A easy dude please mark me as brainlyst
Light travelling in air striked a flat piece of uniformly thick glass at an incident of 60. If the index refraction of the glass is 1.50 what is the angle of refraction in the glass
Answer:
30.81°
Explanation:
Angle of refraction= sin⁻¹ 0.5122 = 30.81°.
Hope this helped!
PLEASE MARK ME AS BRAINLIEST!
Vector B~ has x, y, and z components of 7.1,
8.2, and 8.4 units, respectively.
Calculate the magnitude of B~ .
Can someone please help me?
Answer:
The magnitude of vector B is 13.7
Explanation:
Given that,
Component of vector B is
[tex]x=7.1[/tex]
[tex]y=8.2[/tex]
[tex]z=8.4[/tex]
We need to calculate the magnitude of vector B
Using given data
[tex]B=\sqrt{x^2+y^2+z^2}[/tex]
Put the value into the formula
[tex]B=\sqrt{(7.1)^2+(8.2)^2+(8.4)^2}[/tex]
[tex]B=13.7[/tex]
Hence, The magnitude of vector B is 13.7
If the net force acting on an object is 100 N to the left, Are the forces unbalanced or balanced?
Answer: STOP LOOKING FOR ANSWERS OMG:(((((9
Explanation:
How do astronomers determine the surface temperature of stars? *
Answer:
To the extent that Stellar spectra look like blackbodies, the temperature of a star can also be measured amazingly accurately by recording the brightness in two different filters. To get a stellar temperature: Measure the brightness of a star through two filters and compare the ratio of red to blue light.
Explanation:
On planet Gazorpzorp, an object from rest at a height of 3.00m takes 968 ms to reach the ground. What is the acceleration due to gravity on Gazorpzorp. Please show your work if possible.
Answer:
[tex]a=6.4\ m/s^2[/tex]
Explanation:
Given that,
Initial velocity, u = 0
It falls from a height of 3 m and it takes 968 ms to reach the ground.
We need to find the acceleration due to gravity on Gazorpzorp. Using second equation of motion to find a. It can be given by :
[tex]s=ut+\dfrac{1}{2}at^2[/tex]
u = 0 (at rest) and a = g
[tex]s=\dfrac{1}{2}at^2\\\\a=\dfrac{2s}{t^2}\\\\a=\dfrac{2(3)}{(968\times 10^{-3})^2}\\\\a=6.4\ m/s^2[/tex]
So, the acceleration due to gravity on Gazorpzorp [tex]6.4\ m/s^2[/tex].
5.16 An insulated container, filled with 10 kg of liquid water at 20 C, is fitted with a stirrer. The stirrer is made to turn by lowering a 25-kg object outside the container a distance of 10 m using a frictionless pulley system. The local acceleration of gravity is 9.7 m/s2. Assume that all work done by the object is transferred to the water and that the water is incompressible. A. Determine the work transfer (kJ) to the water. B. Determine the increase in internal energy (kJ) of the water. C. Determine the final temperature ( C) of the water. HINT: Assume that the temperature change is small enough that a constant value of the specific heat is a good approximation. D. Determine the heat transfer (kJ) from the water required to return the water to its initial temperature. Turns, Stephen R.. Thermodynamics (p. 294). Cambridge University Press. Kindle Edition.
Answer:
a) W=2.425kJ
b) [tex]\Delta E=2.425kJ[/tex]
c) [tex]T_f=20.06^{o}C[/tex]
d) Q=-2.425kJ
Explanation:
a)
First of all, we need to do a drawing of what the system looks like, this will help us visualize the problem better and take the best possible approach. (see attached picture)
The problem states that this will be an ideal system. This is, there will be no friction loss and all the work done by the object is transferred to the water. Therefore, we need to calculate the work done by the object when falling those 10m. Work done is calculated by using the following formula:
[tex]W=Fd[/tex]
Where:
W=work done [J]
F= force applied [N]
d= distance [m]
In this case since it will be a vertical movement, the force is calculated like this:
F=mg
and the distance will be the height
d=h
so the formula gets the following shape:
[tex]W=mgh[/tex]
so now e can substitute:
[tex]W=(25kg)(9.7 m/s^{2})(10m)[/tex]
which yields:
W=2.425kJ
b) Since all the work is tansferred to the water, then the increase in internal energy will be the same as the work done by the object, so:
[tex]\Delta E=2.425kJ[/tex]
c) In order to find the final temperature of the water after all the energy has been transferred we can make use of the following formula:
[tex]\Delta Q=mC_{p}(T_{f}-T_{0})[/tex]
Where:
Q= heat transferred
m=mass
[tex]C_{p}[/tex]=specific heat
[tex]T_{f}[/tex]= Final temperature.
