When a charge is placed on a metal sphere, it ends up in equilibrium at the outer surface. Use this information to determine the electric field of +3.0 µC charge put on a 5.0-cm aluminum spherical ball at the following two points in space:
a. A point 1.0 cm from the center of the ball (an inside point).
b. A point 10 cm from the center of the ball (an outside point)
Answer:
a
The value at an inside point is Zero
b
The electric field is [tex]E = 2.7*10^{6} \ N/C[/tex]
Explanation:
From the question we are told that
The magnitude of the charge is [tex]q = 3.0 \mu C = 3.0 *10^{-6} \ C[/tex]
The radius of the spherical ball is [tex]r = 5.0 \ mm = 0.005 \ m[/tex]
Generally according to law postulated by Gauss the magnitude of charge enclosed inside a conducting material is zero which implies that the electric field inside the spherical ball is zero
Generally the electric field out side the spherical ball is mathematically represented as
[tex]E = \frac{kq}{ a^2}[/tex]
Here a is the position outside the spherical ball that is been considered and the value is [tex]a = 10 \ cm = \frac{10}{100} = 0.1 \ m[/tex]
and k is the coulombs constant with value
[tex]k = 9*10^{9}\ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}.[/tex]
=> [tex]E = \frac{ 3 *10^{-6} * 9*10^9 }{ (0.1)^2}[/tex]
=> [tex]E = 2.7*10^{6} \ N/C[/tex]
Two bodies A and B with 8 Kg and 24 kg respectively move
directly to one another for head
of
a speed of 15 m/s
and 15 m/s respectively If e=0.5 for colliding
bodies, determine the following
(1) Their velocities after collision
(2) The loss of K.E of colliding bodies.
Points A (-5,6), B (2,-2), and C (-6,-3) are placed in three different quadrants of a Cartesian coordinate system. Convert each set of Cartesian coordinates to polar coordinates. The angle should be reported as a positive angle, in degrees, from the positive x-axis. (0° < θ < 360°). What is the θ coordinate for point A? Recall this should be in units of degrees.
Answer: A ([tex]\sqrt{61}[/tex],309.8°)
B (2[tex]\sqrt{2}[/tex], 315°)
C ([tex]3\sqrt{5}[/tex], 26.56°)
Explanation: To transform rectangular coordinates into polar coordinates use:
[tex]r=\sqrt{x^{2}+y^{2}}[/tex] and [tex]\theta=tan^{-1}(\frac{y}{x})[/tex]
For point A:
[tex]r=\sqrt{(-5)^{2}+6^{2}}[/tex]
[tex]r=\sqrt{61}[/tex]
[tex]\theta=tan^{-1}(\frac{6}{-5})[/tex]
[tex]\theta=tan^{-1}(-1.2)[/tex]
[tex]\theta=-50.2[/tex]°
Point A is in the II quadrant, so we substract the angle for 360° since it is in degrees:
[tex]\theta=360-50.2[/tex]
[tex]\theta=[/tex] 309.8°
Polar coordinates for point A is ([tex]\sqrt{61}[/tex], 309.8°)
For point B:
[tex]r=\sqrt{2^{2}+(-2)^{2}}[/tex]
[tex]r=\sqrt{8}[/tex]
[tex]r=2\sqrt{2}[/tex]
[tex]\theta=tan^{-1}(\frac{-2}{2} )[/tex]
[tex]\theta=tan^{-1}(1)[/tex]
[tex]\theta=-45[/tex]°
Point B is in IV quadrant, so:
[tex]\theta=360-45[/tex]
[tex]\theta=[/tex] 315°
Polar coordinates for point B is ([tex]2\sqrt{2}[/tex], 315°)
For point C:
[tex]r=\sqrt{(-6)^{2}+(-3)^{2}}[/tex]
[tex]r=\sqrt{45}[/tex]
[tex]r=3\sqrt{5}[/tex]
[tex]\theta=tan^{-1}(\frac{-3}{-6} )[/tex]
[tex]\theta=tan^{-1}(0.5)[/tex]
[tex]\theta=[/tex] 26.56°
Polar coordinates for point C is ([tex]3\sqrt{5}[/tex], 26.56°)
Why is math so hard?
Answer:
its just built different
I could give you a big, fancy scientific answer, but it often comes down to the fact that the way that most instructors teach math is not the way the student(s) can most easily understand it.
