Answer:
a
Explanation:
the answer is a because it is feasible
on a horizontal axis whose unit is the meter, a linear load ranging from 0 to 1 ma a linear load distribution = 2 nC / m.
determine the modulus of the electric field created by the previous loaded bar at the point A of abscissa 2m (we have to find the relation between l, which is the distance between the elementary bar and the point A and x which sweeps the segment [0: 1]
Answer:
The correct answer is - 8.99N/C
Explanation:
[tex]dE=k\dfrac{dq}{x^2}\\ dq=\lambda{dx}\\ \lambda=2nC/m\\ dq=2dxnC\\ dE=k\dfrac{2dx}{x^2}\\ E=2k\int_1^2\dfrac{dx}{x^2}\\ E=2k(\frac{-1}{x})_1^2=k\times10^{-9}N/C\\ E=8.99\times10^9\times10^{-9}N/C\\ E=8.99N/C\\dE=k[/tex]
Let A^=6i^+4j^_2k^ and B= 2i^_2j^+3k^. find the sum and difference of A and B
Explanation:
Let [tex]\textbf{A} = 6\hat{\textbf{i}} + 4\hat{\textbf{j}} - 2\hat{\textbf{k}}[/tex] and [tex]\textbf{B} = 2\hat{\textbf{i}} - 2\hat{\textbf{j}} + 3\hat{\textbf{k}}[/tex]
The sum of the two vectors is
[tex]\textbf{A + B} = (6 + 2)\hat{\textbf{i}} + (4 - 2)\hat{\textbf{j}} + (-2 + 3)\hat{\textbf{k}}[/tex]
[tex] = 8\hat{\textbf{i}} + 2\hat{\textbf{j}} + \hat{\textbf{k}}[/tex]
The difference between the two vectors can be written as
[tex]\textbf{A - B} = (6 - 2)\hat{\textbf{i}} + (4 - (-2))\hat{\textbf{j}} + (-2 - 3)\hat{\textbf{k}}[/tex]
[tex]= 4\hat{\textbf{i}} + 6\hat{\textbf{j}} - 5\hat{\textbf{k}}[/tex]
What do scientists use to determine the temperature of a star?
Answer:
Measure the brightness of a star through two filters and compare the ratio of red to blue light. Compare to the spectra of computer models of stellar spectra of different temperature and develop an accurate color-temperature relation.
In many cartoon shows, a character runs of a cliff, realizes his predicament and lets out a scream. He continues to scream as he falls. If the physical situation is portrayed correctly, from the vantage point of an observer at the foot of the cliff, the pitch of the scream should be Group of answer choices
Answer:
Increasing until terminal velocity is reached
Explanation:
Provided the scream is a constant pitch at the source, Doppler effect will make the pitch increase as the velocity of the source towards the listener increases.
find the rate of energy radiated by a man by assuming the surface area of his body 1.7m²and emissivity of his body 0.4
The rate of energy radiated by the man is 3.86 x [tex]10^{-8}[/tex] J/s. [tex]m^{2}[/tex].
The amount of energy radiated by an object majorly depends on the area of its surface and its temperature. The is well explained in the Stefan-Boltzmann's law which states that:
Q(t) = Aeσ[tex]T^{4}[/tex]
where: Q is the quantity of heat radiated, A is the surface area of the object, e is the emmisivity of the object, σ is the Stefan-Boltzmann constant and T is the temperature of the object.
To determine the rate of energy radiated by the man in the given question;
[tex]\frac{Q(t)}{T^{4} }[/tex] = Aeσ
But A = 1.7 m², e = 0.4 and σ = 5.67 x [tex]10^{-8}[/tex] J/s.
So that;
[tex]\frac{Q(t)}{T^{4} }[/tex] = 1.7 * 0.4 * 5.67 x [tex]10^{-8}[/tex]
= 3.8556 x [tex]10^{-8}[/tex]
= 3.86 x [tex]10^{-8}[/tex] J/s. [tex]m^{2}[/tex]
Thus, the rate of energy radiated by the man is 3.86 x [tex]10^{-8}[/tex] J/s. [tex]m^{2}[/tex].
Learn more on energy radiation of objects by visiting: https://brainly.com/question/12550129
A 2.0 kg puck is at rest on a level table. It is pushed straight north with a constant force of 5N for 1.50 s and then let go. How far does the puck move from rest in 2.25 s?
