Answer: that one is wrong i did it on flocabulary its false
Explanation:
Since electromagnetic radiations do not require a material medium for their propagation and can travel through vacuum, the statement that electromagnetic radiation cannot travel through a vacuum is False, option B.
What is an electromagnetic radiation?!An electromagnetic radiation is a radiation which results from.the interaction of the electric and magnetic fields.
Electromagnetic radiations do not require a material medium for their propagation, and hence can travel through vacuum.
An example of electromagnetic radiation is radio waves.
Hence, the statement that electromagnetic radiation cannot travel through a vacuum is False, option B.
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A vertical force is applied to a block of mass m that lies on a floor. What happens to the magnitude of the normal force on the block from the floor as magnitude F is increased from zero if force is (a) downward and (b) upward
Answer:
The answer is "[tex]\bold{+9.0 \ \frac{m}{s^2}}[/tex]"
Explanation:
Its rate (in respect of time) was its derivative of its specified function x, as well as the speed is the derivative of the speed. Therefore, in Newton's second legislation [tex]a=2c- 3(2.0)(2.0)t[/tex] is used:
[tex]F=(2.0 \ kg)a=40c -24t[/tex] (understood by SI units).
At [tex]t=3.0s, the \ F=-36 \ N[/tex] is told.
Therefore, the solution of c can be used to solve [tex]F=-36\ N=4.0c-24(3.0)[/tex]. This results in [tex]c=+9.0 \ \frac{m}{s^2}[/tex].
A block of mass, m, is pushed up against a spring, compressing it a distance x, and is then released. The spring projects the block along a frictionless horizontal surface, giving the block a speed v. The same spring projects a second block of mass 4m, giving it a speed 3v. What distance was the spring compressed in the second case?
Answer:
x₂ =6*x₁
Explanation:
In absence of friction, along a horizontal surface, all the elastic potential energy stored in the spring becomes kinetic energy, so we can write the following expression for the first case:[tex]\frac{1}{2}*k*x_{1}^{2} = \frac{1}{2}*m*v^{2} (1)[/tex]
In the second case, we have a mass 4m, which reaches to a speed 3v due to the energy stored in the spring.We can write the same equation than for (1) replacing m by 4m and v by 3v, as follows:[tex]\frac{1}{2}*k*x_{2} ^{2} = \frac{1}{2} * 4*m*(3*v)^{2} =\frac{1}{2}*4*m*9*v^{2} = \frac{1}{2}* 36 m*v^{2} (2)[/tex]
Dividing (2) over (1) on both sides, and rearranging terms, we get:[tex]x_{2} ^{2} = 36 * x_{1} ^{2}[/tex]
Taking square roots on both sides, we get:x₂ = 6*x₁So, in this case, the spring was compressed six times the distance that was compressed in the first case.g First find the magnitude of the force F on a positive charge q in the case that the velocity v⃗ (of magnitude v) and the magnetic field B⃗ (of magnitude B) are perpendicular. Express your answer in terms of v, q, B, and other quantities given in the problem statement.
Answer:
The magnitude of the force F on a positive charge q in the case that the velocity v and the magnetic field B are perpendicular is [tex]F = q\cdot v\cdot B[/tex], measured in newtons.
Explanation:
From classical theory on Magnetism, the vectorial form of the magnetic force on a particle is given by:
[tex]\vec F = q\cdot \vec v \times \vec B[/tex] (Eq. 1)
Where:
[tex]\vec F[/tex] - Magnetic force, measured in newtons.
[tex]q[/tex] - Electric charge, measured in coulombs.
[tex]\vec v[/tex] - Velocity of the particle, measured in meters per second.
[tex]\vec B[/tex] - Magnetic field, measured in tesla.
By definition of cross product, we get that magnitude of magnetic force on a positive charge [tex]q[/tex] is:
[tex]F = q\cdot v\cdot B \cdot \sin \theta[/tex] (Eq. 2)
Where:
[tex]v[/tex] - Speed of the particle, measured in meters per second.
[tex]B[/tex] - Magnitude of the magnetic field, measured in tesla.
[tex]\theta[/tex] - Angle between the velocity of the particle and magnetic field, measured in sexagesimal degrees.
