel plates. d is the distance between the two conductive plates. This is shown below. To calculate the total capacitance of multiple capacitors connected in this way, sum the individual capacitances.
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4. Betsy wants to use her own weight to lift a 1,500-newton box. She weighs 500 newtons. Suggest input and output arm lengths that would allow Betsy to lift the box with a lever. Draw a lever and label the input and output arms with the lengths and forces.
Answer:
Explanation:
Given:
F₁ = 500 N
F₂ = 1.500 N
____________
L₁ - ?
L₂ - ?
Moment rule:
M₁ = M₂
F₁*L₁ = F₂*L₂
F₂ / F₁ = 1.500 / 500 = 3
L₁ / L₂ = 3
L₁ = 3*L₂
If L₂ = 0.5 m
then L₁ = 3*0.5 = 1.5 m
a 30.0 kg child starting from rest slides down a water slide with the vertical height of 10.0 m what is the child speed (a )halfway down the slides vertical distance and (b) 3/4 of the way down
We know that
• The mass is m = 30.0 kg.
,• The vertical height is h = 10.0 m.
(a)We have to use the conservation of energy theorem, which states that mechanical energy is constant all the time. Also, halfway down means a height of 5.0 m. It's important to know that at the top the total energy is potential, while halfway is distributed as kinetic and potential, the expression below shows this
[tex]E_{p1}=E_{k1}+E_{p2}[/tex]Then, using the definition of each energy, we have
[tex]mgh_1=\frac{1}{2}mv^2+mgh_2[/tex]Now, we use the given values to find the speed.
[tex]\begin{gathered} \text{mgh}_1=m(\frac{1}{2}v^2+gh_2) \\ gh_1=\frac{1}{2}v^2+gh_2 \\ 9.81m/s^2\cdot10m=\frac{1}{2}v^2+9.81m/s^2\cdot5m \\ 98.1m^2/s^2=\frac{1}{2}v^2+49.05m^2/s^2 \\ 98.1m^2/s^2-49.05m^2/s^2=\frac{1}{2}v^2 \\ 2\cdot49.05m^2/s^2=v^2 \\ v=\sqrt[]{98.1m^2/s^2} \\ v\approx9.9m/s \end{gathered}[/tex]Therefore, the speed of the child halfway down is 9.9 meters per second.(b)In this case, we just have to use as the second height of the equation the magnitude 2.5 meters because that's 3/4 of the way down. So, let's use the same process and expression
[tex]\begin{gathered} gh_1=\frac{1}{2}v^2+gh_2 \\ 9.81m/s^2\cdot10m=\frac{1}{2}v^2+9.81m/s^2\cdot2.5m \\ v=\sqrt[]{2(98.1m^2/s^2-24.53m^2/s^2)} \\ v\approx12.1m/s \end{gathered}[/tex]Therefore, the speed of the child 3/4 of the way down is 12.1 meters per second.Hint: first convert the given final speed (in mph) to find ft/s, then solve for the distance and acceleration.
ANSWER
[tex]\begin{gathered} Dis\tan ce=149.56ft \\ Acceleration=4.14ft/s^2 \end{gathered}[/tex]EXPLANATION
First, we have to find the acceleration of the bicyclist.
To do that, we apply the formula:
[tex]a=\frac{v-u}{t}[/tex]where u = initial velocity
v = final velocity
t = time
Let us convert the speed given to feet per second. To do that, multiply the speed by 1.46667:
[tex]\begin{gathered} 24\cdot1.46667 \\ \Rightarrow35.2mph \end{gathered}[/tex]The bicyclist started from rest, so the initial speed is 0 ft/s.