[tex]T_{0}[/tex]= initial temperature.
So we can solve the forula for the final temperature so we get:
[tex]T_{f}=\frac{\Delta Q}{mC_{p}}+T_{0}[/tex]
So now we can substitute the data we know:
[tex]T_{f}=\frac{2 425J}{(10000g)(4.1813\frac{J}{g-C})}+20^{o}C[/tex]
Which yields:
[tex]T_{f}=20.06^{o}C[/tex]
d)
For part d, we know that the amount of heat to be removed for the water to reach its original temperature is the same amount of energy you inputed with the difference that since the energy is being removed this means that it will be negative.
[tex]\Delta Q=-2.425kJ[/tex]
Equipotential lines are lines with equal electric potential (for example, all the points with an electric potential of 5.0 V). Using the plot tool that comes with voltmeter (pencil icon) make two equipotential lines at r=0.5 m and r=1.5 m. Enable electric field vectors in the simulation. Put an electric field sensor at different points on the equipotential line and note the direction of the electric field vector. What can you conclude about the direction of the electric field vector in relation to the equipotential lines?Equipotential lines are lines with equal electric potential (for example, all the points with an electric potential of 5.0 V). Using the plot tool that comes with voltmeter (pencil icon) make two equipotential lines at r=0.5 m and r=1.5 m. Enable electric field vectors in the simulation. Put an electric field sensor at different points on the equipotential line and note the direction of the electric field vector. What can you conclude about the direction of the electric field vector in relation to the equipotential lines?
Answer:
the equipotential lines or surface are perpendicular to the electric field
Explanation:
In this exercise you are asked to explain the relationship between the direction of the equipotential lines and the electric field.
Electric field lines go from a positive charge to a negative charge, the specific shape of these lines depends on the charge distribution, let's look at two specific cases:
* in the case of point charge the lines are radial
* In the case of electric plates, they are lines perpendicular to the plates.
Equipotential surfaces are related to the elective field by
Ex = - dV / dx
Ey = -dV / dy
Ez = -dV / dz
E = Exi ^ + Ey j ^ + Ezk ^
remember electric potential is a scalar and the electric field is a vector quantity, therefore the equipotential lines or surface are perpendicular to the electric field. Specifically for
* point charge the equipotential surfaces are concentric with the charge
* For electrical plates equipotential lines are parallel to the plates.
Emmy kicks a soccer ball up at an angle of 45° over a level field. She watches the ball's trajectory and notices that it lands, two seconds after being kicked, about 20 m away to the north. Assume that air resistance is negligible, and plot the horizontal and vertical components of the ball's velocity as a function of time. Consider only the time that the ball is in the air, after being kicked but before landing. Take "north" and "up" as the positive x ‑ and y ‑directions, respectively, and use g≈10g≈10 m/s2 for the downward acceleration due to gravity
Let [tex]u[/tex] be the initial velocity of the soccer ball at an angle of inclination of [tex]\theta_0[/tex] with the positive x-axis.
Given that:
[tex]\theta_0=45^{\circ}[/tex]
The horizontal distance covered by the projectile=20 m
Time of flight, [tex]t_f=2[/tex] seconds
Acceleration due to gravity, [tex]g= 10 m/s^2[/tex] downward.
As "north" and "up" as the positive x ‑ and y ‑directions, respectively.
So, [tex]g= -10 m/s^2[/tex]
As the acceleration due to gravity is in the vertical direction, so the horizontal component of the initial velocity remains unchanged.
The x-component of the initial velocity, [tex]u_x=u\cos\theta_0[/tex].
The horizontal distance covered by the projectile [tex]= u_x\times t_f[/tex]
[tex]\Rightarrow u_x\times t_f=20[/tex]
[tex]\Rightarrow u_x\times 2=20[/tex]
[tex]\Rightarrow u_x=10[/tex] m/s
So, the horizontal component of the velocity is 10 m/s which is constant and the graph has been shown in the figure (i).