If velocity and magnetic field are perpendicular, then (Eq. 2) is reduced into this form: ([tex]\theta = 90^{\circ}[/tex])
[tex]F = q\cdot v\cdot B[/tex]
draw a roller coaster track that starts with one large hill, then is followed
by a valley and another, smaller hill.
Draw a cart in four positions on the track as outlined below.
i. Draw the first cart at the top of the first hill. Label it A.
ii. Draw the second cart going down the first hill into the valley. Label it B.
iii. Draw the third cart at the bottom of the valley. Assume that the height of the cart in
this position is zero. Label it C.
iv. Draw the last cart at the top of the second, smaller hill. Label it D.
okay, this is the part that I don't know lol. ill give you brianlist or whatever its called:
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:
At the top of the hill the cart has maximum potential energy and zero kinetic energy. At the valley the cart has maximum kinetic energy and zero potential energy.
According to the principle of conservation of mechanical energy when an object changes position, the mechanical of the object is always conserved.
An object has maximum potential energy at maximum height.When the object is at lowest point, it has maximum kinetic energy.The sum of kinetic energy and potential energy of the object at any point is the total mechanical energy.
M.A = P.E + K.E
At the top of the hill the cart has maximum potential energy and zero kinetic energy. At the valley the cart has maximum kinetic energy and zero potential energy.
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What is the drift speed of the electrons when there is a voltage drop of 118.8 V across the wire? Assume there is one conduction electron per atom.
Answer:
Explanation:
If e be the charge on the electron and v be its drift speed after undergoing voltage drop of 118.8 V
eV = 1/2 m v²
1.6 x 10⁻¹⁹ x 118.8 = 1/2 x 9.1 x 10⁻³¹ v²
v² = 41.775 x 10¹²
v = 6.46 x 10⁶ m /s
drift speed of electron = 6.46 x 10⁶ m /s
A ballistic pendulum is a device often used for measuring the velocity of a projectile. Assume to projectile fully embeds itself in the pendulum. If the projectile mass is 25 grams, the pendulum mass is 2.5 kg, and the vertical displacement of the pendulum after the collision is 12 cm, find the velocity of the projectile just prior to the collision.
Answer:
Explanation:
pendulum rises upto height of .12 m
velocity of projectile and pendulum after collision = √ ( 2gH )
= √ ( 2 x 9.8 x .12 )
= 1.53 m / s
Applying conservation of momentum during collision
mv = (m + M ) X V
m and M are mass of projectile and pendulum and v and V are velocity of projectile before and after collision .
.025 x v = ( 2.5 + .025 ) x 1.53
v = 154.5 m /s
A bullet is fired directly upward with a speed of 330 m/s. How far above the ground will the bullet be in 65 s
Answer:
d = 21450 m
Explanation:
Given that,
The speed of a bullet, v = 330 m/s
We need to find the distance covered by the bullet in 65 s.
We know that,
Speed of an object = distance covered divided by time taken.
It means,
d = vt
d = 330 m/s × 65 s
d = 21450 m
So, the bullet will cover a distance of 21450 m.
Why is force a vector
Answer:
A force vector is a representation of a force that has both magnitude and direction.
The velocity of an object as a function of time is given by v = (12.5t –7.2t2) + (4.3t3). What is its acceleration as a function of time?
Answer:
a(t) = 12.5 -10.4t + 12.9t²Explanation:
The equation is not well written:
let the velocity of an object as a function of time be given by
v = (12.5t - 7.2t²) + (4.3t³)
v(t) = 12.5t - 7.2t² + 4.3t³
acceleration is the change in velocity with respect to time:
a(t) = d(v(t))/dt
a(t) = 12.5 - 2(7.2)t^2-1 + 3(4.3)t^3-1
a(t) = 12.5 -10.4t + 12.9t²
Hence the acceleration as a function of time t is expressed as;
a(t) = 12.5 -10.4t + 12.9t²
Help please, thank you.
