Answer:
The new Coulomb force is q₁q₂/9πε₀r²
Explanation
The coulomb force between the two charges q₁ and q₂ at a distance r in air is given by F = q₁q₂/4πε₀r².
Now, let us assume the material of dielectric constant κ = 9 is placed between them on the side of the q₁ charge. The value of its effective charge is now q₃ = q₁/κ at a distance of d = r/2 from the q₂ charge.
Since we have air between q₂ and q₃, the coulomb force between them is
F' = q₂q₃/4πε₀d²
= q₂(q₁/κ)/4πε₀(r/2)²
= 4q₂q₁/κ4πε₀r²
= 4/κ(q₂q₁/4πε₀r²)
= 4/9 × (q₂q₁/4πε₀r²)
= q₁q₂/9πε₀r²
So, the new Coulomb force is q₁q₂/9πε₀r²
a uniform rod of length 1.5m is placed over a wedge at 0.5m from one end .a force of 100 N is applied at its one end near the wedge to keep it horizontal .find weight of rod and reaction of wedge
Explanation:
The rod is uniform, so the center of gravity is at the center, or 0.75 m from the end. The wedge is 0.5 m from the end, so the center is 0.25 m from the wedge.
Sum the torques about the wedge (it may help to draw a diagram first). Take counterclockwise to be positive.
∑τ = Iα
W (0.25 m) − (100 N) (0.50 m) = 0
W = 200 N
Sum the forces in the y direction.
∑F = ma
F − 100 N − 200 N = 0
F = 300 N
A car is traveling to the right with a speed of 2.0\,\dfrac{\text m}{\text s}2.0 s m 2, point, 0, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction on an icy road when the brakes are applied. The car slows down with constant acceleration for 3.0\,\text m3.0m3, point, 0, start text, m, end text until it comes to a stop. How long does it take the car to slide to a stop?
Answer:
It took the car 3 second to come to full stop
Explanation:
When the car is driving to the right at a speed of 2 m/s, and suddenly brakes giving the car a constant negative acceleration making it travel 3 meters after it stops, we can use the following kinematic equation to solve for the acceleration applied by the brakes:
[tex]v=v_0-a\,t\\0=2-a\,t\\a=\frac{2}{t}[/tex]
and we use this result in the following kinematic equation for the distance traveled to find the time it took it to stop:
[tex]x-x_0=v_0\,t-\frac{1}{2} a\,\,t^2\\3=2\,t -\frac{1}{2} (\frac{2}{t}) \,\,t^2\\3=2\,t-t\\3=t[/tex]
therefore, it took the car 3 seconds to come to a full stop.
The rhinestones in costume jewelry are glass with index of refraction 1.50. To make them more reflective, they are often coated with a layer of silicon monoxide of index of refraction 2.00. What is the minimum coating thickness needed to ensure that light of wavelength 480 nm and of perpendicular incidence will be reflected from the two surfaces of the coating with fully constructive interference?
Answer:
60
Explanation:
According to the given question, the computation of minimum coating thickness is shown below:-
The condition for constructive interference is
[tex]2t_{min} = (m + \frac{1}{2} )\times \frac{\lambda}{^nmateral}[/tex]
[tex]= (0 + \frac{1}{2} )\times \frac{\lambda}{^nmateral}[/tex]
[tex]t = \frac{\lambda}{4n}[/tex]
Now we will put the values to the above formula to reach the answer
[tex]= \frac{480nm}{4\times 2.0}[/tex]
= 60
Therefore we simply applied the above formula to determine the minimum coating thickness
The minimum coating thickness needed to ensure that light of wavelength 480 nm and of perpendicular incidence will be reflected from the two surfaces of the coating with fully constructive interference is 60.
Given that, the index of refraction of glass is 1.50 and the index of refraction of silicon monoxide is 2.00. The wavelength light of 480 nm.
If [tex]\lambda {c}[/tex] is the wavelength in the coating and λ is the wavelength in vacuum, then [tex]\lambda_{c} = \dfrac {\lambda}{n}[/tex] where n is the index of refraction of the coating.
