Answer:
10m
Explanation:
assuming the ball was at 0 m/s before, after 5 seconds the ball would be traveling at 10m
Consider the following scenario. A car for which friction is not negligible accelerates from rest down a hill, running out of gasoline after a short distance (see below). The driver lets the car coast farther down the hill, then up and over a small crest. He then coasts down that hill into a gas station, where he brakes to a stop and fills the tank with gasoline. Identify the forms of energy the car has, and how they are changed and transferred in this series of events.
Answer:
1) U-> K +W
2) K -> W
Explanation:
In this exercise, care must be taken as they indicate that the friction force (rubbing) is not negligible.
1 part at the top of the hill the car has gravitational potential energy, which is transformed in a part into kinetic energy and another part into heat by the work of the friction force that opposes the movement.
2 part when the other hill rises it loses kinetic energy that is transformed into gravitational potential energy and part in heat due to the work of the friction force on this hill.
3rd part in the last descent all the gravitational potential energy is transformed into kinetic energy and the application of the brakes is transformed into heat due to the negative lock of the friction force and filled with gasoline that has chemical energy that can be transformed in the engine.
On the computer, we can only approximate the true Fourier transform because of the need for both time and frequency sampling. Because of the limited frequency resolution, the frequency of a cosine may not line up exactly at a frequency sample, but vou'll still get a peak around that frequency. Assuming that vou sample the signal with fs-8000 Hz, and take its Fourier transform using an FFT with 4096 points, what is the value (in Hz) of the FFT frequency bins closest to the two frequencies of the sinusoids (941 and 1336 Hz)
Answer:
hello your question is incomplete below is the missing part
Consider the sum of sinusoids de (t) = sin(24(941)t) + sin(27(1336)t)
answer : Value of DFT frequency = 684 Hz
Explanation:
Given data:
sample signal ( Fs ) = 8000 Hz
Fourier transform using an FFT with 4096 points
Determine the value of DFT frequency bins closest to the two frequencies of the sinusoids ( 941 and 1336 Hz ) ( in HZ )
fs = 8000 Hz
4096 parts are divided with a gap of Fs/N
attached below is the detailed solution
Mr. Voytko wants to know how high in meters he can lift an 0.3 kg apple with 7.35 joules?
Answer:
the height above the ground through Mr. Voytko lifted the apple is 2.5 m.
Explanation:
Given;
energy of Mr. Voytko, E = 7.35 J
mass of the apple, m = 0.3 kg
Apply the principle of conservation of energy.
Energy of Mr. Voytko = Potential energy of the apple due to its height above the ground.
E = mgh
where;
h is the height above the ground through Mr. Voytko lifted the apple.
g is acceleration due to gravity = 9.8 m/s²
h = E / (mg)
h = 7.35 / (0.3 x 9.8)
h = 2.5 m
Therefore, the height above the ground through Mr. Voytko lifted the apple is 2.5 m.
A cylinder containing the air comprises the systemm. Cycle is completed as follows : (i) 82000 N-m of work is done by the piston on the air during compression stroke and 45 kJ of heat are rejected to the surroundings. (ii) During expansion stroke 100000 N-m of work is done by the air on the piston. Calculate the quantity of heat added to the system؟?
Answer & Explanation:
1 N-m = 1 Joule
So 82 kJ of energy put into the system during (i).
45 kJ of heat leaves the system, so 82 kJ - 45 kJ = 37 kJ is remaining.
(ii) requires 100 kJ of energy but only 37 kJ is available, so 100 kJ - 37 kJ = 63 kJ of heat energy must be added to the system.
The heat given would be equal to the heat emitted from the system and by providing some external source of energy the volume or temperature of the system may increase.
The amount of heat added to the system is 63kJ.
