three different states
Water is known to exist in three different states; as a solid, liquid or gas. Clouds, snow, and rain are all made of up of some form of water. A cloud is comprised of tiny water droplets and/or ice crystals, a snowflake is an aggregate of many ice crystals, and rain is just liquid water.
Answer:
liquid water gas water vapour solid iceCathode ray tubes (CRTs) used in old-style televisions have been replaced by modern LCD and LED screens. Part of the CRT included a set of accelerating plates separated by a distance of about 1.40 cm. If the potential difference across the plates was 23.0 kV, find the magnitude of the electric field (in V/m) in the region between the plates.
Answer:
E = 1.64 x 10⁶ V/m
Explanation:
The electric field in the region between the plates can be given by the following formula:
[tex]E = \frac{\Delta V}{d}[/tex]
where,
E = Electric Field = ?
ΔV = Poetential Difference across the plates = 23 KV = 23000 V
d = distance between plates = 1.4 cm = 0.014 m
Therefore, using these values in the equation, we get:
[tex]E = \frac{23000\ V}{0.014\ m}[/tex]
E = 1.64 x 10⁶ V/m
Help me please I have other ones like this too on my page please help!
Suppose a rocket in space is accelerating at 1.5 m/s2. If, at a later time, the rocket quadruples its thrust (i.e., net propelling force), what is the new acceleration?
The acceleration of a moving object is equal to
Answer:
Acceleration = Δv/Δt or change in velocity over change in time
Explanation:
I need help please .
Answer:
option 5
Explanation:
because all u do is have to add them up
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
In medieval warfare, one of the greatest technological advancement was the trebuchet. The trebuchet was used to sling rocks into castles. You are asked to study the motion of such a projectile for a group of local enthusiast planning a medieval war reenactment. Unfortunately an actual trebuchet had not been built yet, so you decide to first look at the motion of a thrown ball as a model of rocks thrown by a trebuchet. Specifically, you are interested in how the horizontal and the vertical components of the velocity for a thrown object change with time. 1. Make a large rough sketch of the trajectory of the ball after it has been thrown. Draw the ball in at least five different positions; two when the ball is going up, two when it is going down, and one at its maximum height. Label the horizontal and vertical axes of your coordinate system.
2. On the sketch, draw and label the expected acceleration vectors of the ball (relative sizes and directions) for the five different positions. Decompose each acceleration vector into its vertical and horizontal components.
3. On the sketch, draw and label the velocity vectors of the object at the same positions you chose to draw your acceleration vectors. Decomposes each velocity vector into its vertical and horizontal components. Check to see that the changes in the velocity vector are consistent with the acceleration vectors.
4. Looking at the sketch, how does someone expect the ball's horizontal acceleration to change with time? Could you give a possible equation giving the ball's horizontal acceleration as a function of time? Graph this equation. If there are constants in your equation, what kinematic quantities do they represent? How would someone determine these constants from the graph?
5. Looking at the sketch, how does someone expect the ball's horizontal velocity to change with time? Is it consistent with the statements about the ball's acceleration from the previous question? Could you give a possible equation for the ball's horizontal velocity as a function of time? Graph this equation. If there are constants in the equation, what kinematic quantities do they represent? How would someone determine these constants from the graph?
6. Could you give a possible equation for the ball's horizontal position as a function of time? Graph this equation. If there are constants in the equation, what kinematic quantities do they represent? How would someone determine these constants from the graph? Are any of these constants related to the equations for horizontal velocity or acceleration?
7. Repeat questions 4-6 for the vertical component of the acceleration, velocity, and position. How are the constants for the acceleration, velocity and position equations related?
Answer:
2) a_y= -g 3) vₓ=constant v_y = v_{oy} - g t, 4) vₓ = v₀ₓ - ax t
5) changes the horizontal speed, should change range
7) changes the vertical speed change the maximum height
Explanation:
1) After reading your long writing, we are going to solve the exercise, in the attachment you can see the different vectors.