Thus, the minimum coating thickness [tex]T_{min}[/tex] can be calculated as,
[tex]2T_{min} = (m+\dfrac {1}{2})\times\dfrac {\lambda} {n}\\\\T_{min} = (m+\dfrac {1}{2}) \times\dfrac {\lambda} {2n}[/tex].
For the constructive interference, m = 0.
So [tex]T_{min} = \dfrac {\lambda} {4n}[/tex]
[tex]T_{min} = \dfrac {480\rm nm} {4\times2.0}[/tex]
[tex]T_{min} = 60[/tex]
The minimum coating thickness needed to ensure that light of wavelength 480 nm and of perpendicular incidence will be reflected from the two surfaces of the coating with fully constructive interference is 60.
For more details, follow the link given below.
https://brainly.com/question/15541641.
Halley is standing outside on a cloudy day. When she hears thunder, she goes back inside so that she doesn't get caught in
a storm. Is this an example of inference or prediction? Explain
Answer:
Inference
Explanation:
An inference involves the application of logic to progress from a premise to a conclusion or logical consequence on the basis of the evidence or known fact. Inference is a process of thought that be divided into a deduction and an induction aspect.
In the given question Halley, by standing outside was able to deduce the sound of thunder she is then able by inductive reasoning from the fact that storms are usually preceded by and accompany lightening, conclude that there is a storm coming.
Forensic toxicologist analyze and identify drugs that are confiscated from criminals
True
False
The radius of a small ball is around 3.79747 cm. The radius of a basketball is about 3.16 times larger. What is the ratio of the surface areas of the small ball and a basketball? 2. What is the ratio of their volumes?
Explanation:
The ratio of the areas is the square of the ratio of the radii.
A/A = 3.16² = 9.99
The ratio of the volumes is the cube of the ratio of the radii.
V/V = 3.16³ = 31.6
A narrow beam of light containing red (660 nm) and blue (470 nm) wavelengths travels from air through a 1.00 cm thick flat piece of crown glass and back to air again. The beam strikes at an incident angle of 30 degrees. (a) At what angles do the two colors emerge
Answer:
The color blue emerges at 19.16° and the color red emerges at 19.32°.
Explanation:
The angle at which the two colors emerge can be calculated using the Snell's Law:
[tex]n_{1}sin(\theta_{1}) = n_{2}sin(\theta_{2})[/tex]
Where:
n₁ is the refractive index of the incident medium (air) = 1.0003
n₂ is the refractive index of the refractive medium:
blue light in crown glass = 1.524
red light in crown glass = 1.512
θ₁ is the angle of the incident light = 30°
θ₂ is the angle of the refracted light
For the red wavelengths we have:
[tex] \theta_{2} = arcsin(\frac{n_{1}sin(\theta_{1})}{n_{2}}) = arcsin(\frac{1.0003*sin(30)}{1.512}) = 19.32 ^{\circ} [/tex]
For the blue wavelengths we have:
[tex] \theta_{2} = arcsin(\frac{n_{1}sin(\theta_{1})}{n_{2}}) = arcsin(\frac{1.0003*sin(30)}{1.524}) = 19.16 ^{\circ} [/tex]
Therefore, the color blue emerges at 19.16° and the color red emerges at 19.32°.
I hope it helps you!
A concave mirror has a radius of curvature of 80cm. If the real image of an object is formed at 1m, how far from the mirror is the object?
Answer:
66.7 cm
Explanation:
The focal length is half the radius of curvature, or f = 40 cm.
The image is real, so i = +100 cm.
The distance to the object is therefore:
1/o + 1/f = 1/i
1/o + 1/40 = 1/100
o = -66.7 cm
The object is 66.7 cm left of the mirror.
what is the frequency and wavelength of a remote control? PLEASE HURRY
Answer:
940 nanometers wavelength, and about an 300 mHz - 400 gHz frequency
2. Adelia holds a shiny steel spoon with its back (convex surface) facing her eyes at a distance
of approximately 30 cm. She sees an upright image of herself. However, when the spoon
is changed so that the front (concave surface) of the spoon is facing her eyes, an inverted
image is observed.