Now, [tex]u\cos(45^{\circ})=10[/tex] [as [tex]u_x=u\cos\theta_0[/tex]]
[tex]\Rightarrow u=10\sqrt{2}[/tex] m/s.
The vertical component of the initial velocity,
[tex]u_y= u\sin\theta_0[/tex]
[tex]\Rightarrow u_y=10\sqrt{2}\sin(45^{\circ})[/tex]
[tex]\Rightarrow u_y=10[/tex] m/s
Let v be the vertical component of the velocity at any time instant t.
From the equation of motion,
[tex]v=u+at[/tex]
where u: initial velocity, v: final velocity, a: constant acceleration, and t: time taken to change the velocity from u to v.
In this case, we have [tex]u=u_y, a= -10 m/s^2[/tex].
So at any time instant, t.
[tex]v=u_y+(-10)t[/tex]
[tex]\Rightarrow v=10-10t[/tex]
The vertical component of the velocity, v, is the function of time and related as [tex]v=10-10t[/tex].
This is a linear equation.
At 2 second, the vertical component of the velocity
v=10-10x2=-10 m/s.
The graph has been shown in figure (ii).
Type one to two paragraphs describing the changes in potential and kinetic energy of the cart.
Be sure to discuss how the potential and kinetic energy of the cart changes at each of the four
positions along the track, and explain why these changes occur.
Answer:
The cart starts out in position A with high potential energy, low kinetic energy, and some thermal energy. As the cart progresses into position B, the kinetic energy begins to increase and the potential energy begins to decrease; as the thermal energy increases as thermal energy from the track is transferred to the cart through friction. Once the cart reaches position C, it has high kinetic energy, low potential energy, and some thermal energy. At position D, the cart has high potential energy, low kinetic energy, and some thermal energy. The potential energy throughout is correlated to gravity, the kinetic energy is correlated to momentum, and the thermal energy is correlated to friction. The potential energy is at its maximum during position A, and its minimum at position C; the kinetic energy is at its maximum during position C, and its minimum at position A; the thermal energy is at its maximum during position B, and its minimum at position A.
Explanation:
Answer:
Cart A starts out on the top of the hill, it has great potential energy, low kinetic energy, and some thermal energy. Cart B is going down the hill which increased the kinetic energy and the thermal energy while the potential energy is decreasing. Cart C is at the bottom of the hill, having high kinetic energy, low potential energy and some thermal energy. Cart D is at the top of a smaller hill, giving it high potential energy, low kinetic energy and little thermal energy.
These changes occur because when an object is at a high place, it will have high potential energy, and low kinetic energy, and as the height goes down, the kinetic energy increases and the potential energy will decrease.
Explanation:
here's my version of answer
Write a paragraph or two (no more than 1 page) describing the misconception and then explain the correct physics by identifying the forces or physics concepts involved in simple terms, like you are talking to your 3rd grade sibling.
Start with a question (will water fall out of the bucket when I stop rotating it?)
Give the common answer that is incorrect (what does your intuition tell you should happen?).
Explain what happens and why in simple terms (try practicing on an elementary student to make sure they understand).
Use diagrams and sketches in your explanation and avoid using physics terms like rotational, centripetal, normal forces, tension, friction, acceleration, velocity, vectors, buoyancy, energy, kinetics, potential, heat capacity, etc.
Answer:
1.) Everything that moves, will eventually come to a stop. Rest is the “natural” state of all objects
Of all physics misconceptions, this is the most common. Even the great philosopher Aristotle, included it into his most important contribution to the field, his famous Laws of Motion. But now we know it is wrong because Newton’s First Law of Motion tells us that “everything at rest will stay at rest, and everything in motion will stay in motion, unless acted upon by an external force.”
The first statement seems reasonable enough, but the second part is a little bit murky. The reason this confusion persists boils down to the fact that we are unable to identify the force that stops all motion, which is friction. Friction is a force that acts between two objects that are in contact and are moving relative to each other. When we roll a ball, it stops because of the frictional force acting between it and the floor.
2.) A continuous force is needed for continuous motion
This misconception is a direct consequence of the first one. While this is true, if you are, for example, pushing a grocery cart in a supermarket, again this is only because there is friction involved. The force you apply to keep an object moving is only to counteract the frictional force. If you were to throw a rock on outer space, it would travel with a constant velocity forever, unless it hits something, of course. This is because space is mostly empty (it has trace elements of gas and dust throughout), and there would not be any frictional force acting on that rock.