Therefore, the acceleration is:
[tex]\begin{gathered} a=\frac{35.2-0}{8.5} \\ a=4.14ft/s^2 \end{gathered}[/tex]Now, to find the distance traveled, apply the formula:
[tex]s=ut+\frac{1}{2}at^2[/tex]Therefore, the distance traveled is:
[tex]\begin{gathered} s=(0\cdot8.5)+\frac{1}{2}\cdot4.14\cdot8.5^2 \\ s=\frac{1}{2}\cdot4.14\cdot8.5^2 \\ s=149.56ft \end{gathered}[/tex]Monochromatic light of 605 nm falls on a single slit of width 0.095 mm. The slit is located 85 cm from a screen. How far from the center is the first dark band?540 mm5,400 mm0.5 mm5.4 mm
The wavelength of the monochromatic beam is given as,
[tex]\lambda=605\text{ nm}[/tex]The value of the width of the slit is,
[tex]d=0.095\text{ mm}[/tex]The distance of the slit from the screen is,
[tex]D=85\text{ cm}[/tex]Thus, the distance of the dark band from the center is,
[tex]y=\frac{(2m+1)\lambda D}{d}[/tex]For the first dark band,
m=0,
Substituting all the known values,
The value of the distance of the dark band becomes,
[tex]\begin{gathered} y=\frac{605\times10^{-9}\times85\times10^{-2}}{0.095\times10^{-3}} \\ y=540000\times10^{-8}\text{ m} \\ y=5.4\text{ mm} \end{gathered}[/tex]Thus, the first dark band is 5.4 mm away from the center of the screen.
Hence, the correct answer is 5.4 mm.
- If we start an experiment with 191.3 g of a substance, how much should we end with?
More than 191.3 g
191.3 g
Less than 191.3g PLEASE NOW
If we start an experiment with 191.3 g of a substance, we should be ending with 191.3 g
We know that matter cannot be created out of nothing. So that is why the number of atoms of products must be equal to the number of atoms of reactants. That is why a chemical reaction is written as a balanced reaction equation. This is the law of conservation of mass.
During a combustion reaction or during decomposition reaction, the products mass might seem to be different than that of reactants either higher or lower. This is due to the involvement of environment. So while considering mass environment should also be included.
Therefore, if we start an experiment with 191.3 g of a substance, we should be ending with 191.3 g
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Increasing the mass wouldIncreasing the mass would_______can increase or decrease the value of the gravitational force depending on the initial valuehave no effect on the gravitational forceincrease the gravitational forcedecrease the gravitational force
We will have that:
Increasing the mass would increase the gravitational force.
-%deviation from expected value (show calculations) in relation between frequency and radius.
When an object moves in a circular trajectory with a uniform speed, a centripetal force is needed to keep it in that trajectory. In the experiment, the hanging mass is used to keep the tension of the string constant. The tension is equal to the weight of the hanging mass, and it is responsible for keeping the rubber stopper in a circular trajectory.
The centripetal acceleration of a particle in a circular trajectory with radius r and frequency f is:
[tex]a_c=(2\pi f)^2r[/tex]On the other hand, if the mass of the hanging object is M and the mass of the rubber stopper is m, according to Newton's Second Law of Motion, the relationship between the tension of the string and the centripetal acceleration is:
[tex]\begin{gathered} \Sigma F=ma_c \\ \\ \Rightarrow T=m(2\pi f)^2r \\ \\ \Rightarrow Mg=m(2\pi f)^2r \end{gathered}[/tex]If we isolate f from that expression, we get a theoretical relationship between the radius and the frequency:
[tex]\begin{gathered} \Rightarrow(2\pi f)^2=\frac{Mg}{mr} \\ \Rightarrow2\pi f=\sqrt{\frac{Mg}{mr}} \\ \\ \therefore f=\frac{1}{2\pi}\sqrt{\frac{Mg}{mr}} \end{gathered}[/tex]To find the expected value for the frequency for different values of r, replace M=200g, m=15.3g, g=9.81m/s^2 and the different values of r:
[tex]\begin{gathered} f_{10cm}=\frac{1}{2\pi}\sqrt{\frac{(200g)(9.81\frac{m}{s^2})}{(15.3g)(0.10m)}}\approx5.70Hz \\ \\ f_{15cm}=\frac{1}{2\pi}\sqrt{\frac{(200g)(9.81\frac{m}{s^2})}{(15.3g)(0.15m)}}\approx4.65Hz \\ \\ f_{20cm}=\frac{1}{2\pi}\sqrt{\frac{(200g)(9.81\frac{m}{s^2})}{(15.3g)(0.20m)}}\approx4.03Hz \\ \\ f_{25cm}=\frac{1}{2\pi}\sqrt{\frac{(200g)(9.81\frac{m}{s^2})}{(15.3g)(0.25m)}}\approx3.60Hz \\ \\ f_{30cm}=\frac{1}{2\pi}\sqrt{\frac{(200g)(9.81\frac{m}{s^2})}{(15.3g)(0.30m)}}\approx3.29Hz \end{gathered}[/tex]On the other hand, we can find the experimental measurements for the frequency depending on the radius using the information provided in the table.
