Answer:
Explanation:
The balloon would require a time of
t = d/v = 13.5/ (23.6cos38) = 0.7259...s
to travel the horizontal distance.
the vertical position relative to the throw point at that time is
h = 0 + (23.6sin38)(0.7259) + ½(-9.8)(0.7259²)
h = 7.9652...
so as long as the adjacent building is at least 8.0 m higher than the student position, the balloon is in the air for 0.726 s.
If the building is shorter than 8.0 m above the student, the balloon will land on the building roof and will be in the air for a longer period of time
A car's bumper is designed to withstand a 6.12 km/h (1.7-m/s) collision with an immovable object without damage to the body of the car. The bumper cushions the shock by absorbing the force over a distance. Calculate the magnitude of the average force on a bumper that collapses 0.210 m while bringing a 810 kg car to rest from an initial speed of 1.7 m/s.
Answer:
5572.8 N
Explanation:
Applying,
F = ma.............. Equation 1
Where F = Force, m = mass of the car, a = acceleration.
We can find a by applying,
v² = u²+2as............. Equation 2
Where v = final velocity, u = initial velocity, a = acceleration, = distance.
From the question,
Given: v = 0 m/s (come to rest), u = 1.7 m/s, s = 0.210 m
Substitute these value into equation 2
0² = 1.7²+2×0.21×a
a = -1.7²/(2×0.21)
a = -2.89/0.42
a = -6.88 m/s²
Also given: m = 810 kg
Substitute these value into equation 1
F = 810(-6.88)
F = -5572.8 N
Hence the force on the bumber is 5572.8 N
You have a 1 W light bulb in your lab. It puts out light of only 1 frequency. The wavelength of this light is 500nm. you set up a detector with a surface area of 1 square centimeter facing the light source at a distance of 100m.
Required:
a. Find the number of photons hitting the detector every second.
b. What is the maximum E field of the E M wave hitting the detector?
c. What is the maximum value of the B field of this E M wave?
d. How far away would you have to place the detector to only receive 1 photon per second from the light bulb?
Answer:
a) # _photon = 2.5 10¹⁸ photons / s, b) E = 10⁻² N / C, c) B = 3 10⁻¹¹ T
d) r= 2 10⁹ m
Explanation:
a) Let's solve this exercise in part, let's start by finding the energy of each photon using the Planck relation
E₀ = h f
c = λ f
E₀ = h c /λ
E₀ = 6.63 10⁻³³⁴ 3 10⁸/500 10⁻⁹
E₀ = 3.978 10⁻⁻¹⁹ J
Let's use a direct ratio rule to find the number of photons
#_foton = E / Eo
#_fototn = 1 / 3.978 10⁻¹⁹
# _photon = 2.5 10¹⁸ photons / s
b) The intensity received by the detector is related to the electric field
I = E²
Let's look for the intensity that the detector receives, suppose that the emission is shapeless throughout the space
I = P / A
P = I A
Let's use index 1 for the point on the bulb and index 2 for the point on the detector.
The area of a sphere is
A = 4π r²
P = I₁ A₁ = I₂ A₂
I₁ r₁² = I₂ r₂²
I₂ = I₁ r₁²/r₂²
I₂ = I₁ 1 / 100²
I₂ = I₁ 10⁻⁴
we must know the intensity at the output of the bulb suppose that I₁ = 1 J
I₂ = 10⁻⁴ J
let's look for the electric field
E =√I
E = √10⁻⁴
E = 10⁻² N / C
c) for the calculation of the magnetic field we use that the field is in phase
E / B = c
B = E / c
B = 10⁻² / 3 10⁸
B = 3 10⁻¹¹ T
d) Let's use a direct proportions rule if we fear 2.5 10¹⁸ photons in an area A = 4π R² where R = 100 m how many photons are there in the area of the detector r = 1 cm, A’= 10⁻⁴ m²
#_photons = 2.5 10¹⁸ A_detector / A_sphere
#_photons = 2.5 1018 10-4 / 4π 10⁴
#_photons = 2 10⁹ photons in the detector area
for the number of photons to decrease to 1, the radius of the sphere must be 2 10⁹ m
You and your friends find a rope that hangs down 11 m from a high tree branch right at the edge of a river. You find that you can run, grab the rope, swing out over the river, and drop into the water. You run at 2.0 m/s and grab the rope, launching yourself out over the river.