Answer:
a. skin and mucous membranes
hope this helps you
❤❤❤❤❤❤❤
itz Indian heart
An athlete at the gym holds a 3.5kg steel ball in his hand. His arm is 70cm long and has a mass of 4.0kg.What is the magnitude of the torque about his shoulder if he holds his arm straight out to his side, parallel to the floor?What is the magnitude of the torque about his shoulder if he holds his arm straight, but 55 degrees below horizontal?
Answer:
A. 38 nm
B. 22 nm
Explanation:
This question has given us m, mass of the ball as 3.5kg
M = The mass of arms = 4kg
R = length of arms as 70cm
70 cm = 0.7m
0.7m/2 = 0.35m
rmg + Rmg
g = 9.8m/s
= 0.7(3.5)(9.8)+(0.35)(4)(9.8)
= 24.01 + 13.72
= 37.72
~38
B.
55 degrees below horizontal
Cos 55⁰ = 0.5736
= 0.7(3.5)(9.8)(0.5736)+(0.35)(4)(9.8)(0.5736)
= 13.772136+7.869792
= 21.642
~22Nm
The magnitude of the torque about his shoulder, if he held his arm straight is 21.64Nm
The formula for calculating the magnitude of the torque about his arm is expressed as:
[tex]\tau = (F_1r + F_2R) cos\theta[/tex]
F1 and F2 are the forces
[tex]\theta[/tex] is the given angle
F1 = 3.5 * 9.8 = 34.3 N
r = 70cm = 0.7m
F2 = 4.0 * 9.8 = 39.2N
R = 0.7/2 = 0.35m
Substitute the given values into the formula:
[tex]\tau = (34.3(0.7) + (39.2)(0.35)) cos55^0\\\tau = (24.01+13.72)cos55\\\tau = 37.73cos55\\\tau=21.64Nm[/tex]
Hence the magnitude of the torque about his shoulder, if he held his arm straight is 21.64Nm
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Starting at rest at the top of a ski slope, you travel down the hill reaching a velocity of 20 m/s after 13 seconds. Once you reach the bottom of the hill you travel at a constant velocity of 20 m/s for the next 7 seconds. Sketch a quantitative (with numbers) position vs time, velocity vs time, and acceleration vs time graph of the skier's motion. PLEASE SHOW WORK AND I WILL MARK YOU THE BRAINLIEST
Answer:
The graphs are attached
Explanation:
We are told that he started at rest at rest and travelled down the hill reaching a velocity of 20 m/s after 13 seconds.
Acceleration is gotten from;
v = u + at
a = (v - u)/t
a = (20 - 0)/13
a = 20/13 m/s² or 1.54 m/s²
Distance in this period is gotten from;
v² = u² + 2as
s = (v² - u²)/2s
s = (20² - 0²)/(2 × 20/13)
s = 400/(40/13)
s = 130 m
We are told that after reaching the bottom of the hill, he travelled at a constant velocity of 20 m/s for the next 7 seconds.
At constant velocity, acceleration is 0.
Thus,distance in this period is;
s1 = vt = 20 × 7 = 140 m
I've attached the graphs
34)
Elements are composed of
A)
many type of atoms.
B)
the same type of atom.
C)
many different compounds.
D)
the same type of compound.
If I am driving down the highway going north at 50 miles per hour, and another car is driving south at 75 miles per hour. How fast is the car coming toward me?
Answer:
V₂₋₁ = 50 [miles/h]
Explanation:
In order to solve this problem we must use the concept of relative velocities.
Where:
V1 = 50 [miles/h]
V2 = 75 [miles/h]
V₂₋₁ = Relative velocity among the two cars.
V₂₋₁ = 75 - 25 = 50 [miles/h]
That is, the car that goes at 50 miles per hour, sees how the second car approaches at 50 miles per hour.
A ball rolls at a speed of 4 m/s off a flat surface which is a height 1.3m above the ground. how long (in seconds) is the ball in the air before it hits the ground
Which is the dependent variable and what is the trend in the graph?
Experimental Solubility Data for a Sugar
140
130
120
110
Solubility (g/100g H,0)
100
90
80
0
10
50
60
20 30 40
Temperature (*C)
Solubility; decreasing
Solubility; increasing
Temperature; decreasing
Temperature; increasing
Answer: Solubility; decreasing
Explanation:
A charged particle is injected at 109 m/s into a 0.0691‑T uniform magnetic field perpendicularly to the field. The diameter of its orbit is measured and found to be 0.0427 m. What is the charge–to–mass ratio of this particle?