(a) Explain this situation.
(b) Why is an upright image not seen on the front surface of the spoon at that distance?
Answer:
(a) The convex mirror image, is always upright at all positions, while images formed by concave mirrors are always inverted when the object distance from the mirror is more than the mirrors focal length.
(b) An upright image is not seen for object at a distance from a concave mirror further than the focal length of the mirror, which is the spoon in the question
Therefore, the location of her eyes of approximately, 30 cm, from the mirror is more than the mirror's focal length
Explanation:
What is CGS unit of weight
Answer:
[tex] \boxed{dyne}[/tex]Explanation:
The weight of a body is defined as the force with which it is attracted towards the centre of a planet. A spring balance measures it. It's unit is Newton ( N ) in the SI system and dyne in the CGS system.
Hope I helped!
Best regards!
Weight is force.
The CGS unit of force is the dyne.
The SI unit of force is the Newton.
Which atomic number corresponds to the element which is most stable?
01
03
28
36
Answer:
Which atomic number corresponds to the element which is most stable?
01
03
28
36
36 is your answer.
what is the function of eye lens of the human eye
Answer:
Lens of the human Eye is a important and one of that most complex sense organ.
Explanation:
Lens of the human eye it helps that in visualizing light and color perception and objects , glance on the human eye structure and function.
Sense organs are much pretty similar to the camera they help us see the objects clear.
A human eye is the 2.3 cm in diameter and all filled some fluid, and there are following parts in eye:- cornea, Retina , Lens ,Pupil , optic nerves.
cornea :- the cornea is the first transparent part of is called cornea, enters the light through the cornea.
Retina :-it is light sensitive layer that consists of nerve cells,then transmitted to the brain through nerves.
Lens:- behind the pupil there is a transparent structure called lens,it shape focus light on the retina.
Pupil:-it control the value of light that enters the human eye.
Optic nerves is are the two types :- (1) cones (2) Rods .
Answer:
It causes Light Rays to converge on the retina
Explanation:
I took the K12 Test
Air contains 78.08% nitrogen, 20.095% oxygen, and 0.93% argon. a sample of air occupies 8 l at 273 k and applies a pressure of 2.45 atm to its container. the sample of air is heated to 550 k, at which time it takes up 11.8 l. what is the partial pressure of nitrogen after the air has been heated
Answer:
the partial pressure of nitrogen after the air has been heated is 2.61atm
Explanation:
Using combined gas law which is the combination of Boyle's Law and Charles' Law, and it gives The relationship between pressure, volume, and absolute temperature
PV/T= K
P1₁V₁/T₁=P₂V2₂/T₂
P₂=final pressure= ?
P₁= initial pressure=2.45 atm
.T₁= initial temperature= 273k
T₂= final temperature= 550 K,
We can make P₂ subject of the formula then we have
P₂=P₁V₁T₂/T₁T₂
=( 2.45×8×550)/(273×11.8)
P2=3.35atm
Now we can calculate the partial pressure of N₂
N₂= 3.35×(78.08/10)
=2.61atm
Therefore, the partial pressure of nitrogen after the air has been heated is 2.61atm
Help!!! Line B touches the circle at a single point. Line A extends through the center of the circle.
I. What is line b in reference to the circle?
2. How large is the angle between lines A and B?
Answer:
If I understand correctly. Line B is parallel to the circle. Also, the angle is less than 90.
The size of the circle determines. The diameter should not be fixed either.'Circle' is characterized as "a two-dimensional geometric figure comprising of the set of all those points in a plane that are at equal distance from the center."
1). Line B would be considered as the 'tangent' in association with the given circle.
2). The angle lying among the lines A and B would be of 90° i.e. right angle.
Given that,
A circle with line A drawn through its center
Line B moves while touching the edge of the circle and intersects line A outside the circle.