3.) An object is hard to push because it is heavy
This is one of the most common misconceptions because it’s something we see and feel everyday. While a heavy object is really hard to push, it is not because of its weight, but because of its inertia or mass. Inertia is an objects resistance to change in motion. It is important to note that inertia is resistance to “change motion” rather than just motion itself. When, I was a kid, I imagined that it would be easy to carry and push massive objects when in outer space, but not surprisingly, my younger self was wrong.,
With that said… Since these objects still have mass despite being weightless, this mass represents the object’s inertia.
4.) Planets revolve around the sun because they are pushed by gravity
We have to remember that gravity — the weakest of the four fundamental forces — is an attractive force. The reason why planets revolve around the Sun can be chalked up to the fact that the planets were already spinning within the protoplanetary disk encircling a young Sun. Gravity merely keeps the planets in orbit around the Sun, but it isn’t necessarily the one thing pushing the planets along their orbital plane.
5.) Heavier objects fall faster than lighter ones
This misconception is already debunked long ago by Galileo on his experiment when he dropped two objects with different masses on the Leaning Tower of Pisa. He has shown on that experiment that objects move downward with the same acceleration.
Again, the problem comes from not being able to identify another force that is involved, which is air resistance. All objects moving through air, and hence, all falling objects, experience air resistance. This force is proportional to the area of the object in the direction of motion. Usually, this force is negligible, but for light objects — with weight comparable to the air resistance, like a feather — it will have a big effect. This is ultimately confirmed by the famous hammer and feather drop experiment on the moon.
6.) There is no gravity in outer space
There is gravity in outer space, it is just weaker than what we experience here on Earth. Astronauts that are orbiting the Earth don’t experience gravity because they are free-falling (yes, you read that right). All satellites, including the moon and the planets, are in a constant state of freefall.
They just also have a tangential velocity with their free fall, that is why they don’t crash to what they are orbiting. When something is in free fall, it becomes weightless. This is why Kate Upton can do a photo shoot in zero gravity here on Earth. The plane that they are riding in actually went into free fall to do that.
7.) Planets move in circular orbits around the Sun
Planets actually move in elliptical orbits around the sun (with the Sun being the focus of the ellipse). This is actually the first of Kepler’s Three Laws of Planetary Motion, which deals with precisely how planets orbit the Sun.
One misconception deals with our seasons. Some might wrongly come to the conclusion that Earth’s proximity to the Sun dictates the seasons (summer is when Earth is closest to the Sun and winter is when it’s farther away), but that’s not entirely true. In reality, our seasons are caused by the tilt of Earth’s axis.
A locomotive engine pulls a train with a constant force comment on whether its acceleration will increase or decrease if more coaches are added to the train
Explanation:
the acceleration will be unchanged according to newton second law of motion
A locomotive engine pulls a train with a constant force then its acceleration will increase if more coaches are added to the train because the mass of the system would increase
What is Newton's second law?Newton's Second Law states that The resultant force acting on an object is proportional to the rate of change of momentum.
Force = mass ×acceleration
As given in the problem if a locomotive engine pulls a train with a constant force
As given the force is constant but if the number of coaches increases the mass of the train increase to balance the constant force acceleration will decrease.
Thus, if a locomotive engine pulls a train with a constant force then its acceleration will increase if more coaches are added to the train because the mass of the system would increase.
Learn more about Newton's second law, here refer from the link,
brainly.com/question/13447525
#SPJ2
A 90kg woman and a 60kg boy are standing at rest on a frictionless frozen lake. The boy pushes the woman with a 40N horizontal force. Calculate the acceleration of the woman
3600 m/s2
0.66 m/s2
26.7 m/s2
0.44 m/s2
30 m/s2
1.5 m/s2
What happens to them?
The woman will move slowly, and the boy will move faster in the opposite direction.
The woman and the boy will move together with equal speeds.
The woman will move away, but the boy will remain at rest.
The boy will move away, but the woman will remain at rest.
The woman will move slowly, and the boy will move faster in the same direction.
The woman and the boy will move apart at equal speeds.
