Answer:
Road A- dry
Road B- mud
Road C- wet
Explanation:
Surface conditions do affect the ease and speed with which a skateboarder can move, on a muddy surface, the tyres of the skate boards finds it difficult to establish adequate fictional force between the skates trees and the traveling surface. Hence, the muddy surface presents a very slippery travel ground for the skate, hence leading the to skateboarder needing to apply caution.
The speed on a wet surfave is height as the amount of firece that will be applied in other to accelerate is very small. The surface is wet and hence serves as a lubricant between the contact surface.
The dry road also has a high speed but lower than a wet surface, frictional force is high here and this tend to slow the skateboarder down except in sloppy terrains.
A pulley has a mechanical advantage of 1. What does this tell you about the size and direction of the input and output forces?
Answer:
The number of input force is the same as output. ... If it equals once, then both numbers are equal making it the same.Explanation:
Does this helps
Answer:
The number of input force is the same as output. Formula for MA (Mechanical Advantage) is Input Force/Output Force. If it equals once, then both numbers are equal making it the same. In order to raise MA, you must lower efficiency, something you learn around grade 8. Good luck!
P.S. Direction is the same for both, meaning if you pull something, the object you pull will come towards you.
When it comes to the movement of air, friction
A. increases with altitude.
B. is greater near the ground surface.
C. diminishes turbulence.
D. is responsible for weaker winds aloft.
Answer: When it comes to the movement of air, friction is greater near the ground surface.
Explanation:
A resistance in motion observed by an object while on another object is called friction.
For example, a vehicle moving on road will have friction between its tires and the road.
Friction is more near the ground surface rather than away from the ground surface.
Thus, we can conclude that when it comes to the movement of air, friction is greater near the ground surface.
A ball is thrown horizontally at a speed of 24 meters per second from the top of a cliff. If the ball hits the ground 6.0 seconds later, approximately how high is the cliff? ( EASY QUESTION.. PLZZ HELPPP MEEE I WILL MARK YOU THE BRAINLIEST PLZZ)
Answer:
144 meters
Explanation:
the ball is thrown with a speed of 24 meters per second right so if the ball reaches the ground in 6 seconds. the hight of the cliff must be S=v.t
S (height cliff)=24m/s×6s=144
A pilot drops a package from a plane flying horizontally at a constant speed. Neglecting air resistance, when the package hits the ground the horizontal location of the plane will Group of answer choices be behind the package. be over the package. be in front of the package depend of the speed of the plane when the package was released.
Answer:
The location of helicopter is behind the packet.
Explanation:
As the packet also have same horizontal velocity as same as the helicopter, and also it has some vertical velocity as it hits the ground.
The horizontal velocity remains same as there is no force in the horizontal direction. The vertical velocity goes on increasing as acceleration due to gravity acts.
So, the helicopter is behind the packet.
help me with this question
Explanation:
Let's set the x-axis to be parallel to the and positive up the plane. Likewise, the y-axis will be positive upwards and perpendicular to the plane. As the problem stated, we are going to assume that m1 will move downwards so its acceleration is negative while m2 moves up so its acceleration is positive. There are two weight components pointing down the plane, [tex]m_1g \sin \theta[/tex] and [tex]m_2g \sin \theta[/tex] and two others pointing up the plane, the two tensions T along the strings. There is a normal force N pointing up from the plane and two pointing down, [tex]m_1g \sin \theta[/tex] and [tex]m_2g \sin \theta[/tex]. Now let's apply Newton's 2nd law to this problem:
x-axis:
[tex]m1:\:\:\:\displaystyle \sum_i F_i = T - m_1g \sin \theta = - m_1a\:\:\:\:(1)[/tex]
[tex]m2:\:\:\:\displaystyle \sum_i F_i = T - m_2g \sin \theta = m_2a\:\:\:\:(2)[/tex]
y-axis:
[tex]\:\:\:\displaystyle \sum_i F_i = N - m_1g \cos \theta - m_2g \cos \theta = 0[/tex]
Use Eqn 1 to solve for T,
[tex]T = m_1(g \sin \theta - a)[/tex]
Substitute this expression for T into Eqn 2,
[tex]m_1g \sin \theta - m_1a - m_2g \sin \theta = m_2a[/tex]
Collecting all similar terms, we get
[tex](m_1 + m_2)a = (m_1 - m_2)g \sin \theta[/tex]
or
[tex]a = \left(\dfrac{m_1 - m_2}{m_1 + m_2} \right)g \sin \theta[/tex]
A 4-kW resistance heater in a water heater runs for 3 hours to raise the water temperature to the desired level. Determine the amount of electric energy used in both kWh and kJ.