Required:
How long must you hang on if you want to drop into the water at the greatest possible distance from the edge?
Answer:
if you want to drop into the water at the greatest possible distance from the edge, you must hang for 1.662s.
Explanation:
The time period of the oscillation is,
[tex]T = 2\pi \sqrt{ \frac{I} {g }[/tex]
[tex]T = 2\pi \sqrt{\frac{11}{9.8} } \\\\T= 6.65 s[/tex]
This would be the time taken for the person to move from.
The duration of time he hangs over the river be one-fourth of the time period.
Here,
[tex]t= \frac{T}{4} \\\\t=\frac{6.65}{4}\\\\t = 1.662 s[/tex]
Consider a box with two gases separated by an impermeable membrane. The membrane can move back and forth, but the gases cannot penetrate the membrane. The left side is filled with gas A and the right side is filled with gas B. We will assume that equipartition applies to both gases, but gas A has an excluded volume due to large molecules so its entropy has a different formula.
SA=NAkln(VA+ bNA)+f(UA,NA)
SB=NBkln(VB)+f(UB,NB)
Required:
If NA= 1 moles, NB = 2 moles, the total volume of the box is 1 m3, and b= 4 × 10-4 m3/mole, then find the equilibrium value of VA by maximizing the total entropy.
Answer:
The answer is "[tex]0.3336\ m^3[/tex]"
Explanation:
Using the Promideal gas law:
[tex]P_A=P_B\\\\P_A(V_A-\eta_A b)= \eta_A RT......(1)\\\\P_B V_B=\eta_B \bar{R}T........(2)\\\\From (1) \zeta (2)\\\\[/tex]
[tex]\frac{\eta_A}{V_A-\eta_A b}=\frac{\eta B}{V B}\\\\ \frac{V A- \eta_A b}{V B}=\frac{\eta A}{\eta B }\\\\ \frac{V A-b}{V B}=\frac{1}{2}\\\\V A+V B=1\\\\V B =1- V A\\\\\frac{V A-b}{1-V A}=\frac{1}{2}\\\\2V A-2b=1-V A\\\\3 V A=1+2b\\\\V A=\frac{1+2b}{3}\\\\[/tex]
[tex]=\frac{1+2(4\times 10^{-4})}{3}\\\\=0.3336\ m^3[/tex]
The equilibrium value of Va is 0.3336 m³.
Ideal gas lawThe equilibrium value of Va is determine by applying ideal gas law as shown below;
Pressure of gas A = Pressure of gas B
Pa = Pb
Pa(Va - nab) = naRT----(1)
PbVb = nbRT -----(2)
Solve equation (1) and (2)
[tex]\frac{P_b}{RT} = \frac{n_b}{V_b} \\\\\frac{P_b}{P_a(V_a- n_ab)/n_a} = \frac{n_b}{V_b}\\\\\frac{n_a}{V_a - n_ab} = \frac{n_b}{V_b} \\\\\frac{V_a - n_ab}{V_b} = \frac{n_a}{n_b} \\\\\frac{V_a - b}{V_b} = \frac{1}{2}[/tex]
Va + Vb = 1
Vb = 1 - Va
[tex]\frac{V_a - b}{1 - V_a} = \frac{1}{2}[/tex]
2Va - 2b = 1 - Va
3Va = 1 + 2b
[tex]V_ a = \frac{1 + 2b}{3} \\\\V_a = \frac{1 + (2 \times 4\times 10^{-4})}{3} \\\\V_a = 0.3336 \ m^3[/tex]
Thus, the equilibrium value of Va is 0.3336 m³.
Learn more about equilibrium value here: https://brainly.com/question/22569960
F=(4i+3j)N acts on an object of mass m=2k.g and drags it by moving the object from origion to x=5m. Find the workdone on the object and the angle between the force and the displacement
Answer:
nnnjjdndbsnnshfhhgbfbdbdh
How can I solve the following?
