Answer:
True. it is restricted to the orbits of certain radii around the nucleus.
The position can change, but it must be in one of these orbits
Explanation:
In this exercise some affirmations are given and you must select which ones are correct, for this we review the Bohr atomic model that has the following postulates:
* the orbits are circular
* Only certain orbits are stable, stationary state
* the radiation emitted is the difference in energy between two stable orbits
* the size of the orbit is given by the quantization of the angular momentum
L = n [tex]\hbar[/tex]
When reviewing the different statements, the correct one is:
* it is restricted to the orbits of certain radii around the nucleus.
The position can change, but it must be in one of these orbits
When light passes through
an
object, it's called
A. transparent
B. absorption
C. refraction
D. reflection
What happens during
convection?
A. Heat is transferred from warmer
objects to cooler objects by direct contact.
B. Heat is transferred from warmer
objects to cooler objects by moving fluids.
C. Heat is transferred from warmer
objects to cooler objects through empty
space.
Answer:
B
Explanation:
5. Widely spaced isobars indicate
high winds
winds oriented north to south
light winds
variable winds
Answer:
Light Winds
Explanation:
Anyone know how to do this???
Answer:
World War 1 was caused by entangled alliances, nationalism, imperialism, and major
advancements in military technology. Does the Treaty of Versaille address those issues?
Explain your answer using facts. (5 points)
What flows in electricity?
A. Electrons
B. Protons
C. Neutrons
Answer:
A.
Explanation:
Electrons
Plaskett's binary system consists of two stars that revolve in a circular orbit about a center of mass midway between them. This statement implies that the masses of the two stars are equal (see figure below). Assume the orbital speed of each star is |v with arrow| = 230 km/s and the orbital period of each is 15.5 days. Find the mass M of each star. (For comparison, the mass of our Sun is 1.99 1030 kg.)
Answer:
[tex]1.554\times 10^{32}\ \text{kg}[/tex]
Explanation:
M = Mass of each star
T = Time period = 15.5 days
v = Orbital velocity = 230 km/s
G = Gravitational constant = [tex]6.674\times 10^{-11}\ \text{Nm}^2/\text{kg}^2[/tex]
Radius of orbit is given by
[tex]R=\dfrac{vT}{2\pi}[/tex]
We have the relation
[tex]\dfrac{Mv^2}{R}=\dfrac{GM^2}{(2R)^2}\\\Rightarrow M=\dfrac{4Rv^2}{G}\\\Rightarrow M=\dfrac{4\dfrac{vT}{2\pi}v^2}{G}\\\Rightarrow M=\dfrac{2v^3T}{\pi G}\\\Rightarrow M=\dfrac{2\times 230000^3\times 15.5\times 24\times 60\times 60}{\pi\times 6.674\times 10^{-11}}\\\Rightarrow M=1.554\times 10^{32}\ \text{kg}[/tex]
The mass of each star is [tex]1.554\times 10^{32}\ \text{kg}[/tex]
our battery has died and your friends push your vehicle so you can kick-start the engine. You and the vehicle have a combined mass of 1600 kg. If your friends do 6000 J of work and one-third of that is dissipated by friction, how fast is your vehicle traveling?
Answer:2.23 m/s
Explanation:
Given
Mass of person and vehicle is [tex]m=1600\ kg[/tex]
Total work done is [tex]W_t=6000\ J[/tex]
Friction consumes one-third of the energy
The remaining two-third is consumed to increase the kinetic energy
[tex]\Rightarrow \dfrac{2}{3}\times 6000=\dfrac{1}{2}\times 1600\times v^2\\\\\Rightarrow 4000\times 2=1600\times v^2\\\\\Rightarrow v=\sqrt{5}\ \approx 2.23\ m/s[/tex]
PLEASE HELP ASAPPPPP
Answer:
link
Explanation:
ww.comhwhelp
Why does the principal of lateral continuity work?
Answer:
Explanation:
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The principle of lateral continuity works, because they are continuous. Rocks that are alike but were separated by an erosional feature, can be originally continuous.
A 1200kg car is moving down a 30° hill. The driver applies the
brakes at a time that the car's speed is 12m/s. What constant force F
must result if the car is to stop after travelling 100m?
how can you Make different objects using blocks
Please help me!!!!!!!!!