Answer:
d = 6.32 m
Explanation:
Given that,
The mass of a puck, m = 2 kg
It is pushed straight north with a constant force of 5N for 1.50 s and then let go.
We need to find the distance covered by the puck when move from rest in 2.25 s.
We know that,
F = ma
[tex]a=\dfrac{F}{m}\\\\a=\dfrac{5}{2}\\\\a=2.5\ m/s^2[/tex]
Let d is the distance moved in 2.25 s. Using second equation of motion,
[tex]d=ut+\dfrac{1}{2}at^2\\\\d=0+\dfrac{1}{2}\times 2.5\times (2.25)^2\\\\d=6.32\ m[/tex]
So, it will move 6.32 m from rest in 2.25 seconds.
Three wires are connected at a branch point. One wire carries a positive current of 18 A into the branch point, and a second wire carries a positive current of 7 A away from the branch point. Find the current carried by the third wire into the branch point.
Answer:
The current in third branch is 11 A.
Explanation:
incoming current in one branch = 18 A
outgoing current in the other branch = 7 A
let the current in the third branch is i.
According to the Kirchoff's fist law in electricity
incoming current = out going current
18 = 7 + i
i = 11 A
The current in third branch is 11 A.
ai là người phát hiện trái đất hình cầu đầu tiên ?
Answer:
Can't understand the language
What is the percentage of the population that wanted both the swimming pool and the soccer complex? Use your knowledge
of the addition rule and the Venn diagram to answer.
Answer:
The percentage of people who wanted both the swimming pool and the soccer complex is 0.6 + 0.6 – 0.95 = 0.25. This can also be seen in the Venn diagram.
Explanation:
Edmentum
What are the examples of pulley? Plz tell the answer as fast as possible plz.
Answer:
elevators
Theatre system
construction pulley
lifts
Answer:
elevator,cargo lift system
Trình bày những hiểu biết của em về đại lượng vận tốc dài, vận tốc góc(định nghĩa, công thức, ý nghĩa, đơn vị, loại đại lượng).
how can you convert galvanometer into ammeter?
Answer:
A galvanometer is converted into an ammeter by connecting a low resistance in parallel with the galvanometer.
Explanation:
This low resistance is called shunt resistance S. The scale is now calibrated in ampere and the range of the ammeter depends on the values of the shunt resistance.
The masses of two heavenly bodies are 2×10‘16’ and 4×10 ‘22’ kg respectively and the distance between than is 30000km. find the gravitational force between them ? ans. 2.668× 10-9N
[tex]F = 5.93×10^{13}\:\text{N}[/tex]
Explanation:
Given:
[tex]m_1= 2×10^{16}\:\text{kg}[/tex]
[tex]m_2= 4×10^{22}\:\text{kg}[/tex]
[tex]r = 30000\:\text{km} = 3×10^7\:\text{m}[/tex]
Using Newton's universal law of gravitation, we can write
[tex]F = G\dfrac{m_1m_2}{r^2}[/tex]
[tex]\:\:\:\:=(6.674×10^{-11}\:\text{N-m}^2\text{/kg}^2)\dfrac{(2×10^{16}\:\text{kg})(4×10^{22}\:\text{kg})}{(3×10^7\:\text{m})^2}[/tex]
[tex]\:\:\:\:= 5.93×10^{13}\:\text{N}[/tex]
Sometimes the units for an electric field are written as N/C, while other times the units are written as V/m, using dimensional analysis show that N/C is equal to V/m.
a. True
b. False
Answer:
N/C = V/m.
Explanation:
The SI unit of electric field is N/C. Sometimes the units are written as V/m.
We know that,
1 V = 1 J/C
Using dimensional analysis,
The dimensional formula for Joules is [M¹L² T⁻²].
The dimensional form of coulomb is [M⁰ L⁰ T¹ I¹].
So,
J/C = [M¹L² T⁻³I¹] ...(1)
The dimensional formula of Newton is [M¹ L¹ T⁻²]
The dimensional form of coulomb is [M⁰ L⁰ T¹ I¹].
N/C= [M¹L² T⁻³I¹] ....(2)
From (1) and (2) it is clear that the N/C is equal to V/m.
point charges q1=50 uc and q2=-25 uc are placed 1 m apart. what is the force on a third chare q3=2 uc placed midway between q1 and q2? where must q3 of the preceding problem be placed so that the net force on it is zero?