The energy can be estimated as:
Given,
Work done by piston = 82000 NmHeat rejected in surrounding = 45 kJWork done during expansion stroke = 100000 NmQuantity of added heat = ?During compression stroke:
Work done by the piston [tex]\rm (W_{1-2})[/tex] = - 82000Nm or - 82kJHeat rejected to the system [tex]\rm (Q_{1-2})[/tex] = - 45kJWe know that,
[tex]\rm Q_{1-2} = (U_{2} - U_{1}) + W[/tex]
Therefore,
[tex]\begin{aligned}-45 &= \rm (U_{2} - \rm U_{1}) + (-82)\\\\\rm (U_{2}-U_{1}) &= 37\rm (U_{2} - U_{1}) = 37\; kJ \end{aligned}[/tex] (equation 1 )
During Expansion system:
Work done by the piston [tex]\rm (W_{2-1})[/tex] = 100000 Nm or 100 kJ
Now putting values in the equation:
[tex]\begin{aligned}\rm Q_{2-1} &= \rm U_{1} - U_{2} + W\\\\&=\rm (U_{1} - U_{2}) + W\end{aligned}[/tex]
Substituting value from equation 1:
[tex]\begin{aligned}\rm Q_{2-1} &= - 37+100\rm \;kJ\\&= 63\rm \; kJ\end{aligned}[/tex]
Therefore, 63kJ of energy is added to the system.
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PLEASE HELP ASAP WILL GIVE BRAINLIEST TO WHOEVER ANSWERS FIRST!!!
What 2 forces would be responsible for exerting forces to the left on a cyclist that is already in motion moving to the right?
(Select 2 of the choices below)
A. gravity
B. normal force
C. air resistance
D. friction
Answer:
Friction and Air resistence
Explanation:
i already passed this grade years ago...
The forces responsible for exerting forces to the left on a cyclist are -
Air resistanceFrictionWhat is air resistance?Air resistance describes the forces that are in opposition to the relative motion of an object as it passes through the air. We can write air resistance as -
[tex]$F_{D}=\frac{1}{2} \rho v^{2} C_{D} A[/tex]
where -
F{D} = drag
{ρ} = density of fluid
{v} = speed of the object relative to the fluid
C{D} = drag coefficient
{A} = cross sectional area
Given is to find what 2 forces would be responsible for exerting forces to the left on a cyclist that is already in motion moving to the right.
The forces responsible for exerting forces to the left on a cyclist are -
Air resistanceFrictionTherefore, the forces responsible for exerting forces to the left on a cyclist are -
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A squirrel drops an acorn onto the head of an unsuspecting dog. The acorn falls 20 m before it lands on the dog. Ignore air resistance. How many seconds did the acorn fall?
Answer:
2 seconds
Explanation:
From the question,
Applying the equation of motion for a body falling under gravity
s = ut+1/2gt²................................. Equation 1
Where, s = hieght of fall, u = initial velocity, t = time, g = acceleration due to gravity.
Given: s = 20 m, u = 0 m/s, t = ?, g = 10 m/s²
Substitute these values into equation 1
10 = 0(t²)+1/2(10×t)
10 = 5t
Solve for t
5t/5 = 10/5
t = 2 seconds
man is walking due east at the rate of of 4kmph and the rain is falling 30° east of vertical with a velocity of 6kmph the velocity of rain relative to the man will be?
Answer:
No answer
Explanation:
no explanation
231 91 Pa has__neutrons
Answer:
140
Explanation:
a. Using the ideas of electric field and force, explain what would happen to an electron if released from rest at r=2.0m?
b. Would the electron released from rest move to a region of higher electrical potential or lower electrical potential?
c. Would the electron released from rest move such that the system would have higher potential energy or lower potential energy?
if a body of mass m is placed on earth ,what is the amount of potential energy possessed by it (g:-9.8m/s
Answer:
mgh
Explanation:
Assume the height of the body is 1.8m.
The gravity?of the body is G=mg
the height of the gravity center is about 0.9m
E=mgh
=m*9.8m/s*0.9m
= 8.82mJ
if the
gravitational force f=GM1M2/r2 derive the dimention of G
Answer:
M-1 L3 T-2
Explanation:
Where,
M = Mass
L = Length
T = Time
Derivation
From Newton’s law of gravitation,
Force (F) = [GM1M2] × r-2
Gravitational Constant (G) = F × r2 × [Mm]-1 . . . . (1)
Since, Force (F) = Mass × Acceleration = M × [LT-2]
∴ The dimensional formula of force = M1 L1 T-2 . . . . (2)
On substituting equation (2) in equation (1) we get,
Gravitational Constant (G) = F × r2 × [Mm]-1
Or, G = [M1 L1 T-2] × [L]2 × [M]-2 = [M-1 L3 T-2].
Therefore, the gravitational constant is dimensionally represented as M-1 L3 T-2.