We know, radius of the orbit is given by :
[tex]r=\dfrac{mv}{qB}[/tex]
So, ratio is given by :
[tex]\dfrac{q}{m}=\dfrac{v}{Br}\\\\\dfrac{q}{m}=\dfrac{109\ m/s}{0.0691 \ T \times 0.0427\ m}\\\\\dfrac{q}{m}=36942.01 \ C/kg\\\\\dfrac{q}{m}=3.69\times 10^{4}\ C/kg[/tex]
Therefore, the charge–to–mass ratio of this particle is [tex]3.69\times 10^{4}\ C/kg[/tex] .
Hence, this is the required solution.
A 60 kg cyclist approaches the bottom of a gradual hill at a speed of 11 m/s. The hill is 5.0 m high, and the cyclist estimates that she is going fast enough to coast up and over it without peddling. Using conservation of energy, find the speed of the cyclist when she reaches the top of the hill (ignoring air resistance and friction). g
Answer:
9.91m/s
Explanation:
According to conservation of energy, the potential energy at the top of the hill is equal to the kinetic energy along the hill.
PE = KE
mgh = 1/2mv²
m is the mass of the cyclist
g is the acceleration due to gravity
h is the height of the hill
v is the speed of the cyclist
From the formula:
gh = v²/2
Given
g = 9.81m/s²
h = 5.0m
Substitute into the formula:
9.81(5.0) = v²/2
9.81*5*2 = v²
98.1 = v²
v = √98.1
v = 9.91m/s
Hence the speed of the cyclist when she reaches the top of the hill is 9.91m/s
Which statement is true regarding eligibility for becoming pharaoh?
People believed that the gods selected the pharaoh.
A woman could be selected as pharaoh.
The pharaoh’s title had to pass to a royal son.
The title of pharaoh was passed to a military commander.
Answer:
A woman could be selected as pharaoh.
Explanation:
Most pharaohs were men, but there were some exceptions, such as queen Hatshepsut.
Answer:
A
Explanation:
A Category 1 hurricane with a radius of 70km has sustained tangential winds of about 33 m/s. What is the acceleration of a particle on th edge
of the hurricane.
Answer:
[tex]a_c=0.016 \ m/s^2[/tex]
Explanation:
Centripetal acceleration: Is the acceleration experienced by an object while in a uniform circular motion. It points toward the center of rotation and is perpendicular to the linear velocity v and has the magnitude:
[tex]\displaystyle a_c=\frac{v^2}{r}[/tex]
Where v is the linear velocity and r is the radius of the circle.
It's given a hurricane with a radius r=70 Km = 70,000 m has a tangential wind velocity of v=33 m/s. The acceleration of a particle lying on the edge of the hurricane is:
[tex]\displaystyle a_c=\frac{33^2}{70,000}[/tex]
[tex]\boxed{a_c=0.016 \ m/s^2}[/tex]
At an instant when a soccer ball is in contact with the foot of the player kicking it, the horizontal or x component of the ball's acceleration is 810 m/s2 and the vertical or y component of its acceleration is 910 m/s2. The ball's mass is 0.40 kg. What is the magnitude of the net force acting on the soccer ball at this instant?
Answer:
487.13N
Explanation:
According to Newton's second law;
F = ma
F is the net force
m is the mass of the object
a is the acceleration
First we need to find the resultant of the acceleration
a =√810²+910²
a = √656,100+828,100
a =√1,484,200
a = 1218.28m/s²
Ball mass = 0.40kg
Next is to get the magnitude of the net force
F = 0.4 × 1218.28
F = 487.31N
Hence the magnitude of the net force acting on the soccer ball at this instant is 487.13N
Determine the xx component of force at point CC in the pipe assembly. Neglect the weight of the pipe. Take F1F1
Answer:
Hello your question has some missing part and diagram attached below is the missing part and diagram
Determine the x component of force at point C in the pipe assembly. Neglect the weight of the pipe. Take F1 = { 348 i - 427 j} lb and F2 = {-307 j + 122 k} lb.