Answer:
d = -1 m
The negative sign indicates that the charge from that force of the space of the two spheres.
Explanation:
That is a problem of electric forces, given by Coulomb's law
F = [tex]k \frac{ q1q2}{r^2}[/tex]
We use that charges of the same sign repel and charges of different signs do not attract, so the net force is
∑ = F₁₃ + F₂₃
F_ {net} = [tex]k \frac{q_1q_3}{r_{13}^2} + k \frac{q_2q_3}{ r_{23}^}[/tex]
a) the charge is placed at the midpoint between the other two
r₁₃ = r₁₂ = R = ½ m = 0.5mF_ {net} =[tex]\frac{k}{R^2 } \ q3 ( q1+q2)[/tex]
calculate us
F_ {net} = 9 10⁹ / 0.5² 2 10⁻⁶ (50 -25) 10⁻⁶
F_ {net} = 1,800 N
b) where must be placed q3 so that the force is zero
for this case the charge q3 is outside the spheres
∑ F = 0
F₁₁₃ = F₂₃
k q_1 / r_{13}² = k q₂ q₃ / r₂₃²
q₁/ r₁₂² = q₂ / r₂₃²
suppose the distance
r₁₂ = d
the he other sphere is
r₂₃ = d + 1
we substitute
q₃ / d² = q₂ / (d + 1) ²
(d + 1) ² = q₂ / q₃ d²
d² (1 - q₂/ q₃) + 2d + 1 = 0
we solve the equation of a second
d = [-2 + [tex]\sqrt{2^2 - 4 1 ( 1+25/50}[/tex] ] / 2
d = -2 /2
d = -1 m
The negative sign indicates that the charge from that force of the space of the two spheres.
Accommodation of the eye refers to its ability to __________. see on both the brightest days and in the dimmest light see both in air and while under water move in the eye socket to look in different directions focus on both nearby and distant objects
Answer:
to adjust from distant to the near objects
Explanation:
The process of accommodation is achieved by changing in the shape and position of the eye ball. Just like adjusting the lens of the camera.Answer:
The ability of eye lens to change the focal length of eye lens is called accommodation power of eye.
Explanation:
The human eye is the optical instrument which works on the refraction of light.
The ability of eye lens to change its focal length is called accommodation power of eye.
The focal length of eye lens is changed by the action of ciliary muscles.
When the ciliary muscles are relaxed then the thickness of lens is more and thus the focal length is small. When the ciliary muscles is stretched, the lens is thin and then the focal length is large.
A spherical, concave shaving mirror has a radius of curvature of 0.983 m. What is the magnification of a person's face when it is 0.155 m from the vertex of the mirror (answer sign and magnitude)
Answer:
Magnification = 1
Explanation:
given data
radius of curvature r = - 0.983 m
image distance u = - 0.155
solution
we get here first focal length that is
Focal length, f = R/2 ...................1
f = -0.4915 m
we use here formula that is
[tex]\frac{1}{v} + \frac{1}{u} + \frac{1}{f}[/tex] .................2
put here value and we get
[tex]\frac{1}{v} = \frac{1}{0.155} - \frac{1}{4915}[/tex]
v = 0.155 mso
Magnification will be here as
m = [tex]- \frac{v}{u}[/tex]
m = [tex]\frac{0.155}{0.155}[/tex]
m = 1Answer:
The magnification is 1.5.
Explanation:
radius of curvature, R = - 0.983 m
distance of object, u = - 0.155 m
Let the distance of image is v.
focal length, f = R/2 = - 0.492 m
Use the mirror equation
[tex]\frac{1}{f}=\frac{1}{v}+\frac {1}{u}\\\\\frac{-1}{0.492}=\frac{1}{v}-\frac{1}{0.155}\\\\\frac{1}{v}=\frac{1}{0.155}-\frac{1}{0.492}\\\\\frac{1}{v}=\frac{0.492-0.155}{0.155\times 0.492}\\\\\frac{1}{v}=\frac{0.337}{0.07626}\\ \\v = 0.226 m[/tex]
The magnification is given by
m = - v/u
m = 0.226/0.155
m = 1.5
A mass weighing 4 lb stretches a spring 4in. Suppose the mass is given an additional in displacement downwards and then released. Assuming no friction and no external force, the natural frequency W (measured in radians per unit time) for the system is? (Recall that the acceleration due to gravity is 32ft/sec2).
a) None of the other alternatives is correct.
b) W = v2 3
c)w=212
d) w = 4/6
e) w=213
Answer:
4√6 rad/s
Explanation:
Since the spring is initially stretched a length of x = 4 in when the 4 lb mass is placed on it, since it is in equilibrium, the spring force, F = kx equals the weight of the mass W = mg.