Suppose that A’, B’ and C’ are at rest in frame S’, which moves with respect to S at speed v in the +x direction. Let B’ be located exactly midway between A’ and C’. At t’=0 a light flash occurs at B’ and expands outward as spherical wave.
1. According to an observer in S’, do the wave fronts arrive at A’ and C’ simultaneously?
2. According to an observer in S, do the wavefronts arrive at A’ and C’ simultaneously?
3. If you answered no to either 1. or 2., what is the difference in their arrival times and at which point did the front arrive first?
Answer:
1) an observer in B 'sees the two simultaneous events
2)observer B sees that the events are not simultaneous
3) Δt = Δt₀ /√ (1 + v²/c²)
Explanation:
This is an exercise in simultaneity in special relativity. Let us remember that the speed of light is the same in all inertial systems
1) The events are at rest in the reference system S ', so as they advance at the speed of light which is constant, so it takes them the same time to arrive at the observation point B' which is at the point middle of the two events
Consequently an observer in B 'sees the two simultaneous events
2) For an observer B in system S that is fixed on the Earth, see that the event in A and B occur at the same instant, but the event in A must travel a smaller distance and the event in B must travel a greater distance since the system S 'moves with velocity + v. Therefore, since the velocity is constant, the event that travels the shortest distance is seen first.
Consequently observer B sees that the events are not simultaneous
3) let's calculate the times for each event
Δt = Δt₀ /√ (1 + v²/c²)
where t₀ is the time in the system S' which is at rest for the events
How much heat is released when 30g of water at 96 degree C cools to 26 degrees C. The specific heat of water is 4.1 cal/g Degrees C
Answer:
I think it is Q
=
−
53
,
796.6
J
Explanation:
I note a 100% sure ok
A deductive argument is always a stronger argument than an inductive argument
O True
False
A Moving to another question will save this
A skier accelerates from a stop at a rate of 3 m/s2 for 3 seconds. What is her speed at the end of that time?
Answer:
9 m/s
Explanation:
Given:
v₀ = 0 m/s
a = 3m/s²
t = 3 s
Find: v
v = at + v₀
v = (3 m/s²) (3 s) + 0 m/s
v = 9 m/s
What is the horizontal acceleration of a ball that is launched horizontally with a velocity of 5.6 m/s?
O -9.8 m/s2
O 0 m/s2
O 5.6 m/s2
O 9.8 m/s2
Answer:
A
Explanation:
edge 2020
The horizontal acceleration of a ball that is launched horizontally with a velocity of 5.6 m/s will be zero. Option B is correct.
What is projectile motion?The motion of an item hurled or projected into the air, subject only to gravity's acceleration, is known as projectile motion.
The item is known as a projectile, and the course it takes is known as a trajectory. Falling object motion is a simple one-dimensional kind of projectile motion with no horizontal movement.
The projectile's motion is divided into two parts: horizontal and vertical motion.
The horizontal acceleration of a ball that is launched horizontally with a velocity of 5.6 m/s will be zero.
Hence, option B is correct.
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In ancient wars, heavy rocks are thrown horizontally from the wall as weapons. f the initial speed of a rock is 3 m/s , how many meters will it travel horizontally from a wall 25 meters abov the ground after 2 seconds(neglect air resistance, g=9.8m/s2)
Answer:
The horizontal distance traveled by the rock from the wall after 2 seconds is 6m.
Explanation:
Given;
initial speed of the rock, u = 3 m/s
height of the wall, h = 25 m
time of travel, t = 2 seconds
The horizontal distance is determined as follows;
X = ut + ¹/₂gt²
where;
g is acceleration due to gravity, horizontal distance is not affected by gravity, g = 0
X = ut
X = 3 x 2
X = 6 m
Therefore, the horizontal distance traveled by the rock from the wall after 2 seconds is 6m.
what is a plane mirror ?
Answer:
A plain mirrior is a mirrior with flat reflective surface.
hope it is helpful for you.
Two forces,
vector F 1 = (5.90î − 5.60ĵ) N
and
vector F 2 = (4.65î − 5.55ĵ) N,
act on a particle of mass 2.10 kg that is initially at rest at coordinates
(−1.75 m, +4.15 m).