Determine the y component of the force at point C in the pipe assembly.
Determine the x component of the force at point C in the pipe assembly.
Determine the x component of the moment at point C in the pipe assembly.
Determine the y component of the moment at point C in the pipe assembly.
Determine the z component of the moment at point C in the pipe assembly.
answer: attached below is the detailed solution
Explanation:
Attached below is the detailed solution
I need help with this science work.
Answer:
ok what is it?
Explanation:
Answer:how can I help?
Explanation:
The graph below shows the average monthly temperatures at one location on earth over a three-year period.
Which was the highest record temperature?
A:25°F
B:30°F
C:70°F
D:80°F
2. A 60 kg crate is being lowered from a loading dock to the floor of a warehouse by a thick rope. The rope exerts a constant upward force of 500 N as it is descending to the floor. a) Draw a free-body diagram for the crate. b) Calculate the acceleration (if any) of the crate.
Answer:
Acceleration =8.33m/s^2Explanation:
Step one:
Kindly find attached a sketch of the free body diagram
Given data
mass m=60kg
Force F=500N
a=?
Step two:
From Newton's second law "An object’s acceleration equals the vector sum of the forces acting on it, divided by it mass".
mathematically
F=ma
Step three:
Substituting we have
500=60*a
a=500/60
a=8.33m/s^2
A 0.1-kg meter stick is supported at both ends by strings, and there is a 200-g mass attached to it at the 70 cm mark. What is the force exerted by the string connected at the 0 cm mark, if the system is in static equilibrium
Answer:
The force exerted by the string connected at the 0 cm mark is 1.078 N
Explanation:
Given;
mass of the meter stick, m = 0.1 kg
weight of the meter stick, = 0.1 kg x 9.8 m/s² = 0.98 N
the weight attached at 70 cm mark = 0.2 kg x 9.8 m/s² = 1.96 N
0cm--------------------------50cm-----------70cm-------------100cm
↑ ↓ ↓ ↑
F₁ 0.98N 1.96N F₂
Take the moment about F₂: clockwise moment = anticlockwise moment
F₁(100 - 0) = 0.98(100 -50) + 1.96(100 -70)
100F₁ = 49 + 58.8
100F₁ = 107.8
F₁ = 107.8 / 100
F₁ = 1.078 N
Therefore, the force exerted by the string connected at the 0 cm mark is 1.078 N
A 3.3 kg block of copper at a temperature of 74°C is dropped into a bucket containing a mixture of ice and water whose total mass is 1.2 kg. When thermal equilibrium is reached the temperature of the water is 8°C. How much ice was in the bucket before the copper block was placed in it? (Neglect the heat capacity of the bucket.
Answer:
0.228 kilograms of ice were in the bucket before the copper block was placed in it.
Explanation:
In this case, we assume that water inside the bucket was in the form of ice at a temperature of 0 ºC and an atmospheric pressure of 101.325 kilopascals, It is also known that block of copper is cooled down whereas ice is melted and heated up until thermal equilibrium is reached. If ice-block system is an isolated system, then First Law of Thermodynamics depicts the following model:
[tex]-Q_{b} +Q_{l,w}+Q_{s, w} = 0[/tex]
[tex]Q_{b} = Q_{l,w}+Q_{s,w}[/tex] (Eq. 1)
Where:
[tex]Q_{b}[/tex] - Heat released by the block of copper, measured in kilojoules.
[tex]Q_{l,w}[/tex] - Latent heat received by water, measured in kilojoules.
[tex]Q_{s,w}[/tex] - Sensible heat received by water, measured in kilojoules.
By definitions of sensible and latent heat, we expand the equation as follows:
[tex]m_{b}\cdot c_{b}\cdot (T_{b,o}-T) = m_{w}\cdot [L_{f}+c_{w}\cdot (T-T_{w,o})][/tex] (Eq. 2)
Where:
[tex]m_{b}[/tex] - Mass of the block of copper, measured in kilograms.
[tex]c_{b}[/tex] - Specific heat of copper, measured in kilojoules per kilogram-degree Celsius.