Steve is planning his annual Spring Break road trip. He pulls out his map and draws out his route to visit the five locations that he has planned for this year. They go in a counterclockwise loop and he ends up at home, where he started, just in time to start classes again. Whenever he is on the road he travels a constant 60 miles/hour. When Steve adds up the total distance traveled, as measured by his odometer, and divides it by the time that his trip took, he has measured what quantity?
a. His average velocity.
b. His average speed.
c. His instantaneous velocity.
d. His instantaneous speed.
Steve’s average velocity for the whole trip is:______
a. greater than 60 miles/hour.
b. equal to 60 miles/hour.
c. less than 60 mile/hour, but greater than zero.
d. exactly zero.
Answer:
Part 1
Steve is measuring his average speed
Part 2
Average velocity is equal to 60 miles per hour
Explanation:
Part 1
Average velocity is equal to total distance travelled divided by total time taken. It also takes into consideration the change of direction through out the journey.
Hence, Steve is measuring his average speed
Option A is correct
Part 2
Average velocity is equal to 60 miles per hour only because velocity is a vector quantity
Option B is correct
4. Draw conclusions: What is the minimum energy required to break the egg?
.
Answer:
0.25 J
Explanation:
The strength of the egg shell, the size of the egg, and the force used to break it are just some of the variables that affect how much energy is needed to crack an egg. When an object hits the egg with an impact energy of 12–26 mJ, cracks occur.
What is energy?The capacity of a system or object to do work is called energy, which is a fundamental term in physics. Kinetic energy, potential energy, heat energy, electromagnetic energy, and nuclear energy are just a few of the different forms of energy.
While potential energy is the energy possessed by an object as a result of its position or position, kinetic energy is the energy of motion. While electromagnetic energy is energy carried by electromagnetic waves like light, thermal energy is energy related to the temperature of a substance. The energy stored in the nucleus of an atom is called nuclear energy.
Therefore, when an object hits the egg with an impact energy of 12–26 mJ, cracks occur.
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If a car is moving to the left with constant velocity, one can conclude that :
a. there must be no forces exerted on the car.
b. the net force exerted on the car is directed to the left.
c. the net force exerted on the car is zero.
d. there is exactly one force exerted on the car.
Answer:
b. the net force exerted on the car is directed to the left.
Explanation:
Applying Newton's second law of motion, the car will move in the direction of the applied force. If the applied forces are on different directions, the car will move in the direction of the greater force (net force).
Therefore, if a car is moving to the left with constant velocity, one can conclude that the net force exerted on the car is directed to the left.
The correct option is "b. the net force exerted on the car is directed to the left"
1. If a wave has a wavelength of 5.5m and a frequency of 45hz, what is its speed?
Answer:
By using the most simple velocity equation, velocity = distance / time, meaning the speed would be 247.5 meters per second.
Write your paragraph explaining why astronomers' minds are more important than the instruments they use? It has to be a paragraph and please make it quick I’m being timed!!
Answer:
Ill just give you some quick pointers
Explanation:
-in order to create the instruments, they need to use their mind
-their minds and creativity are the base and foundation of anything really
-the brain is the most complex part of your body so much more complicated than any intrument
A 69.0 kg ice skater moving to the right with a velocity of 2.61 m/s throws a 0.22 kg snowball to the right with a velocity of 25.2 m/s relative to the ground. (a) What is the velocity of the ice skater after throwing the snowball
Answer:
0.08m/s
Explanation:
Given data
M1= 69kg
v1= 2.61m/s
M2= 0.22kg
v2= 25.2m/s
Before snowball is thrown:
Total mass of skater + snowball = 69+ 0.22 = 69.22kg
Total Momentum of skater + snowball = mv = 69.22 x 2.61 = 180.7 kgm/s
After snowball is thrown:
Let's call the velocity of the skater V.
Total momentum = momentum of skater + momentum of snowball
=69.22V + (5.544)
= 69.22V + 5.544
So:
180.7 = 69.22V+5.544
180.7- 5.544= 69.22V
175.156= 69.22V
V= 175.156/69.22
V = 2.53m/s
The total momentum after catching the snowball is mV or:
(69.0 + 0.22) x V
So:
5.544= 69.22V
V= 5.544/69.22
V=0.08m/s
The velocity of the ice skater after throwing the snowball is 0.08m/s
The masses of astronauts are monitored during long stays in orbit, such as when visiting a space station. The astronaut is strapped into a chair that is attached to the space station by springs and the period of oscillation of the chair in a friction-less track is measured.