Since 'Tangent' is described as a 'straight-line touching the curve at one point without crossing it,' thus, line B would be considered as the tangent of the circle as it touches the circle's curve once. The angle that is formed at the intersection of the line A and B at the curve of the circle would be a right angle because they are perpendicular to one another. Thus, the perpendicular lines A and B form an angle of 90°.Learn more about 'circle' here:
brainly.com/question/11833983
A capacitor C is fully charged by connecting it to battery of V Volt. Then it is disconnected from battery. If the separation between the plates is halved, what will happen to (i) Charged stored by capacitor (ii) Potential difference across it. (iii) Electric filed strength between the plates. (iv) Energy stored by the capacitor.
Answer:
Explanation:
i )
When it is disconnected with the battery , the charge stored in it becomes fixed . When the plate distance becomes half , its capacitance becomes twice from C to 2C . Let charge stored in it at the time of disconnection from battery be Q . Let plate separation reduces from d to d / 2
So charged stored in it will remain unchanged .
ii )
Potential difference = charge / capacitance
in the first case potential difference = Q / C
in the second case potential difference = Q / 2C
So potential difference becomes half .
iii ) electric field = potential diff / plate separation
in the first case electric field = Q / (d x C )
in the second case electric field = 2 Q / (d x 2C)
= Q / (d x C )
So electric field remains unchanged .
iv)
energy stored in first case = Q² / 2C
In the second case energy stored = Q² / 2x2C
so energy stored becomes half .
Gravity is affected by the ?
Answer:
Mass and distance
Explanation:
The force of gravity the masses exert on each other. If one of the masses is doubled, the force of gravity between the objects is doubled. increases, the force of gravity decreases. Hence, answer is mass and distance
A ship sets out to sail to a point 123 km due north. An unexpected storm blows the ship to a point 112 km due east of its starting point. (a) How far and (b) in what direction (as an angle from due east, where north of east is a positive angle) must it now sail to reach its original destination
Answer:
Explanation:
We shall represent the point of destination and point of new starting point in vector form .
i and j will represent east and north .
point of destination = 123 j
starting point = 112 i
distance between the two
D = 123 j - 112 i
magnitude of D = √ ( 123² + 112² )
= 166.35 km
Direction of D
Tanθ = - 123 / 112
= - 1.09
θ = 132.54 degree from east in positive angle .
A 1 kg object accelerated at a constant rate of 5m/s? Estimate the net force needed to accelerate the object
Answer:
5 N.
Explanation:
Data obtained from the question include the following:
Mass (m) of object = 1 kg
Acceleration (a) = 5 m/s²
Force (F) =?
Force is simple defined as the product of mass and acceleration. Mathematically, it is expressed as:
Force (F) = mass (m) x acceleration (a)
F = ma
With the above formula, we can obtain the net force needed to accelerate the object as follow:
Mass (m) of object = 1 kg
Acceleration (a) = 5 m/s²
Force (F) =?
F = ma
F = 1 x 5
F = 5 N.
Therefore, the net force needed to accelerate the object is 5 N.
Calculate the Schwarzschild radius (in kilometers) for each of the following.1.) A 1 ×108MSun black hole in the center of a quasar. Express your answer using two significant figures.2.) A 6 MSun black hole that formed in the supernova of a massive star. Express your answer using two significant figures.3.) A mini-black hole with the mass of the Moon. Express your answer using two significant figures.4.) Estimate the Schwarzschild radius (in kilometers) for a mini-black hole formed when a superadvanced civilization decides to punish you (unfairly) by squeezing you until you become so small that you disappear inside your own event horizon. (Assume that your weight is 50 kg.) Express your answer using one significant figure.
Answer:
(I). The Schwarzschild radius is [tex]2.94\times10^{8}\ km[/tex]
(II). The Schwarzschild radius is 17.7 km.