Answer:
12kWhr
Explanation:
Energy = Power * Time
Power = 4kW
Time = 3hrs
Substitute into the formula
Energy used up = 4kW * 3hrs
Energy used up = 12kWhr
Find the final velocity if the initial velocity of 8 m/s with an acceleration of 7 m/s2 over a 3 second interval?
I don't know about it your answer will give another people
Answer: Let the final velocity be v.
Given,
Initial velocity(u)=8m/s
Acceleration(a)=7m/s2
Time(t)=3 sec
Then,
v=u+at
=8+7*3 m/s
=29m/s
Therefore, the final velocity is 29m/s.
why are you teachers regarded as professionals
Answer:
coz teaching is their profession.
On topographic maps, contour lines that are farther apart indicate what ?
Answer:
if I am correct, they indicate less steep terrain. think of it as the steeper the terrain the closer together the lines would be. hope that makes sense for you guys.
Answer:
gentle slopes
Explanation:
A 2890-lb car is traveling with a speed of 58 mi/hr as it approaches point A. Beginning at A, it decelerates uniformly to a speed of 18 mi/hr as it passes point C of the horizontal and unbanked ramp. Determine the total horizontal force F exerted by the road on the car just after it passes point B.
Answer:
4592.57 lb
Explanation:
The missing diagram for this question is attached in the image below.
Given that:
the weight of the car = 2890 lb
At point A, the speed of the car [tex](V_A)[/tex] = 58 mi/hr
At point C, the speed of the car [tex](V_C)[/tex] = 18 mi/hr
To ft/s:
[tex](V_A)[/tex] = 58 mi/hr × 5280 ft/1 mi × 1 hr/3600 s
[tex](V_A)[/tex] = 85.07 ft/s
[tex](V_C)[/tex] = 18 mi/hr × 5280 ft/1 mi × 1 hr/3600 s
[tex](V_C)[/tex] = 26.4 ft/s
Between A to C, the total distance is;
[tex]S_{AC} = S_{AB}} + S_{BC} \\ \\ S_{AC} = 331 + \dfrac{\pi r}{2} \\ \\ S_{AC}= 331 + \dfrac{\pi \times 207}{2} \\ \\ S_{AC} = 656.154 \ ft[/tex]
Now, we need to determine the deceleration of the car using the formula:
[tex]V_C^2 = V_A^2 + 2 aS_{AC}[/tex]
[tex]26.4^2 = 85.07^2 + 2 a (654.154)[/tex]
[tex]696.96 = 7236.9049+ 2 a (654.154)[/tex]
[tex]696.96-7236.9049 = 2 a (654.154)[/tex]
[tex]-6539.9449 = 2 a (654.154)[/tex]
[tex]a= \dfrac{-6539.9449} {2(654.154)}[/tex]
a = -4.99 ft/s²
The velocity of the car as it passes via B
[tex]v_B^2 = v_A^2 + 2aS_{AB}[/tex]
[tex]v_B^2 = 85.07^2 + 2(-4.99 \times 331)[/tex]
[tex]v_B =\sqrt{ 85.07^2 + 2(-4.99 \times 331)}[/tex]
[tex]v_B =\sqrt{ 85.07^2 +3303.38}[/tex]
[tex]v_B =\sqrt{ 10540.2849}[/tex]
[tex]v_B =102.67 \ ft/s[/tex]
Along B, the car's acceleration is:
[tex]a_B = \sqrt{a^2 + (\dfrac{v_B^2}{r})^2}[/tex]
[tex]a_B = \sqrt{(-4.99)^2 + \dfrac{102.67^2}{207}^2 }[/tex]
[tex]a_B = 51.17 \ ft/s^2[/tex]
Finally, the total horizontal force F exerted = m[tex]a_B[/tex]
[tex]= (\dfrac{2890}{32.2}) \times 51.17[/tex]
= 4592.57 lb
A 2000-kg truck traveling at a speed of 6.0 m/s slows down to 4.0 m/s along a straight road. What
is the magnitude of the impulse?