Answer:
The acceleration of the woman is 0.44 m/s²
Explanation:
Given;
mass of the woman, m₁ = 90 kg
mass of the boy, m₂ = 60 kg
The force applied by the boy, f₂ = 40 N
The net horizontal force on the woman = 40 N
Apply Newton's second law of motion to determine the acceleration of the woman;
f = ma
a = f / m
a = 40 / 90
a = 0.44 m/s²
Therefore, the acceleration of the woman is 0.44 m/s²
The mass of a fully loaded Boeing 747 is about 4,082,331.33 kg. If it is cruising eastward at a
velocity of 253 m/s, what is its momentum?
some PLEASE help ill give brainliest!!
Answer:
Dang bro what school you in?
EM waves are used for communication with satellites and for observing the Earth from space. True or false
Answer:
true
Explanation:
brief description about construction of mercury thermometer with figure
A car had a constant acceleration of 6m/s? north for 2 seconds. What was thecar's change in velocity if it traveled in a straight line?
A horse slows from a velocity of 9.5 m/s to 5.5 m/s over a distance of 32 m. Find time.
Answer:
8
Explanation:
Finding time= t=change in velocity or distance/acceleration
so 32/9.5-5.5
32/4
8
which of these would most likely be a parts of a lab procedure?
A. write a hypothesis to answer a question
B. write a title at the top of a completed lab report
C. record the time to complete a chemical reaction
D. create a question on the cause of a chemical reaction
Order the circuits according to their power usage, from highest to lowest. All batteries are the same voltage, and all light bulbs have the same resistance. A battery connected to three light bulbs. The positive battery terminal connects to the first bulb. The first bulb connects to the second bulb. The second bulb connects to the third bulb. The third bulb connects to the negative battery terminal. Circuit A A battery connected to two light bulbs. The positive and negative terminals of the battery are connected directly to each bulb. Circuit B A battery connected to two light bulbs. The positive battery terminal connects to the first bulb. The first bulb connects to the second bulb. The second bulb connects to the negative battery terminal. Circuit C A battery connected to three light bulbs. The positive and negative terminals of the battery are connected directly to each bulb. Circuit D Highest power usage Lowest power usage
Answer:
P_a = 3P₀ , P_b = 2 P₀ , P_c = 3 P₀
P_c> P_b> P_a
Explanation:
To determine which circuit consumes more energy than is given by the expression
P = V I
V = I R
I = V / R
P = V² / R
As all circuits have the same battery, the value of the resistance to which the battery is connected determines the consumption
Circuit A
In this circuit the three bulbs are in series so the total resistance is
R_total = 3 R
the power dissipated is
P_a = V² / 3R
if we call
P₀ = V² / R
we substitute
P_a = P₀/3
Circuit B
Two bulbs are connected in series
R_total = 2 R
power is
P_b = V2 / 2R
P_b = P₀/2
Circuit C
The 3 bulbs are connected, but in parallel, the resistance is
1 / R_totak = 1 / R + 1 / R + 1 / R
R_total = R / 3
P _c = V2 3 / R
P_c = 3 Po
By reviewing these results, we can sort the circuits
P_c> P_b> P_a
You have just moved into a new apartment and are trying to arrange your bedroom. You would like to move your dresser of weight 3,500 N across the carpet to a spot 5 m away on the opposite wall. Hoping to just slide your dresser easily across the floor, you do not empty your clothes out of the drawers before trying to move it. You push with all your might but cannot move the dresser before becoming completely exhausted. How much work do you do on the dresser? You have just moved into a new apartment and are trying to arrange your bedroom. You would like to move your dresser of weight 3,500 across the carpet to a spot 5 away on the opposite wall. Hoping to just slide your dresser easily across the floor, you do not empty your clothes out of the drawers before trying to move it. You push with all your might but cannot move the dresser before becoming completely exhausted. How much work do you do on the dresser? W>1.75×104J W=1.75×104J 1.75×104J>W>0J W=0J
Answer:
W = 0J
Explanation:
The work done by the dresser is described as
W = f d (cos θ)
F has been given as the weight of this dresser. And it is 3500 N
d = 0 m
When you put these values into the equation
W = 3500 x 0 x cosθ
W = 0 J
This value tells us that the work done on this dresser is zero. No work has been done. Therefore the last option answers the question.