(a) The period of oscillation of the 10.0 kg chair when empty is 0.750 s. What is the effective force constant of the springs?
(b) What is the mass of an astronaut who has an oscillation period of 2.00 s when in the chair?
(c) The movement of the space station should be negligible. Find the maximum displacement of the 100,000 kg sace station if the astronaut's motion has an amplitude of 0.100 m.
Answer:
a) k = 701.8 N / m, b) m_{ast} = 61.1 kg, c) v ’= -1.3 10⁻⁴ m / s
Explanation:
a) For this exercise let's use the relationship of the angular velocity
w = [tex]\sqrt{ \frac{k}{m} }[/tex]
k = w² m
the angular velocity is related to the period
w = 2π / T
we substitute
k = 4 π² [tex]\frac{m}{T^2}[/tex]
let's calculate
k = 4 π² 10 /0.75²
k = 701.8 N / m
b) now repeat the measurement with an astronaut on the chair
w = [tex]\sqrt{ \frac{k}{m} }[/tex]
where the mass Month the mass of the chair plus the mass of the astronaut
M = m + [tex]m_{ast}[/tex]
M = k / w²
w = 2π / T
let's calculate
w = 2π / 2
w = π rad / s
M = 701.8 /π²
M = 71,111 kg
now we use that
M = m + m_{ast}
m_{ast} = M - m
m_{ast} = 71.111 - 10.0
m_{ast} = 61.1 kg
c) if the astronaut's movement is simple harmonic
x = A cos wt
therefore the speed is
v = [tex]\frac{dx}{dt}[/tex]
v = -Aw sin wt
maximum speed is
v = - Aw
v = 0.100 π
v = 0.31416 m / s
we can suppose that the movement of the space station and the astronaut is equivalent to division of the same
initial instant. Before the move
p₀ = 0
final instant. When the astronaut is moving
p_f = M_station v’+ m_{ast} v
the moment is preserved
p₀ = pf
0 = M__{station} v ’+ m_{ast} v
v ’= - [tex]\frac{m_{ast} }{M_{station} } \ v[/tex]
we substitute
v ’= [tex]\frac{61.1 }{ 100000 } \ 0.31416[/tex]
v ’= -1.3 10⁻⁴ m / s
the negative sign indicates that the station is moving in the opposite direction from the astronaut
A airplane accelerated from 59 m/s to 95 m/s in a distance of 123 meters what was its acceleration in m/s^2, assumed constant?
Answer:
22.54 m/s^2
Explanation:
vf = final velocity = 95 m/s
vi = initial velocity = 59 m/s
d = displacement = 123 m
a = acceleration, unknown
Use this kinematics equation to find a:
vf^2 = vi^2 + 2a*d
95^2 = 59^2 + 2*a*123
22.54 m/s^2 = a
Hope this helps!! :)
We assume the foam plate has a positive charge when rubbed with paper towels.
Lift the pan away from the charged plate using the styrofoam cup. Briefly touch the rim of the pan to neutralize it. Place the neutralized pan on the plate and observe the tape rise. When the pan is on the plate, the rim of the plate has a _____________. This means that the pan base is ________________ charged because the net charge on the pan is __________. You know that this must be the case because as you lift the pan with the cup away from the plate, the tape on the rim goes down.
Answer:
POSITIVE CHARGE, NEGATIVE CHARGE, ZERO
Explanation:
To solve this completion exercise, we must remember that charges of the same sign repel each other and in a metallic object (frying pan) the charge is mobile.
Let's analyze the situation when we touch the pan, the charges are neutralized, therefore when we bring the pan to the plate that has a positive charge, it attracts the mobile negative charges in the pan, until it is neutralized, therefore on the opposite side of the pan. pan (edge with a glued tape) is left with a positive charge; therefore the edge and the tape, which is very light, have positive charges and repel each other.
We must assume that the frying pan is insulated so that the net charge is zero, since the induction process.
Consequently the words to complete the sentence are
When the pan is on the plate, the edge of the plate has a _POSITIVE CHARGE_____.