The magnitude of the impulse of the truck is equal to 4000 Kg.m/s.
What is impulse?Impulse can be described as the integral of a force over the time interval for which it acts. Impulse is also a vector quantity since force is a vector quantity. Impulse can be applied to an object that generates an equivalent vector change in its linear momentum.
The S.I. unit of impulse is N⋅s and the dimensionally equivalent unit of momentum is kg⋅m/s. A resultant force gives acceleration and changes the velocity of an object for as long as it acts.
Given the mass of the truck, m= 2000 Kg
The initial speed of the truck, u = 6 m/s
The final speed of the truck, v = 4 m/s
The change in the linear momentum is equal to the impulse.
I = ΔP = mv - mu
I = 2000 ×4 - 2000 × 6
I = 8000 - 12000
I = - 4000 Kg.m/s²
Therefore, the magnitude of the impulse is 4000 Kg.m/s².
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A 97.0 kg ice hockey player hits a 0.150 kg puck, giving the puck a velocity of 48.0 m/s. If both are initially at rest and if the ice is frictionless, how far does the player recoil in the time it takes the puck to reach the goal 14.5 m away
Answer:
s₁ = 0.022 m
Explanation:
From the law of conservation of momentum:
[tex]m_1u_1 + m_2u_2 = m_1v_1+m_2v_2[/tex]
where,
m₁ = mass of hockey player = 97 kg
m₂ = mass of puck = 0.15 kg
u₁ = u₂ = initial velocities of puck and player = 0 m/s
v₁ = velocity of player after collision = ?
v₂ = velocity of puck after hitting = 48 m/s
Therefore,
[tex](97\ kg)(0\ m/s)+(0.15\ kg)(0\ m/s)=(97\ kg)(v_1)+(0.15\ kg)(48\ m/s)\\\\v_1 = -\frac{(0.15\ kg)(48\ m/s)}{97\ kg} \\v_1 = - 0.074 m/s[/tex]
negative sign here shows the opposite direction.
Now, we calculate the time taken by puck to move 14.5 m:
[tex]s_2 =v_2t\\\\t = \frac{s_2}{v_2} = \frac{14.5\ m}{48\ m/s} \\\\t = 0.3\ s[/tex]
Now, the distance covered by the player in this time will be:
[tex]s_1 = v_1t\\s_1 = (0.074\ m/s)(0.3\ s)[/tex]
s₁ = 0.022 m
what is measured by the ammeter
Answer:
amperes
Ammeter, instrument for measuring either direct or alternating electric current, in amperes. An ammeter can measure a wide range of current values because at high values only a small portion of the current is directed through the meter mechanism; a shunt in parallel with the meter carries the major portion.
Explanation:
hope it helps
Ethyl alcohol is :
a. None of the above
b. Semi polar solvent
c. Polar solvent
d. Non-Polar solvant
Answer:
D. Non- polar solvant
Explanation:
l think that's it
Answer:
I think the answer is D polar solvent
A 5.0-kg mass is placed at (3.0, 4.0) m, and a 6.0-kg mass is placed at (3.0, -4.0) m. What is the moment of inertia of this system of masses about the y-axis?
Answer:
the moment of inertia of this system of masses about the y-axis is 99 kgm²
Explanation:
Given the data in the question;
mass m₁ = 5.0 kg at point ( 3.0, 4.0 )
mass m₂ = 6.0 kg at point ( 3.0, -4.0 )
Now, Moment of inertia [tex]I[/tex] of this system of masses about the y-axis will be;
Moment of inertia [tex]I[/tex]ₓ = mixi²
Moment of inertia [tex]I[/tex] = m₁x₁² + m₂x₂²
we substitute
Moment of inertia [tex]I[/tex] = [ 5.0 × ( 3 )² ] + [ 6.0 × ( 3 )² ]
Moment of inertia [tex]I[/tex] = [ 5.0 × 9 ] + [ 6.0 × 9 ]
Moment of inertia [tex]I[/tex] = 45 + 54
Moment of inertia [tex]I[/tex] = 99 kgm²
Therefore, the moment of inertia of this system of masses about the y-axis is 99 kgm²
Why does the moon appear dark from space?