So, W = F
mg = kx where m = mass = 4lb, g = acceleration due to gravity = 32 ft/s², k = spring constant and x = equilibrium displacement of spring = 4 in = 4 in × 1ft /12 in = 1/3 ft
making k the spring constant subject of the formula, we have
k = mg/x
substituting the values of the variables into the equation, we have
k = mg/x
k = 4 lb × 32 ft/s² ÷ 1/3 ft
k = 32 × 4 × 3
k = 384 lbft²/s²
Now, assuming there is no friction and no external force, we have an undamped system.
So, the natural frequency for an undamped system, ω = √(k/m) where k = spring constant = 384 lbft²/s² and m = mass = 4 lb
So, substituting the values of the variables into the equation, we have
ω = √(k/m)
ω = √(384 lbft²/s² ÷ 4 lb)
ω = √96
ω = √(16 × 6)
ω = √16 × √6
ω = 4√6 rad/s
An 1800-W toaster, a 1400-W electric frying pan, and a 55-W lamp are plugged into the same outlet in a 15-A, 120-V circuit. (The three devices are in parallel when plugged into the same socket.)
a. Will this combination blow the 15-A fuse?
b. What current is drawn by each device?
Being in parallel each device will have an equal voltage drop of 120 V
A. Yes the combination will blow the fuse. See part B for the total current.
B. Toaster = 1800W / 120V = 15A
Frying Pan = 1400W / 120V = 11.67A
Lamp = 55W / 120V = 0.458A
Total amps = 15 + 11.67 + 0.458 = 27.128 Amps
27.128A is greater than 15A so the fuse will blow.
calculate the length of wire.
Answer:
L = 169.5 m
Explanation:
Using Ohm's Law:
V = IR
where,
V = Voltage = 1.5 V
I = Current = 10 mA = 0.01 A
R = Resistance = ?
Therefore,
1.5 V = (0.01 A)R
R = 150 Ω
But the resistance of a wire is given by the following formula:
[tex]R = \frac{\rho L}{A}[/tex]
where,
ρ = resistivity = 1 x 10⁻⁶ Ω.m
L = length of wire = ?
A = cross-sectional area of wire = πr² = π(0.6 mm)² = π(0.6 x 10⁻³ m)²
A = 1.13 x 10⁻⁶ m²
Therefore,
[tex]150\ \Omega = \frac{(1\ x\ 10^{-6}\ \Omega .m)L}{1.13\ x\ 10^{-6}\ m^2}\\\\L = \frac{150\ \Omega(1.13\ x\ 10^{-6}\ m^2)}{1\ x\ 10^{-6}\ \Omega .m}\\\\[/tex]
L = 169.5 m
Michelson and Morley concluded from the results of their experiment that Group of answer choices the experiment was successful in not detecting a shift in the interference pattern. the experiment was a failure since they detected a shift in the interference pattern. the experiment was a failure since there was no detectable shift in the interference pattern. the experiment was successful in detecting a shift in the interference pattern.
Answer:
The results of the experiment indicated a shift consistent with zero, and certainly less than a twentieth of the shift expected if the Earth's velocity in orbit around the sun was the same as its velocity through the ether.
Explanation:
In these formulas, it is useful to understand which variables are parameters that specify the nature of the wave. The variables E0E0E_0 and B0B0B_0 are the __________ of the electric and magnetic fields. Choose the best answer to fill in the blank.
Suppose you exert a force of 314 N tangential to a grindstone (a solid disk) with a radius of 0.281 m and a mass of 84.2 kg What is the resulting angular acceleration of the grindstone assuming negligible opposing friction
Answer:
The angular acceleration is 26.6 rad/s^2.
Explanation:
Force, F = 314 N
radius, r = 0.281 m
mass, m = 84.2 kg
The grindstone is a disc.