(a) What are the components of the particle's velocity at t=10.4s ?
(b) In what direction is the particle moving at t = 10.4 s? (+counterclockwise from the x axis)
(c) What displacement does the particle undergo during the first 10.4 s?
(d) What are the coordinates of the particle at t = 10.4 s? (in x and y meters)
First compute the resultant force F:
[tex]\mathbf F_1=(5.90\,\mathbf i-5.60\,\mathbf j)\,\mathrm N[/tex]
[tex]\mathbf F_2=(4.65\,\mathbf i-5.55\,\mathbf j)\,\mathrm N[/tex]
[tex]\implies\mathbf F=\mathbf F_1+\mathbf F_2=(10.55\,\mathbf i-11.15\,\mathbf j)\,\mathrm N[/tex]
Then use Newton's second law to determine the acceleration vector [tex]\mathbf a[/tex] for the particle:
[tex]\mathbf F=m\mathbf a[/tex]
[tex](10.55\,\mathbf i-11.15\,\mathbf j)\,\mathrm N=(2.10\,\mathrm{kg})\mathbf a[/tex]
[tex]\mathbf a\approx(5.02\,\mathbf i-5.31\,\mathbf j)\dfrac{\rm m}{\mathrm s^2}[/tex]
Let [tex]\mathbf x(t)[/tex] and [tex]\mathbf v(t)[/tex] denote the particle's position and velocity vectors, respectively.
(a) Use the fundamental theorem of calculus. The particle starts at rest, so [tex]\mathbf v(0)=0[/tex]. Then the particle's velocity vector at t = 10.4 s is
[tex]\mathbf v(10.4\,\mathrm s)=\mathbf v(0)+\displaystyle\int_0^{10}\mathbf a(u)\,\mathrm du[/tex]
[tex]\mathbf v(10.4\,\mathrm s)=\left((5.02\,\mathbf i-5.31\,\mathbf j)u\,\dfrac{\rm m}{\mathrm s^2}\right)\bigg|_{u=0}^{u=10.4}[/tex]
[tex]\mathbf v(10.4\,\mathrm s)\approx(52.2\,\mathbf i-55.2\,\mathbf j)\dfrac{\rm m}{\rm s}[/tex]
If you don't know calculus, then just use the formula,
[tex]v_f=v_i+at[/tex]
So, for instance, the velocity vector at t = 10.4 s has x-component
[tex]v_{f,x}=0+\left(5.02\dfrac{\rm m}{\mathrm s^2}\right)(10.4\,\mathrm s)=52.2\dfrac{\rm m}{\mathrm s^2}[/tex]
(b) Compute the angle [tex]\theta[/tex] for [tex]\mathbf v(10.4\,\mathrm s)[/tex]:
[tex]\tan\theta=\dfrac{-55.2}{52.2}\implies\theta\approx-46.6^\circ[/tex]
so that the particle is moving at an angle of about 313º counterclockwise from the positive x axis.
(c) We can find the velocity at any time t by generalizing the integral in part (a):
[tex]\mathbf v(t)=\mathbf v(0)+\displaystyle\int_0^t\mathbf a\,\mathrm du[/tex]
[tex]\implies\mathbf v(t)=\left(5.02\dfrac{\rm m}{\mathrm s^2}\right)t\,\mathbf i+\left(-5.31\dfrac{\rm m}{\mathrm s^2}\right)t\,\mathbf j[/tex]
Then using the fundamental theorem of calculus again, we have
[tex]\mathbf x(10.4\,\mathrm s)=\mathbf x(0)+\displaystyle\int_0^{10.4}\mathbf v(u)\,\mathrm du[/tex]
where [tex]\mathbf x(0)=(-1.75\,\mathbf i+4.15\,\mathbf j)\,\mathrm m[/tex] is the particle's initial position. So we get
[tex]\mathbf x(10.4\,\mathrm s)=(-1.75\,\mathbf i+4.15\,\mathbf j)\,\mathrm m+\displaystyle\int_0^{10.4}\left(\left(5.02\dfrac{\rm m}{\mathrm s^2}\right)u\,\mathbf i+\left(-5.31\dfrac{\rm m}{\mathrm s^2}\right)u\,\mathbf j\right)\,\mathrm du[/tex]
[tex]\mathbf x(10.4\,\mathrm s)=(-1.75\,\mathbf i+4.15\,\mathbf j)\,\mathrm m+\dfrac12\left(\left(5.02\dfrac{\rm m}{\mathrm s^2}\right)u^2\,\mathbf i+\left(-5.31\dfrac{\rm m}{\mathrm s^2}\right)u^2\,\mathbf j\right)\bigg|_{u=0}^{u=10.4}[/tex]
[tex]\mathbf x(10.4\,\mathrm s)\approx(542\,\mathbf i-570\,\mathbf j)\,\mathrm m[/tex]
So over the first 10.4 s, the particle is displaced by the vector
[tex]\mathbf x(10.4\,\mathrm s)-\mathbf x(0)\approx(270\,\mathbf i-283\,\mathbf j)\,\mathrm m-(-1.75\,\mathbf i+4.15\,\mathbf j)\,\mathrm m\approx(272\,\mathbf i-287\,\mathbf j)\,\mathrm m[/tex]
or a net distance of about 395 m away from its starting position, in the same direction as found in part (b).