(III). The Schwarzschild radius is [tex]1.1\times10^{-7}\ km[/tex]
(IV). The Schwarzschild radius is [tex]7.4\times10^{-29}\ km[/tex]
Explanation:
Given that,
Mass of black hole [tex]m= 1\times10^{8} M_{sun}[/tex]
(I). We need to calculate the Schwarzschild radius
Using formula of radius
[tex]R_{g}=\dfrac{2MG}{c^2}[/tex]
Where, G = gravitational constant
M = mass
c = speed of light
Put the value into the formula
[tex]R_{g}=\dfrac{2\times6.67\times10^{-11}\times1\times10^{8}\times1.989\times10^{30}}{(3\times10^{8})^2}[/tex]
[tex]R_{g}=2.94\times10^{8}\ km[/tex]
(II). Mass of block hole [tex]m= 6 M_{sun}[/tex]
We need to calculate the Schwarzschild radius
Using formula of radius
[tex]R_{g}=\dfrac{2MG}{c^2}[/tex]
Put the value into the formula
[tex]R_{g}=\dfrac{2\times6.67\times10^{-11}\times6\times1.989\times10^{30}}{(3\times10^{8})^2}[/tex]
[tex]R_{g}=17.7\ km[/tex]
(III). Mass of block hole m= mass of moon
We need to calculate the Schwarzschild radius
Using formula of radius
[tex]R_{g}=\dfrac{2MG}{c^2}[/tex]
Put the value into the formula
[tex]R_{g}=\dfrac{2\times6.67\times10^{-11}\times7.35\times10^{22}}{(3\times10^{8})^2}[/tex]
[tex]R_{g}=1.1\times10^{-7}\ km[/tex]
(IV). Mass = 50 kg
We need to calculate the Schwarzschild radius
Using formula of radius
[tex]R_{g}=\dfrac{2MG}{c^2}[/tex]
Put the value into the formula
[tex]R_{g}=\dfrac{2\times6.67\times10^{-11}\times50}{(3\times10^{8})^2}[/tex]
[tex]R_{g}=7.4\times10^{-29}\ km[/tex]
Hence, (I). The Schwarzschild radius is [tex]2.94\times10^{8}\ km[/tex]
(II). The Schwarzschild radius is 17.7 km.
(III). The Schwarzschild radius is [tex]1.1\times10^{-7}\ km[/tex]
(IV). The Schwarzschild radius is [tex]7.4\times10^{-29}\ km[/tex]
The energy conservation allows to find the Schwarschild radius for several bodies of different masses are:
1) Black hole quasar is: r = 2.9 10⁸ km
2) Blsck hole supernove is: r = 17.7 km
3) Mini black hole is: r = 1.1 10⁻⁷ km
4) Human body is: r= 7 10⁻²⁹ km
The schwarschild radius is defined as the distance from a black hole center at radius which the escape velocity is equal to the light speed, in some cases it is also called the event horizon.
Let's use Newton's second law where force is the universal law of attraction and acceleration is centripetal.
F = ma
F = [tex]G \frac{Mm}{r^2}[/tex]
Where F is the force, M the mass of the black hole, m the handle of the body, r the radius and v the speed of the body.