To do so, remember that the frequency is equal to the average number of spins from the three trials over the time.
The experimental values for the frequencies are:
[tex]\begin{gathered} f_{10cm-exp}=\frac{35+34+32}{3\times10s}\approx3.37Hz \\ \\ f_{15cm-exp}=\frac{31+29+27}{3\times10s}\approx2.90Hz \\ \\ f_{20cm-exp}=\frac{27+27+25}{3\times10s}\approx2.63Hz \\ \\ f_{25cm-exp}=\frac{25+23+24}{3\times10s}\approx2.40Hz \\ \\ f_{30cm-exp}=\frac{26+23+20}{3\times10s}\approx2.30Hz \end{gathered}[/tex]To find the percent deviation from the expected value, use the following formula:
[tex]\%dev=\frac{f_{exp}-f}{f}\times100\%[/tex]Then, the percent deviations from the expected values are:
[tex]\begin{gathered} \%dev_{10cm}=\frac{3.37Hz-5.70Hz}{5.70Hz}\times100\%\approx-40.9\% \\ \\ \%dev_{15cm}=\frac{2.90Hz-4.65Hz}{4.65Hz}\times100\%\approx-37.6\% \\ \\ \%dev_{20cm}=\frac{2.63Hz-4.03Hz}{4.03Hz}\times100\%\approx-34.7\% \\ \\ \%dev_{25cm}=\frac{2.40Hz-3.60Hz}{3.60Hz}\times100\%\approx-33.3\% \\ \\ \%dev_{30cm}=\frac{2.30Hz-3.29Hz}{3.29Hz}\times100\%\approx-30.1\% \end{gathered}[/tex]
What is the mass of a truck if it produces a force of 13,607 N while accelerating at a rate of 2.9 m/s²?
Newton's Second Law.
We have the following data, provided in the exercise:
Mass (m) = ?
Force (F) = 13607 N
Acceleration (a) = 2.9 m/s²
To calculate the mass, we solve the following formula:
[tex]\boxed{\large\displaystyle\text{$\begin{gathered}\sf \bf{F=m*a \ \ < ====Formula \ clear==== > \ \frac{F}{a}=\frac{m*\not{a}}{\not{a}} } \end{gathered}$}}[/tex][tex]\boxed{\large\displaystyle\text{$\begin{gathered}\sf \bf{\frac{F}{a}=m \ \iff \ Address \ is \ changed \Rightarrow \ m=\frac{F}{a} } \end{gathered}$}}[/tex]
We substitute the data in the clear formula.
[tex]\boxed{\large\displaystyle\text{$\begin{gathered}\sf \bf{m=\frac{13607 \ N}{2.9 \ \frac{m}{s^{2} } } } \end{gathered}$} }[/tex]
We break down the units of Newton = Kg * m/s².
[tex]\boxed{\large\displaystyle\text{$\begin{gathered}\sf \bf{m=\frac{13607 \ Kg*\not{\frac{m}{s^{2}} } }{2.9 \not{\frac{m}{s^{2}} } } } \end{gathered}$} }[/tex]
[tex]\boxed{\large\displaystyle\text{$\begin{gathered}\sf \bf{m=4692.07 \ Kg } \end{gathered}$} }[/tex]
The mass of the truck is: 4692.07 Kilograms (Kg).The steeper the slope of a current vs voltage graph, the _________ the resistance.A. Not enough infoB. LowerC. The sameD. Greater
The slope of the graph current vs voltage can be explained with the next
[tex]\begin{gathered} V=IR \\ I=\frac{V}{R} \end{gathered}[/tex]In this case, current vs voltage means that current is y axis, and voltage is x axis. Based n that, the equation is the second one
[tex]I=\frac{V}{R}[/tex]And the slope is the inverse of the resistance. Dont forget that current - voltage is different to voltage - current
Joshua rode his bicycle from his school to home. The total distance covered in this journey was 800 meters and the total time taken was 7 minutes. What is the average speed of joshuas bicycle in this journey
Answer: 6,857.14m/h (6.86km/h or 1.90m/s)
Explanation:
Ricardo, of mass 72 kg, and Carmelita, who is lighter, are enjoying Lake Merced at dusk in a 39 kg canoe. When the canoe is at rest in
the placid water, they exchange seats, which are 3.2 m apart and symmetrically located with respect to the canoe's center. Ricardo
notices that the canoe moves 46 cm horizontally relative to a pier post during the exchange and calculates Carmelita's mass. What is it?