The torque is given by
torque = force x radius
Torque = 314 x 0.281 = 88.234 Nm
The torque is given by
Torque = Moment of inertia x angular acceleration
[tex]88.234 = 0.5 mr^2 \alpha \\\\88.234 = 0.5\times 84.2\times 0.281\times 0.281\times \alpha \\\\\alpha = 26.6 rad/s^2[/tex]
A uniform magnetic field passes through a horizontal circular wire loop at an angle 15.1° from the normal to the plane of the loop. The magnitude of the magnetic field is 3.35 T , and the radius of the wire loop is 0.240 m . Find the magnetic flux Φ through the loop.
Answer:
0.5849Weber
Explanation:
The formula for calculating the magnetic flus is expressed as:
[tex]\phi = BAcos \theta[/tex]
Given
The magnitude of the magnetic field B = 3.35T
Area of the loop = πr² = 3.14(0.24)² = 0.180864m²
angle of the wire loop θ = 15.1°
Substitute the given values into the formula:
[tex]\phi = 3.35(0.180864)cos15.1^0\\\phi =0.6058944cos15.1^0\\\phi =0.6058944(0.9655)\\\phi = 0.5849Wb[/tex]
Hence the magnetic flux Φ through the loop is 0.5849Weber
A 100 kg man is one fourth of the way up a 4.0 m ladder that is resting against a smooth, frictionless wall. The ladder has mass 25 kg and makes an angle of 56 degrees with the ground. What is the magnitude of the force of the wall on the ladder at the point of contact, if this force acts perpendicular to the wall and points away from the wall
Answer:
[tex]N_f=248N[/tex]
Explanation:
From the question we are told that:
Mass [tex]m=100kg[/tex]
Ladder Length [tex]l=4.0m[/tex]
Mass of Ladder [tex]m_l=25kg[/tex]
Angle [tex]\theta=56 \textdegree[/tex]
Generally the equation for Co planar forces is mathematically given by
[tex]mgcos \theta *2+Mgcos\theta*1 -N_fsin \theta*4=0[/tex]
Therefore
[tex]25*9.81cos 56 *2+100*9.81cos56*1 -N_fsin 56*4=0[/tex]
[tex]N_f=248N[/tex]
In a game of pool, the cue ball moves at a speed of 2 m/s toward the eight ball. When the cue ball hits the eight ball, the cue ball bounces off with a speed of 0.8 m/s at an angle of 20', as shown in the diagram below. Both balls have a mass of 0.6 kg.
a) what is the momentum of the system before the collision
b) what is the momentum after the collision
c) what angle dose the right ball travel after the collision
d) what is the magnitude of the eight balls velocity after the collision
Answer:
a) p₀ = 1.2 kg m / s, b) p_f = 1.2 kg m / s, c) θ = 12.36, d) v_{2f} = 1.278 m/s
Explanation:
For this exercise we define a system formed by the two balls, which are isolated and the forces during the collision are internal, therefore the moment is conserved
a) the initial impulse is
p₀ = m v₁₀ + 0
p₀ = 0.6 2
p₀ = 1.2 kg m / s
b) as the system is isolated, the moment is conserved so
p_f = 1.2 kg m / s
we define a reference system where the x-axis coincides with the initial movement of the cue ball
we write the final moment for each axis
X axis
p₀ₓ = 1.2 kg m / s
p_{fx} = m v1f cos 20 + m v2f cos θ
p₀ = p_f
1.2 = 0.6 (-0.8) cos 20+ 0.6 v_{2f} cos θ
1.2482 = v_{2f} cos θ
Y axis
p_{oy} = 0
p_{fy} = m v_{1f} sin 20 + m v_{2f} cos θ
0 = 0.6 (-0.8) sin 20 + 0.6 v_{2f} sin θ
0.2736 = v_{2f} sin θ
we write our system of equations
0.2736 = v_{2f} sin θ
1.2482 = v_{2f} cos θ
divide to solve
0.219 = tan θ
θ = tan⁻¹ 0.21919
θ = 12.36
let's look for speed
0.2736 = v_{2f} sin θ
v_{2f} = 0.2736 / sin 12.36
v_{2f} = 1.278 m / s
A object of mass 3.00 kg is subject to a force Fx that varies with position as in the figure below. A coordinate plane has a horizontal axis labeled x (m) and a vertical axis labeled Fx (N). There are three line segments. The first segment runs from the origin to (4,3). The second segment runs from (4,3) to (11,3). The third segment runs from (11,3) to (17,0). (a) Find the work done by the force on the object as it moves from x = 0 to x = 4.00 m. J (b) Find the work done by the force on the object as it moves from x = 4.00 m to x = 11.0 m. J (c) Find the work done by the force on the object as it moves from x = 11.0 m to x = 17.0 m. J (d) If the object has a speed of 0.450 m/s at x = 0, find its speed at x = 4.00 m and its speed at x = 17.0 m.