The energy of the gravitational field is
F = [tex]- \frac{dU}{dr }[/tex]
U = [tex]-G \frac{Mm}{r}[/tex]
Let's use conservation of energy
Em₀ = K + U = ½ m v² - [tex]G \frac{Mm}{r}[/tex]
In infinity the energy
Em_f = 0
energy is conserved
Em₀ = Em_f
½ m v² - [tex]G \frac{Mm }{r}[/tex] = 0
r = [tex]\frac{2GM}{v^2}[/tex]
From the definition of the Schwarschild radius this speed is equal to the light speed
v = c
r = [tex]\frac{2GM}{c^2 }[/tex]
They ask to calculate the radius for several cases of different mass, claculate the constant value
V = [tex]\frac{2 \ 6.67 \ 10^{-11} }{(3 \ 10^8) ^2 }[/tex]
V = 1.482 10⁻²⁷
1) A black hole of mass M = 1 10⁸ [tex]M_{sum}[/tex]
The tabulated mass of the sun is [tex]M_{sum}[/tex] = 1.989 10³⁰ kg
Let's substitute
r = 1.482 10⁻²⁷ 1 10⁸ 1.989 10³⁰
r = 2.94 10⁸ km
With two significant figures
r = 2.9 10⁸ km
2) A black hole of mass M = 6 [tex]M_{sum}[/tex]
r = 1.482 10⁻²⁷ 6 1.989 10-30
r = 17.7 km
3) a mini black hole with the mass of the moon
Tabulated mass of the moon M = 7.35 10²² kg
r = 1.482 10⁻²⁷ 7.35 10²²
r = 1.1 10⁻⁷ km
4) A person of M = 50 kg
r = 1.482 10⁻²⁷ 50
r= 7 10-29 km
In conclusion using the conservation of energy we can find the Schwarschild radius for several bodies of different masses are:
1) Black hole quasar is: r = 2.9 10⁸ km
2) Blsck hole supernove is: r = 17.7 km
3) Mini black hole is: r = 1.1 10⁻⁷ km
4) Human body is: r= 7 10⁻²⁹ km
Learn more here: https://brainly.com/question/12647190
A 30-cm long string, with one end clamped and the other free to move transversely, is vibrating in its second harmonic. The wavelength of the constituent traveling waves is:
Answer:
[tex]\lambda[/tex] = 40 cm
Explanation:
given data
string length = 30 cm
solution
we take here equation of length that is
L = [tex]n \times \frac{1}{4} \lambda[/tex] ...............1
so
total length will be here
[tex]L = \frac{\lambda}{2} + \frac{\lambda}{4}\\[/tex]
[tex]L = \frac{3 \lambda }{4}[/tex]
so [tex]\lambda[/tex] will be
[tex]\lambda = \frac{4L}{3}\\\lambda = \frac{4\times 30}{3}[/tex]
[tex]\lambda[/tex] = 40 cm
A space rocket accelerates uniformly from rest to 160ms^-1 upwards in 4.0s, then travels with a constant speed of 160ms^-1 for the next 5.0s.
What is the initial acceleration of the rocket?
Answer:
40 ms¯².
Explanation:
To solve this problem, we shall illustrate the question with a diagram.
The attached photo gives a better understanding of the question.
From the attached photo:
Velocity (v) = 160 ms¯¹
Time (t) = 4 secs.
Acceleration (a) =?
Acceleration (a) = Velocity (v) /time (t)
a = v/t
a = 160/4
a = 40 ms¯²
Therefore, the initial acceleration of the rocket is 40 ms¯².
You are hiking in a canyon and you notice an echo. You decide to let out a yell and notice it took 2 seconds before you heard the echo of your yell. How far away is the canyon wall that reflected your yell
Answer:
d = 343 m
Explanation:
Given that,
You notice that an echo took 2 seconds before you heard the echo of your yell.
We need to find that how far away is the canyon wall that reflected your yell. It means we need to find the distance.
The distance covered by an object is given by :
d = v × t
v is speed of sound in air, v = 343 m/s
The sound took 1 s to reach the wall and 1 s back to you.
It means that,
d = 343 × 1
d = 343 m
So, canyon will reflect your yell at a distance of 343 m
the indiscriminate use of drugs without the doctor's prescription is called
Answer:
drug misuse
Explanation:
this is because the person is using those drugs but indiscriminately meaning without correct judgement or in this context without a prescription therefore without following guidelines and legal standards eg self medication
but it is not abuse because it is being used for its intended purpose just without any guidelines
what is the function of eye lens of human eye
Answer:
the lens is located in the eye. by changing its shape, the lens changes the focal distance of the eye. in other words,it focuses the light rays that pass through it (and onto the retina) in order to create clear images of objects that are positioned
at various distances.
A model rocket blasts off from the ground, rising straight upward with a constant acceleration that has a magnitude of 76.8 m/s2 for 1.99 seconds, at which point its fuel abruptly runs out. Air resistance has no effect on its flight. What maximum altitude (above the ground) will the rocket reach?