(d) See part (c).
What energy is there in climbing stairs
Answer:
There is our potential energy in climbing stairs.
A football field lies so one endzone is to the east and one is to the west. A running back gets the ball and starts running 20o north of east. After running 10 meters, he sees open field and starts running straight east for 60 meters at which point a safety closes in on him. He cuts so he is running 35o east of south. After running another 20 meters he gets in the end zone. Ultimately we want to know what is the vector that points from where he started to where he scored.
Answer:
( 80.87 i - 12.96 j ) m
Explanation:
Running back movements :
20⁰ north of east for 10 meters,
Runs straight east for 60 meters
runs 35⁰ east of south for 20 meters
A) show vector pointing from the starting point to the end ( where he scored )
The final vector displacement : ( 80.87 i - 12.96 j ) m
which is : 81.90 m, 9.10⁰ south of east
attached below is the required diagram
Which of the following statements is true about parenthood in developed nations?
I would say the decision not to have children or to have only one child is becoming more common.
Two wooden crates rest on top of one another. The smaller top crate has a mass of m1 = 21 kg and the larger bottom crate has a mass of m2 = 90 kg. There is NO friction between the crate and the floor, but the coefficient of static friction between the two crates is μs = 0.8 and the coefficient of kinetic friction between the two crates is μk = 0.64. A massless rope is attached to the lower crate to pull it horizontally to the right (which should be considered the positive direction for this problem). 1)The rope is pulled with a tension T = 261 N (which is small enough that the top crate will not slide). What is the acceleration of the small crate? m
Answer:
2.35 m/s²
Explanation:
Given that
Mass of the smaller crate, m₁ = 21 kg
Mass of the larger crate, m₂ = 90 kg
Tensión of the rope, T = 261 N
We know that the sum of all forces for the two objects with a force of friction F and a tension T are:
(i) m₁a₁ = F
(ii) m₂a₂ = T - F, where m and a are the masses and accelerations respectively.
1) no sliding can also mean that:
a₁ = a₂ = a
This makes us merge the two equations written above together as:
m₂a = T - m₁a
If we then solve for a, we would have something like this
a = T / (m₁+m₂)
a = 261 / (21 + 90)
a = 261 / 111
a = 2.35 m/s²
Therefore, the needed acceleration of the small crate is 2.35 m/s²
An object with mass m is given initial velocity to slide across a horizontal plane AB towards a semi circle BCD with radius R.
Between the object and the plane exists a kinetic friction u_k=0.5, but only between the section FB with length R. F is in the middle of A and B. The inside of the circle is smooth.
When the object reached to point C, it applies a force of 3mg on the semi circle.
The object is going to the left.
1. Write an expression for the initial velocity at point A.
The expression for the initial velocity at point A is:
0 = (velocity at point A - 0) / time
Simplifying the equation, we find:
Velocity at point A = 0
The initial velocity at point A is zero, indicating that the object starts from rest before sliding on the horizontal plane AB.
To write an expression for the initial velocity at point A, we need to analyze the forces acting on the object and apply the principles of motion.
Given:
Mass of the object, m
Radius of the semi circle, R
Coefficient of kinetic friction, μ[tex]_k[/tex] = 0.5
Force applied at point C, F = 3mg
The object is initially at rest.