Ricardo notices that the canoe moves 46 cm horizontally relative to a pier post during the exchange and the mass is 49 kg.
What is mass?Mass is defined as a measurement of the fundamental attribute of all stuff, inertia.
Mass is also defined as a physical body's total amount of matter.
As per question,
Let's refer to Ricardo's mass as MR and Carmelita's mass as MC.
Let the two-person system's center of mass, which is believed to be closer to Ricardo, be x meters from the center of the canoe's length L and mass m.
MC(L/2 +x) = mx + MR(L/2-x).
It is derived from the second law of parallel forces, which states that the sum of the moments of mass moving clockwise and counterclockwise is equal.
The canoe's center has now moved twice as far from its initial location.
So, x = 46/2 = 23 cm = 0.23 m
MC = MR (L / 2x) - mx / L / 2 + x
MC = 72 ( 3.2 / 2 - 0.23 ) - 39 x 0.23 / 3.2 / 2 + 0.23
MC = 72 ( 1.6 - 0.23 ) - 39 x 0.23 / 1.6 + 0.23
MC = 72 x 1.37 - 8.97 / 1.83
MC = 98.64 - 8.97 / 1.83
MC = 49 kg
Thus, Ricardo notices that the canoe moves 46 cm horizontally relative to a pier post during the exchange and the mass is 49 kg
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Two ways to modulate a wave are to change the wave's ___ (AM) or the wave's__ (FM).
Answer:
AM - amplitude modulation
FM - frequency modulation
Aristotle's rhetorical triangle tells us that effective messages show awareness of
ethos, pathos, and logos
logic, emotion, and values
speaker, audience, and purpose
thought, voice, and opinion
Aristotle's rhetorical triangle tells us that effective messages show awareness of ethos, pathos, and logos. The correct answer to the question is option A.
What is Aristotle's rhetorical triangle?
Aristotle taught that a speaker's effective messages show awareness of the ethos, pathos, logos triangle, which expresses the idea that a good argument must be ethically, emotionally, and logically sound.
Ethos is a Latin word which means for ethics. This emphasize that that effective messages should show trust. The audience should be able to be confident of the source of information.Pathos talks about to the emotions.Logos is the Latin word for logic. The messages would be verified by the audience. The speaker would ensure there are facts backing up claims.In summary, Aristotle's rhetorical triangle shows that the Logos, Ethos, and Pathos are very important when giving an effective message.
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Explain why planets orbit the sun instead of traveling off into space
Answer:
gravity
Explanation:
The gravity of the Sun keeps the planets in their orbits. They stay in their orbits because there is no other force in the Solar System which can stop them.
An Olympic diver is on a diving platform 8.40 m above the water. To start her dive, she runs off of the platform with a speed of 1.2 m/s in the horizontal direction. What is the diver speed, in m/s, just before she enters the water?m/s
Given data
*The given distance above the water is s = 8.40 m
*The given initial speed of the Olympic diver is u = 1.2 m/s
*The value of the acceleration due to gravity is a = 9.8 m/s^2
The formula for the diver speed, in m/s, just before she enters the water is given by the equation of motion as
[tex]v^2=u^2+2as[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} v^2=(1.2)^2+2\times9.8\times8.40 \\ v=12.88\text{ m/s} \end{gathered}[/tex]Hence, the diver speed, in m/s, just before she enters the water is v = 12.88 m/s
Which of these does sound travel the fastest Through?A. steel SolidB. water LiquidC. Air gasD. Sound travels the same through all mediums
Speed of sound is maximum on solid.
Correct option.