This means that the base of the container is loaded NEGATIVE CHARGE_____ because the net charge of the container is ___ZERO_
Identical balls are dropped from the same initial height and bounce back to half the initial height. In Case 1, the ball bounces off a cement floor, and in Case 2, the ball bounces off a piece of stretchy rubber. Two cases of a ball dropping and bouncing off of a surface. In case one, the ball approaches a cement floor, comes in contact, momentarily comes to rest, and then bounces off with a velocity smaller than the approach velocity. In case 2, the ball approaches a piece of stretchy rubber. When the ball comes in contact with the rubber, it deforms the rubber and comes momentarily to rest. The ball bounces off again with a velocity smaller than the approach velocity, and the rubber regains its original shape. In both cases, the balls are dropped from the same initial height and reach the same final height. In which case is the magnitude of the ball's change in momentum the greatest
Answer:
the impulse must be the same in these two cases F t = m ([tex]\sqrt{2g h_f } - \sqrt{2g h_o}[/tex])
Explanation:
For this exercise we use the relationship between momentum and momentum
I = Δp
F t = m v_f - m v₀
To know the speed we use the conservation of energy
starting point. Highest point
Em₀ = U = m g h
fincla point. Just before the crash
Em_f = K = ½ m v²
energy is conserved
Em₀ = Em_f
m g h = ½ m v²
v = [tex]\sqrt{2gh}[/tex]
we substitute in the impulse relation
F t = m ([tex]\sqrt{2g h_f } - \sqrt{2g h_o}[/tex])
therefore we can see that as in case the initial and final heights are equal, the impulse must be the same in these two cases
Ocean waves crash on the beach at a velocity of 3.5 m/s. If the distance between the crests of each wave is 4 m, find the frequency of the waves.
a. 0.0088 Hz
b. 14.0 Hz
c. 1.14 Hz
d. 0.88 Hz
Answer:
d
Explanation:
velocity=frequency × wavelength
frequency=speed/wavelength
frequency=3.5÷4
=0.875~0.88
The frequency of the waves is (d) 0.88 Hz. So, correct answer is option (d).
What is frequency of wave?The frequency of a sinusoidal wave is the number of full oscillations performed by any wave constituent in a unit of time. According to the definition of frequency, if a body is moving periodically, it has completed one cycle after going through a number of situations or postures and then returning to its initial position. Therefore, frequency is a quantity that describes the rate of oscillation and vibration.
Given parameter,
Velocity of the waves = 3.5 m/s.
distance between the crests of each wave, that is, wavelength of the waves = 4 m.
We know that, for a wave transmission,
velocity of wave =frequency of wave × wavelength of wave
⇒ frequency of wave=speed of wave/wavelength of wave
⇒ frequency of wave =3.5 m/s ÷4m
=0.875 Hz
≈ 0.88 Hz
Hence, the frequency of the waves is 0.88 Hz.
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A 10-kg rock falls from a height of 8-m above the ground. What is the kinetic energy of the rock just before it hits the ground?
Answer: 800
Explanation:
1/2 x m x v^2 = m x g x h
KE = 10 x 10 x 8
KE= 800
The energy of the body by the virtue of its motion is known as the kinetic energy of the body. The kinetic energy of the rock just before it hits the ground will be 784.8 J.
What is kinetic energy?The energy of the body by the virtue of its motion is known as the kinetic energy of the body. It is defined as the product of half of mass and square of the velocity.
According to the law of conservation of energy, energy can not be created nor be destroyed can be transferred from one form to another form.
Kinetic energy= potential energy
Kinetic energy= mgh
Kinetic energy= 10×9.81×8
Kinetic energy=784.8 J
Hence the kinetic energy of the rock just before it hits the ground will be 784.8 J.
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A liquid fueled rocket is red on a test stand. The rocket nozzle has an exit diameter of 30 cm and the combustion gases leave the nozzle at a velocity of 3800 m/s and a pressure of 100 kPa, which is the same as the ambient pressure. The temperature of the gases in the combustion area is 2400 C. Find (a) the temperature of the gases at the nozzle exit plane, (b) the pressure in the combustion area, and (c) the thrust developed. Assume that the gases have a speci c heat ratio of 1.3, and a molar mass of 9. Assume that the ow in the nozzle is isentropic.