Answer:
Explanation:
An impulse results in a change of momentum.
The impulse is the product of a force and a distance. This will be represented by the area under the curve
a) W = ½(4.00)(3.00) = 6.00 J
b) W = (11.0 - 4.00)(3.00) = 21.0 J
c) W = ½(17.0 - 11.0)(3.00) = 9.00 J
d) ASSUMING the speed at x = 0 is in the direction of applied force
½(3.00)(v₄²) = ½(3.00)(0.450²) + 6.00
v₄ = 2.05 m/s
½(3.00)(v₁₇²) = ½(3.00)(0.450²) + 6.00 + 21.0 + 9.00
v₁₇ = 4.92 m/s
If the initial speed is NOT in the direction of applied force, the final speed will be slightly less in both cases.
a. Do the waves made by the two faucets travel faster than the waves made by just one faucet?
b. How do you know this? Describe how the two-faucet wave pattern compares with the one-faucet pattern.
c. Describe what happens to the two-faucet wave pattern as the separation of the faucets is increased.
Answer:
asdasd dsa dasdasd sadas dasd asdasd asd asd dsa asdd 223 aasd ada dasd sa dasd dsaa sd adsd asasd
Explanation:
A planet of mass m = 4.25 x 1024 kg orbits a star of mass M = 6.75 x 1029 kg in a circular path. The radius of the orbits R = 8.85 x 107 km. What is the orbital period Tplanet of the planet in Earth days?
285.3 days
Explanation:
The centripetal force [tex]F_c[/tex] experienced by the planet is the same as the gravitational force [tex]F_G[/tex] so we can write
[tex]F_c = F_G[/tex]
or
[tex]m\dfrac{v^2}{R} = G\dfrac{mM}{R^2}[/tex]
where M is the mass of the star and R is the orbital radius around the star. We know that
[tex]v = \dfrac{C}{T} = \dfrac{2\pi R}{T}[/tex]
where C is the orbital circumference and T is orbital period. We can then write
[tex]\dfrac{4\pi^2R}{T^2} = G\dfrac{M}{R^2}[/tex]
Isolating [tex]T^2[/tex], we get
[tex]T^2 = \dfrac{4\pi^2R^3}{GM}[/tex]
Taking the square root of the expression above, we get
[tex]T = 2\pi \sqrt{\dfrac{R^3}{GM}}[/tex]
which turns out to be [tex]T = 2.47×10^7\:\text{s}[/tex]. We can convert this into earth days as
[tex]T = 2.47×10^7\:\text{s}×\dfrac{1\:\text{hr}}{3600\:\text{s}}×\dfrac{1\:\text{day}}{24\:\text{hr}}[/tex]
[tex]\:\:\:\:\:= 285.3\:\text{days}[/tex]
A spherically mirrored ball is slowly lowered at New Years Eve as midnight approaches. The ball has a diameter of 8.0 ft. Assume you are standing directly beneath it and looking up at the ball. When your reflection is half your size then the mirror is _______ ft above you.
Answer:
The distance between mirror and you is 2 ft.
Explanation:
diameter, d = 8 ft
radius of curvature, R = 4 ft
magnification, m = 0.5
focal length, f = R/2 = 4/2 = 2 ft
let the distance of object is u and the distance of image is v.
[tex]\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\\\\\frac{1}{2}=\frac{1}{v}+\frac{1}{u}\\\\v = \frac {2 u}{u - 2}[/tex]
Use the formula of magnification
[tex]m = \frac{v}{u}\\\\0.5 =\frac { u}{u - 2}\\ \\u - 2 = 2 u \\\\u = -2 ft[/tex]