2) The acceleration vectors are vertical and directed downwards due to the attraction of the Earth (gravity force) this force is constant, on the x axis there is no acceleration
3) the velocity vectors on the x-axis are constant because there are no relationships and the y-axis changes value according to the expression
v_y = v_{oy} - gt
at the point of maximum height, vy = 0 is equal to the maximum height
4) For someone to change the horizontal acceleration we must assume a friction with the air, in this case they relate it would be in the opposite direction to the horizontal speed
In the graph it would be directed to the left, therefore the velocity would be
vₓ = v₀ₓ - ax t
5 and 6) If someone changes the horizontal speed, they should change the range of the shot for greater horizontal speed, the rock goes further.
the equations of motion are
x = v₀ₓ t
y = v_{oy} t - ½ g t²
7) If someone changes the vertical speed change the maximum height, but not the scope of the shot, for higher speed higher maximum height,
the equations of motion are the same.
Experiments carried out on the television show Mythbusters determined that a magnetic field of 1000 gauss is needed to corrupt the information on a credit card's magnetic strip. (They also busted the myth that a credit card can be demagnetized by an electric eel or an eelskin wallet.) Suppose a long, straight wire carries a current of 5.0 A .
How close can a credit card be held to this wire without damaging its magnetic strip?
Answer:
his distance is too small (r = 0.01 mm), therefore the cut can be at any distance
Explanation:
For this exercise let's use the ampere law.
Let's use a cylinder as the circulating surface
∫ B. ds = μ₀ I
in this case the field is circular and ds is circular therefore the angle between them is zero and cos 0 = 1
B 2π r = μ₀ I
r = [tex]\frac{\mu_o I}{2\pi B}[/tex]
The field needed to demagnetize the card is B = 1000 gauss = 0.1 T
r = [tex]\frac{4\pi 10^{-7} 5.0 }{2\pi \ 0.1}[/tex]
r = 2 10⁻⁷ 5.0/0.1
r = 1 10⁻⁵ m
this distance is too small (r = 0.01 mm), therefore the cut can be at any distance
A 150 Kg objects is lifted heights of 12 meters . What is the gravitational poteentain energy of the object.
Explanation:
u7ruey737€*hr7j37j27jw7uw7bwydbe7887yeyhduheyheyy755÷÷+5÷÷8€737€=67577
a potted plant falls from a window sill and is gaining speed. which one of the following statements is true of the plant?
a) its kinetic energy is constant
b) its kinetic energy is increasing
c) its kinetic energy is decreasing
the equation P^xV^yT^z= constant is Boyle law for what is the values of x,y,z
Answer:
x = 1, y = 1 and z = 0
Explanation:
Given equation;
[tex]P^x V^y T^z = constant[/tex]
Boyle's law states that at constant temperature, the volume of a fixed mass of gas is inversely proportional to its pressure.
Mathematically the law is written as;
[tex]PV = constant[/tex]
From the given equation, the values of x, y and z that will match this law is calculated as follows;
[tex]P^1 V^1 T^0 = constant\\\\P^1 = P\\\\V^1 = V\\\\T^0 = 1\\\\P \times V \times 1 = PV = constant\\\\Thus, x = 1, \ y = 1 \ \ and \ z = 0[/tex]
Check out this app! It's millions of students helping each other get through their schoolwork. https://brainly.app.link/qpzV02MawO
Answer:
OK we appreciate your concern.
why is the meteor shower is best observed after midnight?
Answer:
At dawn your location on earth is pointed straight in the direction of the Earth's travel in its orbit. Between midnight and dawn you are moving head-on through the location of the meteors in space, which means that you will, on average, observe more of them.
- public.nrao.edu
Explanation:
hope this helps
2.3 The motion of an object is accelerated
when its speed:
a
decreases
b remains constant
increases.
Answer:
The motion of an object is accelerated when its speed increases.
Which of the following best
describes amplitude?