But why does it appear bright when observed from earth, especially when it is full moon?
Answer:
The moon is actually quite dim.
Explanation:
compared to other astronomical bodies. The moon only seems bright in the night sky because it is so close to the earth and because the trees, houses, and fields around you are so dark at night. In fact, the moon is one of the least reflective objects in the solar system.
Answer:
It reflects the light send from the sun.
Explanation:
If the moon is between you and the sun, you will see the back of it which doesnt reflect light.
12. What type of circuit is the diagram below?
series circuit
parallel circuit
Answer:
parallel circuit
Explanation:
An electric circuit can be defined as an interconnection of electrical components which creates a path for the flow of electric charge (electrons) due to a driving voltage.
Generally, an electric circuit consists of electrical components such as resistors, capacitors, battery, transistors, switches, inductors, etc.
Basically, the components of an electric circuit can be connected or arranged in two forms and these includes;
I. Series circuit
II. Parallel circuit: it's an electrical circuit that has the same potential difference (voltage) across its terminals or ends. Thus, its components are connected within the same common points so that only a portion of current flows through each branch.
Hence, the type of circuit that the above diagram above represents is a parallel circuit.
Answer:
parallel circuit
Explanation:
I got it right on my exam
An irregular shape object has a mass of 19 oz. A graduated cylinder with and initial volume of 33.9 mL. After the object was dropped in the graduated cylinder, it had a volume of 92.8 mL. What is the density of object( g/mL)
Explanation:
m = 19 oz × (28.3 g/1 oz) = 537.7 g
V = 92.8 mL
[tex]\rho = \dfrac{m}{V}= \dfrac{537.7\:g}{92.8\:mL} = 5.79\:\frac{g}{mL}[/tex]
A shooting star is actually the track of a meteor, typically a small chunk of debris from a comet that has entered the earth's atmosphere. As the drag force slows the meteor down, its kinetic energy is converted to thermal energy, leaving a glowing trail across the sky. A typical meteor has a surprisingly small mass, but what it lacks in size it makes up for in speed. Assume that a meteor has a mass of 1.5
Answer:
A. Power generated by meteor = 892857.14 Watts
Yes. It is obvious that the large amount of power generated accounts for the glowing trail of the meteor.
B. Workdone = 981000 J
Power required = 19620 Watts
Note: The question is incomplete. A similar complete question is given below:
A shooting star is actually the track of a meteor, typically a small chunk of debris from a comet that has entered the earth's atmosphere. As the drag force slows the meteor down, its kinetic energy is converted to thermal energy, leaving a glowing trail across the sky. A typical meteor has a surprisingly small mass, but what it lacks in size it makes up for in speed. Assume that a meteor has a mass of 1.5 g and is moving at an impressive 50 km/s, both typical values. What power is generated if the meteor slows down over a typical 2.1 s? Can you see how this tiny object can make a glowing trail that can be seen hundreds of kilometers away? 61. a. How much work does an elevator motor do to lift a 1000 kg elevator a height of 100 m at a constant speed? b. How much power must the motor supply to do this in 50 s at constant speed?
Explanation:
A. Power = workdone / time taken
Workdone = Kinetic energy of the meteor
Kinetic energy = mass × velocity² / 2
Mass of meteor = 1.5 g = 0.0015 kg;
Velocity of meteor = 50 km/s = 50000 m/s
Kinetic energy = 0.0015 × (50000)² / 2 = 1875000 J
Power generated = 1875000/2.1 = 892857.14 Watts
Yes. It is obvious that the large amount of power generated accounts for the glowing trail of the meteor.
B. Work done by elevator against gravity = mass × acceleration due to gravity × height
Work done = 1000 kg × 9.81 m/s² × 100 m
Workdone = 981000 J
Power required = workdone / time
Power = 981000 J / 50 s
Power required = 19620 Watts
Therefore, the motor must supply a power of 19620 Watts in order to lift a 1000 kg to a height of 100 m at a constant speed in 50 seconds.
plz answer the question
Answer:
Ray A - incident ray
Ray B - reflected ray
Kulsum’s TV uses 45 W. How much does it cost her to watch TV for one month (30 days). She watches TV for 4 hours/day during mid-peak time (10.4 cents/kWh).