A) steel Solid
if a child on a skateboard has a total momentum of 86.94kgm/s and a speed of 2.3m/s, calculate the total mass of the child and skateboard
if may 20 is a last quarter moon, when could you expect the first quarter moon to occur?
In the early afternoon or high in the sky around the time, the sun is setting.
The crescent moon rises around noon and sets around midnight. It can be seen near the horizon in the evening or high in the sky as the sun sets. If you live in the northern hemisphere, the first quarter moon appears as the right half of the illuminated surface. Occurs when the moon is at right angles to the sun as seen from Earth.
The right half of the moon appears bright and the left half appears dark. During the period from the new moon to the first quarter moon, the illuminated part of the moon grows larger day by day and continues to grow until the full moon. The first quarter is also called the crescent moon because 50% of its surface is briefly illuminated by the sun. Usually, he takes 3 nights. Each phase of the moon has a profound effect on humans and the earth.
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Descriptions
A. Research that is designed to describe the characteristics or behaviors of a person or particular population
B. The thing the researcher changes in the experiment
C. Research that attempts to test cause and effect between two things
D. The thing that is influenced by another variable in an experiment
E. Research that attempts to assess the relationship between two aspects of human behavior
Fill in the blank with the letter of the description that best matches the term.
Descriptive Studies
PLEASE HELP!! IF SOMEONE KNOWS THE REST PLEASE TELL ME! THANKYOU!
Answer:
I think it is C
Explanation:
1) Prove that for the adiabatic process of gas pVr =const ; r=Cp/Cv *
2) Write down the expression for the number of molecules of N2 gas having speeds in the interval (100m/s ; 102m/s) for one mole of gas at T=300K
3) Demonstrate how to obtain the Boltzmann Distribution of Energies from the Maxwell-Boltzmann distribution of speeds.*
For an adiabatic process of gas PV(^r)= Constant, where r = Cp/Cv , Cp, and Cv are constants for a particular gas.
What is an ideal gas?It is an imaginary gas for which the volume occupies by it is negligible, this gas does not exist in a practical situation and the concept of an ideal gas is only the theoretical one.
Universal Gas Equation for the ideal gas,
PV=nRT
By differentiating on both sides: P.dV + V.dP = n.R.dT…. (1)
For the adiabatic process, the expansion or compression of gas occurs at a very fast rate which makes no heat transfer. Therefore, dQ= 0
A/c to First Law of TD: dQ=dU+ dW, implies dU= -dW=-P.dV ….. (2)
But dU= n.Cv. dT …. (3) Combining 2 and 3, we get: n.dT= -P.dV/Cv …..(3)
Using R= Cp-Cv, the gas law gives: from 1 and 3.
n.dT= (P.dV + V.dP)/(Cp-Cv) = -PdV/Cv
On rearranging terms we get, (dP/P)+((Cp/Cv)*(dV/V))=0
Cp and Cv are constants for a particular gas, hence integrating,
Ln(P)+ r.Ln(V)=constant (replacing Cp/Cv = r or Gamma)
Hence, P.V(^r)= constant is termed an adiabatic equation.
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How many kilometers does the space shuttle have to travel to complete one orbit? In terms of a circle, what is this distance called? Explain.
40,840.7 kilometers the space shuttle have to travel to complete one orbit. In terms of a circle, this distance is termed as the circumference of the circle.
The circumference of a circle is the length measured around its edge. The diameter of a circle is the distance from the center to the outside.
Here we need to find the distance of the space shuttle that completed one circle, ie, the circumference of the orbit. The Circumference or distance covered by the space shuttle can be denoted by [tex]C_{SS}[/tex] and can be calculated by application of the below formula,
[tex]C_{SS}[/tex] = 2πr
where r is the radius of earth ie, 6500Km
Therefore, the equation becomes:
[tex]C_{SS}[/tex] = 2×π×r
= 2×π×6500
=40,840.7
So, the kilometer required to travel is 40,847.7Km
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The graph below shows the number of sunspots observed between 1750 and 2000.
Based on the graph, which of these statements is most likely correct about the periods 1800 to 1850 and 1900 to 1950?
A) There was decrease in global temperatures.
B) There was increase in global temperatures.
C) There was no significant change in solar activity.
D) There was no significant change in global climate.