A. Amplitude is how fast a wave travels.
B. Amplitude is how far a wave moves from its resting
position
C. Amplitude is the resting position of a wave.
Answer:
its b
Explanation:
Amplitude, in physics, the maximum displacement or distance moved by a point on a vibrating body or wave measured from its equilibrium position. It is equal to one-half the length of the vibration path.
how is a trench and a tsunami related? 6-8 sentences
Answer: A tsunami is a very long-wavelength wave of water that is generated by sudden displacement of the seafloor or disruption of any body of standing water. Tsunami are sometimes called "seismic sea waves", although they can be generated by mechanisms other than earthquakes. Tsunami have also been called "tidal waves", but this term should not be used because they are not in any way related to the tides of the Earth. Because tsunami occur suddenly, often without warning, they are extremely dangerous to coastal communities. Ocean trenches are steep depressions in the deepest parts of the ocean [where old ocean crust from one tectonic plate is pushed beneath another plate, raising mountains, causing earthquakes, and forming volcanoes on the seafloor and on land.
Explanation:
1. If airbags reduce the impact force from an accident why has there been questions over their safety?
2. Are airbags the safest option to prevent serious injury or death from a car accident?
Answer:Air bags can leave you in even more injury, From the impact they give
You could end up with a broken nose,arm
concussion
What causes coastal erosion
La erosión costera es la pérdida o desplazamiento de tierra, o la remoción a largo plazo de sedimentos y rocas a lo largo de la costa debido a la acción de olas, corrientes, mareas, agua impulsada por el viento, hielo transportado por el agua u otros impactos de marejadas ciclónicas.
is this correct and why
Answer:
correct ooooooooooooo
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)
Riders in an amusement park ride shaped like a Viking ship hung from a large pivot are rotated back and forth like a rigid pendulum. At each end of the swing the ship hangs motionless for a moment before the ship swings down under the influence of gravity. Assume that this motionless point occurs when the bar connecting the pivot point and the ship is horizontal.
Required:
a. Assuming negligible friction, find the speed of the riders at the bottom of its arc, given the system's center of mass travels in an arc having a radius of 14.0 m and the riders are near the center of mass.
b. What is the centripetal acceleration at the bottom of the arc?
c. Draw a free body diagram of the forces acting on a rider at the bottom of the arc.
d. Find the force exerted by the ride on a 60.0 kg rider and compare it to her weight.
e. Discuss whether the answer seems reasonable.
Answer:
a) v = 16.57 m / s, b) a = 19.6 m / s², d) N = 1.76 10³ N, N / W = 3
Explanation:
This exercise looks interesting, but I think you have some problem with the writing, the questions seem a bit disconnected from the initial text.
Let's answer the questions.
a) For this part we can use energy considerations.
Starting point. The upper part of the trajectory indicates that the arm is horizontally
Em₀ = U = m g h
in this case h = r
Final point. For lower of the trajectory
Em_f = K = ½ m v²
as they indicate that there is no friction
Em₀ = em_f
mgh = ½ m v²
v = [tex]\sqrt{2gh}[/tex]
let's calculate
v = [tex]\sqrt{2 \ 9.8 \ 14.0}[/tex]
v = 16.57 m / s
b) the centripetal acceleration has the formula
a = v² / r
a = 16.57² / 14.0
a = 19.6 m / s²
c) see attached where the diagram is
where N is the normal and w the weight
d) let's use Newton's second law
N-W = m a
N - mg = m ar
N = m (g + a)
let's calculate
N = 60.0 (9.8 + 19.6)
N = 1.76 10³ N
the relationship with weight is
N / W = 1.76 10³/( 60 9.8)
N / W = 3
normal is three times greater than body weight
e) the answer is reasonable since by Newton's first law the body must continue in a straight line, therefore to change its trajectory a force must be applied to deflect it
1. A person kicks a rock off a cliff horizontally with a speed of 20 m/s. It takes 7.0 seconds to hit the
ground, find:
a. height of the cliff
b. final vertical velocity
C. range
D.speed and angle of impact
This problem involved half projectile.