Hi there! :)
[tex]\large\boxed{17.32 m/s}[/tex]
Use the following equation in solving for kinetic energy:
KE = 1/2mv² where:
KE = kinetic energy (J)
m = mass (kg)
v = velocity (m/s)
Plug in the given values:
12,000 = 40v²
Divide both sides by 40:
12,000 / 40 = v²
300 = v²
Take the square root of both sides:
√300 = v
v ≈ 17.32 m/s
A projectile was fired horizontally from a cliff 20m above the ground. If
the horizontal range of the projectile is 40m, calculate the initial velocity
of the projectile.
The initial velocity of the projectile is 19.8m/s
Explanation:
First, find time.
From our kinematics equations:
delta y = Vi•t + (1/2)at^2
rearrange,
t = sqrt[(2•delta y)/a]
t = sqrt[(2•20m)/9.8m/s^2]
t = 2.02s
Next, plug time into new kinematics equation to solve for the Vi in the x direction (horizontal)
delta x = Vi•t + (1/2)at^2
delta x = Vi•t
Rearrange:
Vi = delta x/t
Vi = 40m/2.02s
Vix = 19.8m/s
If you stand at a certain position and count the number of crests that pass by you over a certain amount of time, what are you most likely measuring? Period Speed Wavelength Frequency
Answer:
Wavelength
Explanation:
A vibration can cause a disturbance to travel through a medium, transporting energy without transporting matter. This is what a wave is.
A 4.0 kg circular disk slides in the x- direction on a frictionless horizontal surface with a speed of 5.0 m/s as shown in the adjacent Figure. It collides with an identical disk that is at rest before the collision. The collision is elastic. Disk 1 goes off at an 60 5.0 m/s angle of 60 with respect to the x-direction. Disk 2 g 30 goes off at an angle of 30 with respect to the x-direction. What best describes the speeds of the disks after the collision?
Solution :
Let [tex]$m_1=m_2=4$[/tex] kg
[tex]$u_1 = 5$[/tex] m/s
Let [tex]$v_1$[/tex] and [tex]$v_2$[/tex] are the speeds of the disk [tex]$m_1$[/tex] and [tex]$m_2$[/tex] after the collision.
So applying conservation of momentum in the y-direction,
[tex]$0=m_1 .v_1_y -m_2 .v_2_y $[/tex]
[tex]$v_1_y = v_2_y$[/tex]
[tex]$v_1 . \sin 60=v_2. \sin 30$[/tex]
[tex]$v_2 = v_1 \times \frac{\sin 60}{\sin 30}$[/tex]
[tex]$v_2=1.732 \times v_1$[/tex]
Therefore, the disk 2 have greater velocity and hence more kinetic energy after the collision.
Now applying conservation of momentum in the x-direction,
[tex]$m_1.u_1=m_1.v_1_x+m_2.v_2_x$[/tex]
[tex]$u_1=v_1_x+v_2_x$[/tex]
[tex]$5=v_1. \cos 60 + v_2 . \cos 30$[/tex]
[tex]$5=v_1. \cos 60 + 1.732 \times v_1 \cos 30$[/tex]
[tex]$v_1 = 2.50$[/tex] m/s
So, [tex]$v_2 = 1.732 \times 2.5$[/tex]
= 4.33 m/s
Therefore, speed of the disk 2 after collision is 4.33 m/s
Which statement explains how a chemical reaction depends on collisions
between molecules?
A. The reactants must collide in a certain way for the reaction to
occur.
B. The reaction will occur if the reactants collide in anyway.
C. The reaction will occur if the product collide in anyway.
D. The product must collide in a certain way for the reaction to occur
Answer: A. The reactants must collide in a certain way for the reaction to occur.
Explanation: I did the quiz :)
Answer:
A) The reactants must collide in a certain way for the reaction to
occur.
Explanation:
A.P.E.X
Which letter represents the
Pacific Plate?
Answer:
C
Explanation:
how solar system believed to have been formed?
Explanation:
Our solar system formed about 4.5 billion years ago from a dense cloud of interstellar gas and dust. The cloud collapsed, possibly due to the shockwave of a nearby exploding star, called a supernova. When this dust cloud collapsed, it formed a solar nebula—a spinning, swirling disk of material.