Let's break down the motion into two parts: the motion on the horizontal plane AB and the motion along the semi circle BCD.
1. Motion on the horizontal plane AB:
The only force acting on the object on the horizontal plane is the force of kinetic friction. The frictional force can be calculated using:
Frictional force, f = μ[tex]_k[/tex]* Normal force
The normal force is equal to the weight of the object, which is mg.
Normal force, N = mg
Frictional force, f = μ[tex]_k[/tex] * mg
The frictional force acts in the opposite direction to the motion, so its magnitude is negative. Thus, the net force on the object on the horizontal plane is:
Net force = -f = -μ[tex]_k[/tex]* mg
Using Newton's second law, we can relate the net force to the acceleration:
Net force = mass * acceleration
-μ[tex]_k[/tex] * mg = m * acceleration
The acceleration can be expressed as the rate of change of velocity:
Acceleration = (final velocity - initial velocity) / time
Since the object is initially at rest, the initial velocity is zero.
For more such information on: velocity
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This section provides your reader with background information about why you conducted this experiment and how it was completed. Outline the steps of the procedure in full sentences. It also provides potential answers (your hypothesis/es) relative to what you expected the experiment to demonstrate. This section should be 1–3 paragraphs in length. Plz Help
Answer:
Me and I have to work on my yum new drink I think ima was my birthday I think we can get a good drink or something haha was that I
Explanation: is the red one is not the one that is the one that was killed and I killed the red light and I did the same for me and it is not the proof that you are
I did the test
3. Magnitude of a physical quantity is the number of times a standard quantity is present in it.
Calculate the mass of air in a room if the floor dimensions are 10 mx
12 m and height 4 m. The density of air is 1.26 kgm.
604.8 kg
Explanation:Hi there !
d = m/V => m = d×V
V = l×w×h = 10m×12m×4m = 480 m³
m = 1.26kg/m³×480m³ = 604.8 kg
Good luck !
the circuit consist of three resistors R1=5, R2=3, R3=4 and are connected in series to each other. A voltmeter connected in parallel to resister R2 measures voltage of 6v. what is the net in the current?
Answer:
Current in the circuit= 2 A
Explanation:
Current in a Series Connection
When two or more resistors are connected in series, all of them have the same current, and the sum of their individual voltages is the total voltage applied to the circuit.
We know a voltmeter connected to R2=3Ω measures 6V. Applying Ohm's law:
[tex]V=R.I[/tex]
We can calculate the current through R2:
[tex]\displaystyle I=\frac{6}{3}=2\ A[/tex]
Since the current in R2 is the same in the rest of the resistors, the current of R1 and R3 is also 2 A.
James yap run fast while dribbling the ball as he crosses the middle part of the basketball court he changes his phase to avoid opponent what specific
component of physical fitness did James Yap?
Answer:
The answer is "Speed".
Explanation:
In the question, the Speed is the correct answer because it can be viewed as its rate from, which a length is covered via an object. It is also divided by time into other components of distance. Its SI unit of speed seems to be the meter/second, but still, the kilometer/hour or, in the Us UK, miles per hour is the most basic fundamental of speed in everyday use.
Physical fitness is the body's ability to efficiently work together to enable good health while performing daily living tasks
The correct option is AgilityThe component of physical fitness James Yap applied is agility
The reason agility is the correct physical fitness component is as follows:
Question: The possible options in the question are;
Agility
Speed
Coordination
Agility is the capability to make nimble movement such as starting, accelerating, decelerating, stopping, change ones direction, quickly and easily while proper posture is being maintainedAgility in basketball enables the athlete to restore their center of gravity to a position vertically above their base constantly as their direction of motion is changed both quickly and at average speed during play, requiring coordination and neuromuscular efficiencyTherefore, the component of physical fitness James Yap applied to run fast while dribbling the ball, changing phase (pace), is agility
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When an automobile moves towards a listener, the sound of its horn seems relatively
a. Low pitched (low frequency) b. High Pitched (high frequency)
c. Normal (no change in frequency)
Answer:
Higher Pitched! (b)
Explanation:
The Doppler effect makes it seem high pitched, as the sound waves are mushed together at a higher frequency, while moving away, the sound waves spread out like a lower frequency wave!
something that can not be used up or depleted
Answer:
oxygen
Explanation:
as it is in the air it can't be depleted or used up
Answer:
Deplete definition is - to empty of a principal substance. How to use deplete in a ... to reduce in amount by using up The soil was depleted of minerals. deplete.