Based on the graph, during the periods 1800-1850 and 1900-1950, there was a increase in global temperature.
Sunspots is an activity on the surface of the Sun that manifest as transient, darker-than-the-neighboring-areas spots. These parts on which it occurs have lower surface temperatures because magnetic flux concentrations there prevent convection from occurring.
Sunspot can increase the temperature of earth but by only a margin of about 0.05 to 0.1 Celsius.
As we can see from the graph that during the year 1800-1850 and 1900-1950 that number of sunspots is increasing continuously. So, there will a slight increase will be noticed in the global temperature.
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Two ice skaters stand facing each other at rest on a frozen pond. They push off against one another and the 49 kg skater acquires a speed of 2.10 m/s. If the other skater acquires a speed of 3.81 m/s, what is her mass in kilograms?
The mass of the second skater is equal to 27 Kg, which acquires a speed of 3.81 m/s.
What is law of conservation of linear momentum?According to the law of conservation of momentum, the sum of the momentum of the two objects before and after the collision must be equal.
m₁ u₁ + m₂u₂ = m₁ v₁ + m₂ v₂
where m₁ and m₂ is the mass of the collided bodies, u₁ and u₂ are the initial speed while v₁ & v₂ is their final speed.
The linear momentum can be described as the product of the mass times the velocity of the object. Conservation of momentum is a characteristic held by an object where the total amount of momentum remains the same.
Given the initial speed of the skaters is equal to zero so u₁ = u₂ =0
The mass of the one skater, m₁ = 49 Kg
The first skater acquires the speed, v₁ = 2.10 m/s
The speed acquired by the second skater, v₂ = -3.81 m/s
From the law of conservation of momentum, we can find the mass of another skater:
m₁ u₁ + m₂ u₂ = m₁v₁ + m₂ v₂
49 × 0 + m₂ ×0 = 49 × (2.10) + m₂ × (- 3.81)
m₂ = -49 × (2.10)/ (- 3.81)
m₂ = 27 Kg
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When two coherent waves combine destructively to create a minimum, what is the smallest phase difference possible in Degrees and Radians?
180° or π Radians is the smallest phase difference possible When two coherent waves combine destructively to create a minimum
Two waves of the same wavelength that are entirely out of phase with one another might cancel one another out or interact negatively. When there is a 180 degree or larger radian phase difference, this happens. Only when the two waves have the same amplitude do they completely cancel out. Partially destructive superposition occurs when two waves that are entirely out of phase yet have different amplitudes. The intensity is 0.
Superposition is the phenomenon in which the combined displacement of two combining waves is equal to their vector sum.
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In Young's double-slit experiment, two slits are separated by 5.0 mm and illuminated by light with a wavelength of 480 nm. The screen is 3.0 m from the plane of the slits. Calculate the separation between the eighth bright fringe and the third dark fringe observed with respect to the central bright fringe.
We are given the following information.
Seperation between slits: d = 5.0 mm
Wavelength of light: λ = 480 nm
Distance from the plane of slits: D = 3.0 m
We are asked to calculate the separation between the 8th bright fringe and the 3rd dark fringe observed with respect to the central bright fringe.
The position of the 8th bright fringe is given by
[tex]\begin{gathered} x_n=\frac{n\lambda D}{d} \\ x_8=\frac{8\cdot480\times10^{-9}\cdot3}{5\times10^{-3}} \\ x_8=2.304\times10^{-3}\;m \end{gathered}[/tex]The position of the 3rd dark fringe is given by
[tex]\begin{gathered} x_n=(\frac{2n-1}{2})\frac{\lambda D}{d} \\ x_3=(\frac{2\cdot3-1}{2})\frac{480\times10^{-9}\cdot3}{5\times10^{-3}} \\ x_3=7.2\times10^{-4}\;m \end{gathered}[/tex]Finally, the separation between the 8th bright fringe and the 3rd dark fringe is
[tex]\begin{gathered} x_8-x_3=2.304\times10^{-3}-7.2\times10^{-4} \\ x_8-x_3=1.584\times10^{-3}\;m \end{gathered}[/tex]Therefore, the separation between the eighth bright fringe and the third dark fringe observed with respect to the central bright fringe is 1.584×10⁻³ m.