Answer:
Explanation:
Take note that when the fuel of the rocket is consumed, the acceleration would be zero. However, at this phase the rocket would still be moving up until all the forces of gravity would dominate and change the direction of the rocket. Hence, there will be a need to calculate two distances, one from the ground until the point where the fuel is consumed and from that point to the point where the gravity would change the direction.
Given:
a = 76.8 m/s^2
t = 1.99 s
Solution:
d = vi (t) + 0.5 (a) (t^2)
d = (0) (1.99) + 0.5 (76.8) (1.99)^2
d = 0+38.4×3.9601
d = 152.068m
vf = vi + at
vf = 0 + 76.8 (1.99)
vf = 152.83 m/s (velocity when the fuel is consumed completely)
Then, we calculate the time it takes until it reaches the maximum height.
vf = vi + at
0 = 152.83+(-9.8) (t)
0 = 152.83 + (-9.8) (t)
-152.83 = -9.8t
t = 152.83/9.8 s
t = 15.59s
Then, the second distance
d= vi (t) + 0.5 (a) (t^2)
d = 152.83 (15.59) + 0.5 (-9.8) (15.59^2)
d = 2382.6197- 1190.93
d = 1191.68m
Then, we determine the maximum altitude:
d1 + d2 = 152.068 m + 1191.68m = 1343.748m
in how many ways can five basketball players be placed in three postitions?
Answer:
Well if they playing a game like that
What would happen to the
to the output force (F2)if the area
of the piston1 is made larger than that of piston2? plz help
Answer:
The output force would decrease since exerted force is inversely proportional to contact area.
Explanation:
F ∝ [tex]\frac{1}{A}[/tex] , Where 'F' is the exerted force and 'A' is the contact area.
1. The first stage of stellar formation is the assembly (or formation) of a_______ . 2. Stars form in clouds. 3. When a cloud collapses it into smaller and smaller pieces till those pieces reach stellar masses. 4. As the density increases, the fragments become_______ , which________ the temperature of the cloud and inhibits further fragmentation. 5. The _______then condense into rotating globs of gas that serve as stellar embryos. 6. During this stage the surface temperature_______ and luminosity_________ despite decreasing size. 7. The central temperature and pressure increase as the potential energy is converted to energy due to contraction.
Answer: The keywords to input into the blank spaces were omitted which are ; fragments, increases, decreases, opaque, raises, molecular, protostar, gravitational, thermal.
Please find the answers for the blank spaces in the explanation column
Explanation:
1. The first stage of stellar formation is the assembly (or formation) of a_PROTOSTAR______ . 2. Stars form in MOLECULAR clouds. 3. When a cloud collapses, it FRAGMENTS into smaller and smaller pieces till those pieces reach stellar masses. 4. As the density increases, the fragments become_OPAQUE______ , which___RAISES_____ the temperature of the cloud and inhibits further fragmentation. 5. The __FRAGMENTS_____then condense into rotating globs of gas that serve as stellar embryos. 6. During this stage the surface temperature INCREASES_______ and luminosity__INCREASES_______ despite decreasing size. 7. The central temperature and pressure increase as the GRAVITATIONAL potential energy is converted to energy due to THERMAL contraction.
The above describes the evolution or formation of stars which begins with the collapse of a molecular cloud by gravitational force, this collapse cause the molecular cloud to fragment with the release of heat --Gravitation potential energy causing an increased temperature and pressure of the newly condensing fragment with increasing luminosity known as a Protostar.
Which statement describes one feature of Rutherford's model of the atom?
O The atom is mostly empty space.
O The atom cannot be divided into smaller particles.
O Electron clouds are regions where electrons are likely to be found.
O The electrons are located within the positive material of the nucleus.
Answer:
The atom is mostly empty space.
Explanation:
Ernest Rutherford in the year 1911 developed a model of an atom. In his model the atom was shown that the atom is mostly empty spaces composed of negative electrons which orbits round a positive nucleus just like planets revolving round the sun.
Rutherford came up with this model by performing experiments in 1909 which involves the scattering of alpha particles on a thin gold foil. From this experiments he came up with his model.
Answer:
A!
Explanation:
The atom is mostly empty space.