[tex]T_{b,o}[/tex] - Initial temperature of block of copper, measured in degrees Celsius.
[tex]m_{w}[/tex] - Mass of water, measured in kilograms.
[tex]L_{f}[/tex] - Latent heat of fusion of water, measured in kilojoules per kilogram.
[tex]T_{w,o}[/tex] - Initial temperature of water, measured in degrees Celsius.
[tex]T[/tex] - Final temperature of water-block system, measured in degrees Celsius.
[tex]c_{w}[/tex] - Specific heat of water, measured in kilojoules per kilogram-degree Celsius
And we clear the mass of water of the system:
[tex]m_{w} = \frac{m_{b}\cdot c_{b}\cdot (T_{b,o}-T)}{L_{f}+c_{w}\cdot (T-T_{w,o})}[/tex]
If we know that [tex]m_{b} = 3.3\,kg[/tex], [tex]c_{b} = 0.385\,\frac{kJ}{kg\cdot ^{\circ}C}[/tex], [tex]T_{b,o} = 74\,^{\circ}C[/tex], [tex]T = 8\,^{\circ}C[/tex], [tex]L_{f} = 334\,\frac{kJ}{kg}[/tex], [tex]c_{w} = 4.186\,\frac{kJ}{kg\cdot ^{\circ}C}[/tex] and [tex]T_{w,o} = 0\,^{\circ}C[/tex], then the mass of the ice inside the bucket is:
[tex]m_{w} = \frac{(3.3\,kg)\cdot \left(0.385\,\frac{kJ}{kg\cdot ^{\circ}C} \right)\cdot (74\,^{\circ}C-8\,^{\circ}C)}{334\,\frac{kJ}{kg}+\left(4.186\,\frac{kJ}{kg\cdot ^{\circ}C} \right)\cdot (8\,^{\circ}C-0\,^{\circ}C) }[/tex]
[tex]m_{w} = 0.228\,kg[/tex]
0.228 kilograms of ice were in the bucket before the copper block was placed in it.
A uniform plank 8.00 m in length with mass 30.0 kg is supported at two points located 1.00 m and 5.00 m, respectively, from the left-hand end. What is the maximum additional mass you could place on the right-hand end of the plank and have the plank still be at rest
Answer:
30 kg
Explanation:
Given that a uniform plank 8.00 m in length with mass 30.0 kg is supported at two points located 1.00 m and 5.00 m, respectively, from the left-hand end. What is the maximum additional mass you could place on the right-hand end of the plank and have the plank still be at rest ?
Taking the moment at the middle
Sum of Clockwise moment = sum of anticlockwise moments
Sum of clockwise moment will be:
30 × ( 4 - 1 ) = 30 × 3 = 90Nm
At equilibrium, the sum of anticlockwise moments will be equal to sum of clockwise moment.
90 = 30 × ( 5 - 4 ) + M
90 = 30 × 1 + M
90 = 30 + M
M = 90 - 30
M = 60
Moment = W × distance
Let the minimum distance = 2
60 = 4W
W = 60/2
W = 30 kg
Therefore, the maximum mass will be 30kg
A 27 kg child slides down a playground slide at a constant speed. The slide has a height of 4.0 m and is 7.0 m long. Part A Using the law of conservation of energy, find the magnitude of the kinetic friction force acting on the child.
Answer:
Explanation:
The forces acting along the playground are:
Kinetic friction force Fk and
Wsintheta (weight along the plane)
Since sintheta = opp/hyp
Sintheta = 4/7
Taking the sum of forces along the playground
\sumFx = max
Wsintheta - Fk = max
Since the speed is constant, ax = 0, the equation becomes
Wsin theta - Fk = 0
Fk = Wsintheta
From the question
W = mg
W = 27×9.81
W = 264.87N
Fk = 264.87 × 4/7
Fk = 1059.48/7
Fk = 151.35N
Hence the magnitude of the kinetic friction force acting on the child is 151.35N
Question 3: In one half hour, a car traveled 25.2 km. Its average speed is
50.4 km/min
12.6 km/hr
14 m/s
none of them
Answer:
12.6 km/hr
explanation:
25.2 x [tex]\frac{1}{2}[/tex]