I will mark Brainlyest for the CORRECT ANSWER in % how are apes like humans
Explanation:
Chimpanzees are genetically closest to humans, and in fact, chimpanzees share about 98.6% of our DNA. We share more of our DNA with chimpanzees than with monkeys or other groups, or even with other great apes! We also both play, have complex emotions and intelligence, and a very similar physical makeup.
(I hope that's good) :)
A racecar goes around a 400m track in 58 seconds, what is the speed and velocity?
Answer:
v = 6.89 [m/s]
velocity = 0
Explanation:
To solve this problem we must use the definition of speed which is defined by the following expression of physics, such as the relationship between distance and time.
v = x/t
where:
v = speed [m/s]
x = distance = 400 [m]
t = time = 58 [s]
v = 400/58
v = 6.89 [m/s]
The velocity is a vector, since the racing car moves in circles IE its movement started where it ended, its velocity is zero.
Calculate Gravitational Potential Energy for an object on Earth with a mass of 2 kg and a height of 7 m.
Answer:
137.2J
Explanation:
Ep= mgh
Given,
m=2kgh=7mand we know, g = 9.81 N/kg
Ep= 2 × 9.81 × 7
Ep= 137.2J
3 Physics Questions Please Help
1. Which will hit the ground FASTER (greater ), a pencil that is dropped from the top of the table ( = 0) or a pencil that rolls of the edge of the table ( > 0)? (Assume the table is level and both pencils leave the table at the same instant.)
a. The pencil that rolls off the table will hit faster.
b. The pencil that is dropped will hit faster.
c. Both will hit at the same speed.
d. Can’t tell without knowing how fast the second pencil was rolling.
2. You are throwing a water balloon as hard as you can off the top of the bleachers. You want the balloon to hit the ground with as great a velocity as possible. Which of these actions will help the hit the ground fastest—that is, with greatest ?
a. Throwing the balloon UP at any angle of θ with an initial velocity vmax.
b. Throwing the balloon DOWN at the opposite angle (-θ) with the same initial velocity vmax.
c. Both of these options—choices (a) and (b)—will yield the same .
d. Cannot compare these options without knowing angle θ.
3. Batman, hanging from the top of a low building, sees a criminal a long-distance away pointing a gun over his head at him. (The gun has no sighting tool, and the criminal does not know physics.) Assume Batman reacts just as the criminal fires the gun. (Which of the following should Batman NOT do? Circle any/all BAD MOVES.
a. Stay where he is and hope the criminal will miss.
b. Retract his grapple and hope to dodge the bullet by quickly shooting upward.
c. Let go of the building and hope to dodge the bullet by falling downward.
d. Push off the building and hope to dodge the bullet by moving sideways.
Check: did you see the word NOT in the question?
Answer:
Number one is C.Both
Number two is A. throw upwards
Number three is B. react his grapple... shooting upward
name the four forces in physics?
Answer:
Gravitational
Electrostatic
magnetic
Frictional
gravitational
electrostatic
magnetic
frictional
hope it helps
pls mark as brainliest
What is the acceleration of the object?
m/s2
Answer:
nice app is brainly heheh........
Answer:
The acceleration is 100m
Explanation:
It takes a ball 30 seconds to travel from point A to point B. What is the ball’s speed?
(A + B) / 30 s
(B – A) / 30 s
30 s / (A - B)
30 s / (B – A)
Answer:
Direction from A to B divided by time(30s)
Explanation:
Two rocks, each of mass 72 kg, are positioned 95 m away from each other in deep space. What is the magnitude of the gravitational attraction between them? G=6.674 x 10^-11 N*m^2/kg^2.
3.8 x 10-12N
9.8 x 10-11N
3.0 x 10-10N
3.8 x 10-11N
Answer:
3.8 x 10^-11N
Explanation:
Given
M=72 kg
R=95 m
G=6.67x10^-11
Equation
Fg= 6.67x10^-11((72kg*72)/(95)^2)
Plug into calculator and get
3.8127756 x 10^-11
if a 100-n net force acts upon a 50g car, what will the acceleration of the car be
Answer:
2m/s
Explanation:
The formula for acceleration can be found by [ a = f/m ]
We are given the force [ 100N ] and the mass [ 50g ].
We can use these values to solve for the acceleration.
a = 100/50
a = 2m/s
Best of Luck!