initial velocity, vo = 20 m/s
time of flight, t = 7 s
(a) Simply use the formula to get the height, h:
h = vo*t - (1/2)gt^2
(b) To get the final vertical velocity or terminal velocity (vf), use the formula:
(vf)^2 - (vo)^2 = 2gh
(c) Use the formula find the horizontal distance traveled, R:
R = vo * cos(θ) * t
But since the angle involved with respect to horizontal is zero, and cos(0) = 1, we have
R = vo * t
Hope this helps~ `u`
Jai
You are comparing the beam waste for two different situations with the goal of using the smallest beam waste possible. A Nd-YAG laser system emits light at 532 nm and the beam is 8 mm in diameter. You also have a Ti-sapphire laser that emits at 855 nm and has a beam diameter of 6 mm. Compare the beam waist for both laser systems using a focusing lens with a focal length of 10 mm. Assume the light fills the lenses in each case
Answer:
comparing the beam waist for both lasers ( ratio of the beam waists )
4.536 μm / 2.117 μm = 2.14
Explanation:
Nd-YAG laser system : emits at 532 nm , beam diameter = 8 mm
Ti-sapphire laser system : emits at 855 nm , Beam diameter = 6mm
Comparing the beam waist for both lase systems using a focusing lens
Focal length = 10 mm
assumption : light fills lenses in each laser system
Beam waist radius ( W ) = [tex](\frac{2\beta }{\pi } )(\frac{F}{D} )[/tex]
β = wavelength , D = diameter illuminated , F = focal length
For
Nd-YAG laser system
β = 532 mm , D = 8 mm
hence ( Wn ) = [tex](\frac{2\beta }{\pi } )(\frac{F}{D} )[/tex] = ( 2*532 / π ) ( 10 / 8 ) = 2.117 μm
For
Ti-sapphire laser
β = 855 nm , D = 6 mm
hence ( Wt ) [tex](\frac{2\beta }{\pi } )(\frac{F}{D} )[/tex] = ( 2* 855 ) / π ) ( 50 / 6 ) = 4.536 μm
comparing the beam waist for both lasers ( ratio of the beam waists )
4.536 μm / 2.117 μm = 2.14
The certain forest moon travels in an approximately circular orbit of radius
14,441,566 m with a period of 6 days 10 hr, around its gas giant exoplanet host. Calculate the mass of the exoplanet from this
information. (Units: kilograms)
Answer:
Mass of Exoplanet = 0.58 kg
Explanation:
First, we will calculate the speed of the forest moon:
[tex]speed = v = \frac{Circumference}{time}\\[/tex]
circumference = 2πr = 2π(14441566 m) = 90739035.3 m
time = 6 days 10 hr = (6 days)(24 h/1 day)(3600 s/1 h) + (10 h)(3600 s/1 h)
time = 554400 s
Therefore,
[tex]v = \frac{90739035.3\ m}{554400\ s}\\\\v = 163.67\ m/s[/tex]
We know that the centripetal force on forest moon will be equal to the gravitational force given by Newton's Gravitational Law, as follows:
[tex]Centripetal\ Force = Gravitational\ Force\\\frac{m_{moon}v^2}{r} = \frac{Gm_{moon}m_{exoplanet}}{r^2}\\\\m_{exoplanet} = \frac{v^2r}{G}\\\\m_{exoplanet} = \frac{(163.67\ m/s)^2(14441566)}{6.67\ x\ 10^{-11}\ N.m^2/kg^2}[/tex]
Mass of Exoplanet = 0.58 kg
...........................
Answer: kooi kooi
how r u and thanks for the free points :)
an ice skater is moving across a flat and level skating rink and is speeding up. which one of the following statement is true of the ice skater
a) its potential energy is constant
b) its potential energy is increasing
c) its potential energy is decreasing
Answer:
A
Explanation:
The angular momentum of a system of particles around a point in a fixed inertial reference frame is conserved if there is no net external torque around that point:
d
→
L
d
t
=
0
or
→
L
=
→
l
1
+
→
l
2
+
⋯
+
→
l
N
=
constant
.
Note that the total angular momentum
→
L
is conserved. Any of the individual angular momenta can change as long as their sum remains constant. This law is analogous to linear momentum being conserved when the external force on a system is zero.
A ball is thrown so that its speed increases by 20 m/s in 10 seconds. What is the ball’s acceleration?
Answer: a= 2 m/s²
Explanation: acceleration = change of speed/ time = 20 m/s / 10 s
Please help me !!!!!!!
Answer: I believe that it is 35 Joules.
Explanation:
100 - 65 = 35
:)
Need this homework for today so help 50 point
Need help with questions 2–3 ASAP pls ease help
Answer:
A and D
Explanation:
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