Explanation:
Which of the following statements does not represent assumptions that stage theories share?
A.
Each stage builds on the knowledge of the previous stage.
B.
Each stage corresponds to a specific age.
C.
A child can pass through the stages in any order.
D.
Each stage is qualitatively different than the next.
Each stage corresponds to a specific age. is the statements do not represent assumptions that stage theories share.
What is stage theory?Any speculative construct is used to represent phases or steps in a time-based process, such as a belief that development proceeds in discontinuous phases indicated by functional changes.
Each stage corresponds to a specific age the statements do not represent assumptions that stage theories share.
The assumptions for the stage theory are as follows;
1. Every stage serves as a building block for the next.
2. The phases are followed in the same sequence by everyone.
3. Each step is distinct in terms of quality. It is a change in nature, not merely a change in amount.
Each stage corresponds to a specific age. is the statements do not represent assumptions that stage theories share.
Hence option B is the right option.
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Nuclear power doesn't directly emit greenhouse gases, but it does leave behind dangerous nuclear _______.
Answer:
waste
Explanation:
Nuclear power produces radioactive waste that can harm the human body. The material produced (usually from either U - 235 or U - 238 or some radioactive isotope) can be just as destructive to cells as in outer space.
These wastes don't directly emit greenhouse gases, but the material waste mist be disposed properly or it'll harm humans and other organisms.
While helping an astronomy professor, you discover a binary star system in which the two stars are in circular orbits about the system's center of mass. From their color and brightness, you determine that each star has the same mass as our Sun. The orbital period of the pair is 31.3 days , based on the oscillation of brightness observed as one star occludes (hides) the other. From this information you are able to ascertain the distance between the stars.
Required:
Calculate the distance between the stars.
Answer:
The value is [tex]d = 3.66 *10^{10} \ m [/tex]
Explanation:
From the question we are told that
The orbital period is [tex]T = 31.3 \ days = 31.3 * 86400 = 2704320 \ s[/tex]
Generally for the centripetal force acting one of the stars will be equal to the gravitational force between the stars and this can be mathematically represented as
[tex]F_g = F_c[/tex]
=> [tex]\frac{Gm*m}{d^2} = \frac{mv^2}{r}[/tex]
Here m represents the mass of each star given that they have same mass and this equal to the mass of the sun whose value is
[tex]m = 1.989*10^{30} \ kg[/tex]
d(2 * r) is the diameter of the orbit which is distance between each star
r is the radius of the orbit
G is the gravitational constant with value [tex]G= 6.67*0^{-11} \ m^2/kg \cdot s^2[/tex]
v is the velocity of the stars which can be mathematically represented as
[tex]v = \frac{2 \pi r}{T}[/tex]
So
[tex]\frac{Gmm }{(2 * r)^2} = \frac{m (\frac{2\pi r}{T} )^2}{r}[/tex]
=> [tex]r =[ \frac{Gm T^2}{16 \pi ^2} ]^{\frac{1}{3} }[/tex]
=> [tex]r = [\frac{6.67 *10^{-11} * (1.989 *10^{30}) * 2704320^2}{16 * (3.142)^2} ]^{\frac{1}{3} }[/tex]
=> [tex]r = 1.83 *10^{10} \ m[/tex]
Generally the distance between the star is
[tex]d = 2 * r[/tex]
=> [tex]d = 2 * 1.83 *10^{10} [/tex]
=> [tex]d = 3.66 *10^{10} \ m [/tex]
3) A 10 ohm resistor and a 5 ohm resistor are connected in a series. A 2 A current
runs through the 10 ohm resistor. What is the potential difference in the 5 ohm resistor?
a) 10 V
c) 30 V
b) 20V
d) 40 V
Write one reason why we study motion
Answer:
We study motion becuase it is the best way of determining the best method of doing the work by scrutinizing the motions made by the worker or the machine
Hope this helps :)
1. Small hole is made up of a window
shutter of a room, 12.5m and image
of the car outside the room cast on
the opposite if the image is 6.5m
high and the tree is 32.5cm from
the window. What is the high of
the tree?
Answer:
I don't know sorry I hope you find what your looking