Answer:
1. Temperature= 869.35 K
2. Pressure of combustion = 12994.043 kpa
3. Thrust = 127x10⁶N
Explanation:
this problem has been fully explained in the attachment. please use it to get a clearer explanation of the answer.
1.
The temperature = (273+2400k) - (3800)²/2(4003)
= 2673 - 14440000/8006
= 2673 - 1803.65
= 869.35 K
Approximately 869.4K
2. We first get mach number
= 3800/√1.3(923.8)(869.35)
= 3800/1021.78
= 3.719
Pressure = 100kpa[1+2.07464415]^1.3/0.3
= 12995.043kpa
C. Thrust
Pi/4(3800)²(0.3)²(100x10³)/(923.8)(869.4)
= 12678.621
= 126.781 kN
Thrust is approximately 127kN = 127x10⁶N
How could being mindful in conversation be helpful?
You can see if the person your talking to and being mindfull they will see respect towards you or they see you as a mindful person.It shows a person that you care.
Explanation:
What force is left out of the Quantum Mechanics theory?
Answer:
Quantum mechanics is a key hypothesis in material science that gives a portrayal of the actual properties of nature at the size of iotas and subatomic particles. It is the establishment of all quantum physical science including quantum science, quantum field hypothesis, quantum innovation, and quantum data science.
Explanation:
It is the greatest of issues, it is the littlest of issues. At present physicists have two separate rule books clarifying how nature functions. There is general relativity, which perfectly represents gravity and everything it overwhelms: circling planets, impacting worlds, the elements of the growing universe all in all. That is enormous. At that point there is quantum mechanics, which handles the other three powers – electromagnetism and the two atomic powers. Quantum hypothesis is very proficient at portraying what happens when a uranium molecule rots, or when singular particles of light hit a sun based cell. That is little.
A makeshift sign hangs by a wire that is extended over an ideal pulley and is wrapped around a large potted plant on the
roof as shown in the figure below. When first set up by the shopkeeper on a sunny and dry day, the sign and the pot are in
equilibrium. The mass of the sign is 27.5 kg, and the mass of the potted plant is 67.5 kg.
Plant
sale
today!
(a) Assuming the objects are in equilibrium, determine the magnitude of the static friction force experienced by the
potted plant.
N
(b) What is the maximum value of the static friction force if the coefficient of static friction between the pot and the
roof is 0.707?
N
Answer:I know the answer for B cus I’m doing the same problem. For B, you would only take the coefficient of friction given and then multiply it by the Normal Force, which in this case is the same as the Gravitational Force.
Explanation:
A 3.0-kilogram mass is traveling in a circle of
0.20-meter radius with a speed of 2.0 meters per
second. What is its centripetal acceleration?
(1) 10. m/s
(3) 60. m/s2
(2) 20. m/s2
(4) 6.0 m/s2
Answer:
[tex]a=20\ m/s^2[/tex]
Explanation:
Given that,
The mass of an object, m = 3 kg
The radius of a circle, r = 0.2 m
The speed of the object, v = 2 m/s
We need to find the centripetal acceleration. Its formula is given by :
[tex]a=\dfrac{v^2}{r}\\\\a=\dfrac{(2)^2}{0.2}\\\\a=20\ m/s^2[/tex]
So, the centripetal acceleration is [tex]20\ m/s^2[/tex].
reason why the center of gravity must not be at 50cm
Answer:
hope this helps
hope this is what u want
1. A block with mass 20 kg is
sliding up a plane (Ukinetic=0.3,
inclined at 10°) at a speed of
2 m/s to the right (positive
X-direction). How far does it
go up along the plane before
it comes to rest momentarily?
Answer: 0.435 m
Explanation:
Given
mass m=20 kg
initial speed u=2 m/s
coefficient of kinetic friction [tex]\mu_k=0.3[/tex]
deceleration which opposes the motion is given by
[tex]\Rightarrow a=g\sin \theta+\mu_kg\cos \theta\\\Rightarrow a=g(\sin \theta +\mu_k\cos \theta)[/tex]
[tex]\Rightarrow a=9.8(\sin 10^{\circ}+0.3\times \cos 10^{\circ})\\\Rightarrow a=4.59\ m/s^2[/tex]
using [tex]v^2-u^2=2as[/tex]
[tex]\Rightarrow s=\dfrac{2^2}{2\times 4.59}=0.435\ m[/tex]