Answer:
Total cost = 56.16 cents
Explanation:
Given the following data;
Power = 45 Watts
Time = 4 hours
Number of days = 30 days
Cost = 10.4 cents
To find how much does it cost her to watch TV for one month;
First of all, we would determine the energy consumption of the TV;
Energy = power * time
Energy = 45 * 4
Energy = 180 Watt-hour = 180/1000 = 0.18 Kwh (1 Kilowatts is equal to 1000 watts).
Energy consumption = 0.18 Kwh
Next, we find the total cost;
Total cost = energy * number of days * cost
Total cost = 0.18 * 30 * 10.4
Total cost = 56.16 cents
you are stowing items and come across an aerosol bottle of hairspray.what should you do?
Answer:
below
Explanation:
A car accelerates uniformly from rest to a speed of 55.0 mi/h in 13.0 s. (a) Find the distance the car travels during this time. m (b) Find the constant acceleration of the car. m/s2
Answer:
(a) 159.84 m
(b) 1.89 m/s²
Explanation:
Applying,
(a)
s = (v+u)t/2.................. Equation 1
Where s = distance traveled by the car, u = initial velocity, v = final velocity, t = time.
From the question,
Given: u = 0 m/s ( from rest), v = 55 mi/h = (55/2.237) m/s = 24.59 m/s, t = 13 s
Substitute these values into equation 1
s = (24.59+0)13/2
s = 159.84 m
(b)
Also applying,
a = (v+u)/t................. Equation 2
Where a = acceleration of the car.
substituting into equation 2,
a = (24.59+0)/13
a = 1.89 m/s²
A camera lens with focal length f = 50 mm and maximum aperture f>2
forms an image of an object 9.0 m away. (a) If the resolution is limited
by diffraction, what is the minimum distance between two points on the
object that are barely resolved? What is the corresponding distance
between image points? (b) How does the situation change if the lens is
“stopped down” to f>16? Use λ= 500 nm in both cases
Answer:
The minimum distance between two points on the object that are barely resolved is 0.26 mm
The corresponding distance between the image points = 0.0015 m
Explanation:
Given
focal length f = 50 mm and maximum aperture f>2
s = 9.0 m
aperture = 25 mm = 25 *10^-3 m
Sin a = 1.22 *wavelength /D
Substituting the given values, we get –
Sin a = 1.22 *600 *10^-9 m /25 *10^-3 m
Sin a = 2.93 * 10 ^-5 rad
Now
Y/9.0 m = 2.93 * 10 ^-5
Y = 2.64 *10^-4 m = 0.26 mm
Y’/50 *10^-3 = 2.93 * 10 ^-5
Y’ = 0.0015 m
Select the correct answer.
What are the directions of an object's velocity and acceleration vectors when the object moves in a circular path with a constant speed?
OA. The question is meanimgless, since the acceleration is zero.
ов.
The vectors point in opposite directions.
Oc.
Both vectors point in the same direction.
OD
The vectors are perpendicular,
Answer:
A
Explanation:
If the object is moving at a constant speed, the object isn't accelerating as the velocity doesn't change.
Answer: C.
Explanation: plato users
Determine the magnitude of the minimum acceleration at which the thief can descend using the rope. Express your answer to two significant figures and include the appropriate units.
Answer: hello your question is incomplete below is the missing part
A 69-kg petty thief wants to escape from a third-story jail window. Unfortunately, a makeshift rope made of sheets tied together can support a mass of only 58 kg.
answer:
To 2 significant Figures = 1.6 m/s^2
Explanation:
Calculate the magnitude of minimum acceleration at which the thief can descend
we apply the relation below
Mg - T = Ma --- ( 1 )
M = 69kg
g = 9.81
T = 58 * 9.81
a = ? ( magnitude of minimum acceleration)
From equation 1
a = [ ( 69 * 9.81 ) - ( 58 * 9.81 ) ] / 69
= 1.5639 m/s^2
To 2 significant Figures = 1.6 m/s^2
An object that sinks in water has a mass in air of 0.0675 kg. Its apparent mass when submerged in water is 0.0424 kg. What is the specific gravity SG of the object? What material is the object probably made?