You measure the pressure of the four tires of your car each to be 35.0 pounds per square inch (psi). You then roll your car forward so that each tire is upon a sheet of paper. You outline the surface area of contact between each tire and the paper, which you later measure to be 32.0 square inches. What is the weight of your car
Answer:
Weight of car = 36034.88 lb.ft/s²
Explanation:
We are told the pressure of the four tires of your car is; P = 35.0 psi
Also, the surface area of contact is; A = 32 in²
Thus;
Weight of tires = Pressure × Area
W_tires = 35 × 32
Weight_tires = 1120lb.
To get the weight of the car, we will multiply the tire weight by acceleration due to gravity.
It's value in ft/s² is g = 32.174 ft/s²
Thus;
Weight of car = 1120 × 32.174
Weight of car = 36034.88 lb.ft/s²
Two type of microscopes used to view cells are optical and__ microscopes
options:
laser
Electron
Which of the following would fill in the table where "A" is?
Symbol
Element Name
Atomic Number Mass Number
#nº
#p
#e
Net Charge
Si
Silicon
A (#13)
27
B (#14 14 C (#15)
0
D (#16)
Potassium
18
39
19
18 E (#17)
A. 3
B. 13
C. 14
D. 28
Answer:
.lyrjhg.
Explanation:
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A train crosses 650m long bridge and 800m long platform in 20sec and 30 sec respectively. what is speed of train?
Answer:
29 m/s
Explanation:
Assuming the bridge and the platform are back to back,
Average speed = Total Distance/ Total Time
Avg S = 1,450m/50s = 29 m/s
The average speed of the train is 29 m/s
The average speed is defined as the ratio of total distance traveled and the total time taken.
Total distance = 650 m + 800 m = 1450 m
Total Time = 20 s + 30 s = 50 s
Average speed = total distance/total time
= 1450 / 50
= 29 m/s is the average speed of the train.
Learn more about average speed:
https://brainly.com/question/11753352
Please help I have no idea how to do this
Answer:
So an object with mass is attracted to another object with mass, and the gravitational force is directly proportional to the masses of the two objects, and inversely proportional to the square of the distance between the two objects.
If distance were to increase, than the gravitational force would decrease. If mass were to increase, so would the gravitational force.
Explanation:
Explanation:
Lets take gravitational force(F) and mass(m) and distance (r)
now for a body in contact with the surface of the earth, its mass is also considered(m‘),now the mass of the earth(m") is also considered,then the distance from the body to the center of the earth (r).ie it r because its practically the radius of the earth. is also considered
So by using dimensional analysis ....
we get F a m'•m"/r² ,where a is proportional to.
now since F is directly increaseproportional to m ie. F a m, then an increase in mass of the body increases it's gravitational force(and clearly that makes sense because the bigger you are the stronger you get pulled to the ground)
then we also see that F is inversely proportional to r ie.F a 1/r ,then an increase in the distance between the ground an the object decrease it's gravitational force ( meaning as any object on earth keeps on moving away from the ground the gravitational force between the object and the center of the earth is weak, when it reaches space then the force becomes virtually negligible!)
So to answer the second question, we clearly see that doubling the mass of the body increases the gravitational force between it and the earth
and doubling the distance on the other hand will decrease the attraction between the body and the earth
So a body forcefully projected into the air fights against gravity but its easier as it keeps on getting higher, If it has a greater mass like that of a trail or truck , it will not even probably stay in the air for long , unless its projected with a very high velocity
I hope this helps, and you can ask me any question concerning this via the comments platform.
How much power is required to carry a 35N
package a vertical distance of 18 m if the work on
the package is accomplished in 30 s?
Explanation:
force=35
distance=18
time=30
power=f×d/t
p=35×18/30
p=21
What is the difference between the isotopes Hydrogen-2 and Hydrogen-3?
Answer:
They each have one single proton (Z = 1), but differ in the number of their neutrons. Hydrogen has no neutron, deuterium has one, and tritium has two neutrons. The isotopes of hydrogen have, respectively, mass numbers of one, two, and three.
Explanation:
An airplane in level flight is acted on by four basic forces. Drag is air resistance, lift is the upward force provided by the wings, thrust is the force provided
by the airplane's engines, and weight is the downward force of gravity acting on the airplane.
Lift
Thrust
Drag
Weight
In level flight at constant speed, which pair of forces must be equal?
A concept is the general idea of objects, events, animals, or people based on common features, traits, or characteristics.