Answer:
1. SG
true
=2.689
2. The object is probably some sort of minerals and rocks such as Feldspar, Corals, Beryl, etc.
Explanation:
Given:
mass in the air= 0.0675 kg
mass in water= 0.0424 kg
The specific gravity of the object will be 2.6892. It is the ratio of the density of the given fluid and the standard fluid.
What is density?Density is specified as the mass divided by the volume. It is represented by the unit of measurement as kg/m³.
The mass of the object in air;
m=Vρ₀
m=0.0675 kg
Buoyant force on the object;
B= Vρₐg
For equilibrium;
N+B=m₀g
n=m₀g-Vρₓg
N/g=m₀-Vρₓ
N/g=0.0424 kg
[tex]\rm \frac{V\rho_0}{V\rho_x} =\frac{0.0675 }{m_0-0.0424 \ kg} \\\\ \frac{\rho_0}{\rho_x} =\frac{0.0675}{0.0675-0.0424} \\\\ \frac{\rho_0}{\rho_x} =2.6892[/tex]
Hence, the specific gravity of the object will be 2.6892.
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A 55 kg pole vaulter falls from rest from a height of 5.4 m onto a foam rubber pad. The pole vaulter comes to rest 0.24 s after landing on the pad.
a. Calculate the athlete's velocity just before reaching the pad
b. Calculate the constant force exerted on the pole vaulter due to the collision.
Answer:
a) 10.3 m/s
b) 566 N
Explanation:
[tex]v {}^{2} = {u}^{2} + 2as \\ v {}^{2} = 0 {}^{2} + 2(9.81)(5.4) \\ v = 10.3 \: ms {}^{ - 1} [/tex]
[tex]force \: = \frac{d(mv)}{dt} \\ = 55(10.293) \\ = 566 \: newtons[/tex]
The athelete velocity will be 10.3 and constant force 566 N.
What is velocity?The displacement that an object or particle experiences with respect to time is expressed vectorially as velocity. The meter per second (m/s) is the accepted unit of velocity magnitude (also known as speed).
Alternately, the magnitude of velocity can be expressed in centimeters per second (cm/s). Depending on how many dimensions are included, there are numerous ways to indicate the direction of a velocity vector.
The car's velocity in relation to your body is zero when you are driving. The speed of the car in relation to you if you were to stand by the side of the road is 20 m/s northward.
Therefore, The athelete velocity will be 10.3 and constant force 566 N.
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What is the efficiency of a machine that uses 102 kJ of energy to do 98 kJ of work?
A 3 5m container is filled with 900 kg of granite (density of 2400 3 kg m/ ). The rest of the volume is air, with density equal to 3 1.15 / kg m . Find the mass of air and the overall (average) specific volume
Complete question:
A 5-m³ container is filled with 900 kg of granite (density of 2400 kg/m3). The rest of the volume is air, with density equal to 1.15 kg/m³. Find the mass of air and the overall (average) specific volume.
Answer:
The mass of the air is 5.32 kg
The specific volume is 5.52 x 10⁻³ m³/kg
Explanation:
Given;
total volume of the container, [tex]V_t[/tex] = 5 m³
mass of granite, [tex]m_g[/tex] = 900 kg
density of granite, [tex]\rho _g[/tex] = 2,400 kg/m³
density of air, [tex]\rho_a[/tex] = 1.15 kg/m³
The volume of the granite is calculated as;
[tex]V_g = \frac{m_g}{ \rho_g}\\\\V_g = \frac{900 \ kg}{2,400 \ kg/m^3} \\\\V_g = 0.375 \ m^3[/tex]
The volume of air is calculated as;
[tex]V_a = V_t - V_g\\\\V_a = 5 \ m^3 \ - \ 0.375 \ m\\\\V_a = 4.625 \ m^3[/tex]
The mass of the air is calculated as;
[tex]m_a = \rho_a \times V_a\\\\m_a = 1.15 \ kg/m^3 \ \times \ 4.625 \ m^3\\\\m_a = 5.32 \ kg[/tex]
The specific volume is calculated as;
[tex]V_{specific} = \frac{V_t}{m_g \ + \ m_a} = \frac{5 \ m^3}{900 \ kg \ + \ 5.32\ kg} = 5.52 \times 10^{-3} \ m^3/kg[/tex]