Please select the best answer from the choices provided
ОТ
OF
Answer:
True.
Explanation:
Science can be defined as a branch of intellectual and practical study which systematically observe a body of fact in relation to the structure and behavior of non-living and living organisms (animals, plants and humans) in the natural world through experiments.
A concept is the general idea of objects, events, animals, or people based on common features, traits, or characteristics. It is typically considered to be an abstract idea or general notion that is formed by the mental combination of all the characteristics pertaining to the idea.
Generally, concepts are used in the problem-solving process as it serves as a benchmark or yardstick for the solutions. Some examples of concepts in physics are conservation of energy and momentum, electromagnetic waves, heat, motion and forces, electricity and magnetism, gravity, etc.
Answer:
TRUE
Explanation:
edge
1. Describe the components of the reflex arc
difference between ferromagnetic and antiferromagnetic ..
Answer:
ferromagnetic materials are materials that are highly attracted to magnets while antiferromagnetic materials are materials that are not attracted by magnets
Consider two insulating balls with evenly distributed equal and opposite charges on their surfaces, held with a certain distance between the centers of the balls. Construct a problem in which you calculate the electric field (magnitude and direction) due to the balls at various points along a line running through the centers of the balls and extending to infinity on either side. Choose interesting points and comment on the meaning of the field at those points. For example, at what points might the field be just that due to one ball and where does the field become negligibly small? Among the
things to be considered are the magnitudes of the charges and the distance between the centers of the balls. Your instructor may wish for you to consider the electric field off axis or for a more complex array of charges, such as those in a water molecule.
Answer:
interest point:
1) Point on the left side
2) Point within the radius r₁ of the first sphere
3) Point between the two spheres
4) point within the radius r₂ of the second sphere
5) Right side point
Explanation:
In this case, the total electric field is the vector sum of the electric fields of each sphere, to simplify the calculation on the line that joins the two spheres
We will call the sphere on the left 1 and it has a positive charge Q with radius r1, the sphere on the right is called 2 with charge -Q with radius r2. The total field is
E_ {total} = E₁ + E₂
E_{ total} = [tex]k \frac{Q}{x_1^2} + k \frac{Q}{x_2^2}[/tex]
the bold indicate vectors, where x₁ and x₂ are the distances from the center of each sphere. If the distance that separates the two spheres is d
x₂ = x₁ -d
E total = [tex]k \frac{Q}{x_1^2} - k \frac{Q}{(x_1 - d)^2}[/tex]
Let's analyze the field for various points of interest.
1) Point on the left side
in this case
E_ {total} = [tex]k Q \ ( \frac{1}{x_1^2} - \frac{1}{(x_1 +d)2} )[/tex]
E_ {total} = [tex]k \frac{Q}{x_1^2}[/tex] [tex]( 1 - \frac{1}{(1 + \frac{d}{x_1} )^2 } )[/tex]
We have several interesting possibilities:
* We can see that as the point is further away the field is more similar to the field created by two point charges
* there is a point where the field is zero
E_ {total} = 0
x₁² = (x₁ + d)²
2) Point within the radius r₁ of the first sphere.
In this case, according to Gauus' law, the charge is on the surface of the sphere at the point, there is no charge inside so this sphere has no electric field on its inner point
E_ {total} = [tex]-k \frac{Q}{x_2^2} = -k \frac{Q}{((d-x_1)^2}[/tex]
this expression holds for the points located at
-r₁ <x₁ <r₁
3) Point between the two spheres
E_ {total} = [tex]k \frac{Q}{x_1^2} + k \frac{Q}{(d+x_1)^2}[/tex]
This champ is always different from zero
4) point within the radius r₂ of the second sphere, as there is no charge inside, only the first sphere contributes
E_ {total} = [tex]+ k \frac{Q}{(d-x_1)^2}[/tex]+ k Q / (d-x1) 2
point range
-r₂ <x₂ <r₂
5) Right side point
E_ {total} = [tex]k \frac{Q}{(x_2-d)^2} - k \frac{Q}{x_2^2}[/tex]
E_ {total} = [tex]- k \frac{Q}{x_2^2} ( 1- \frac{1}{(1- \frac{d}{x_2})^2 } )[/tex]- k Q / x22 (1- 1 / (x1 + d) 2)
we have two possibilities
* as the distance increases the field looks more like the field created by two point charges
* there is a point where the field is zero