In (Figure 1), let V = 15.0 V and C1=C2=C3= 24.2 μF.
Part A: How much energy is stored in the capacitor network as shown in (Figure 1)?
Part B: How much energy would be stored in the capacitor network if the capacitors were all in series?
Part C: How much energy would be stored in the capacitor network if the capacitors were all in parallel?
Answer:
Part A - 4.084 mJ
Part B - 0.908 mJ
Part C - 8.168 mJ
Explanation:
Part A: How much energy is stored in the capacitor network as shown in (Figure 1)?
Since capacitors C₂ and C₃ are in series, their equivalent capacitance is C',
1/C' = 1/C₂ + 1/C₃ (Since C₁ = C₂ = C₃ = C)
1/C' = 1/C + 1/C
1/C' = 2/C
C' = C/2
Since C' is in parallel with C₁, the equivalent capacitance for the circuit is C" = C₁ + C' = C + C/2 = 3C/2
C" = 3C/2
The energy stored in the circuit, W = 1/2C"V² where C" = equivalent capacitance = 3C/2 and V = voltage = 15.0 V
W = 1/2C"V²
W = 1/2(3C/2)V²
W = 3CV²/4
since C = 24.2 μF = 24.2 × 10⁻⁶ F
W = 3CV²/4
W = 3 × 24.2 × 10⁻⁶ F (15.0 V)²/4
W = 3 × 24.2 × 10⁻⁶ F × 225 V²/4
W = 16335/4 × 10⁻⁶ FV²
W = 4083.75 × 10⁻⁶ J
W = 4.08375 × 10⁻³ J
W = 4.08375 mJ
W ≅ 4.084 mJ
Part B: How much energy would be stored in the capacitor network if the capacitors were all in series?
If the capacitors are connected in series, their equivalent resistance is C'
and 1/C' = 1/C₁ + 1/C₂ + 1/C₃
Since C₁ = C₂ = C₃ = C
1/C' = 1/C + 1/C + 1/C
1/C' = 3/C
C' = C/3
The energy stored in the circuit, W = 1/2C'V² where C' = equivalent capacitance = C/3 and V = voltage = 15.0 V
W = 1/2C'V²
W = 1/2(C/3)V²
W = CV²/6
since C = 24.2 μF = 24.2 × 10⁻⁶ F
W = CV²/6
W = 24.2 × 10⁻⁶ F (15.0 V)²/6
W = 24.2 × 10⁻⁶ F × 225 V²/6
W = 5445/6 × 10⁻⁶ FV²
W = 907.5 × 10⁻⁶ J
W = 0.9075 × 10⁻³ J
W = 0.9075 mJ
W ≅ 0.908 mJ
Part C: How much energy would be stored in the capacitor network if the capacitors were all in parallel?
If the capacitors are connected in parallel, their equivalent resistance is C'
and C' = C₁ + C₂ + C₃
Since C₁ = C₂ = C₃ = C
C' = C + C + C
C' = 3C
The energy stored in the capacitor network, W = 1/2C'V² where C' = equivalent capacitance = 3C and V = voltage = 15.0 V
W = 1/2C'V²
W = 1/2(3C)V²
W = 3CV²/2
since C = 24.2 μF = 24.2 × 10⁻⁶ F
W = 3CV²/2
W = 3 × 24.2 × 10⁻⁶ F (15.0 V)²/2
W = 3 × 24.2 × 10⁻⁶ F × 225 V²/2
W = 16335/2 × 10⁻⁶ FV²
W = 8167.5 × 10⁻⁶ J
W = 8.1675 × 10⁻³ J
W = 8.1675 mJ
W ≅ 8.168 mJ
A stationary horn emits a sound with a frequency of 228 Hz. A car is moving toward the horn on a straight road with constant speed. If the driver of the car hears the horn at a frequency of 246 Hz, then what is the speed of the car? Use 340 m/s for the speed of the sound
Answer: 26.84 m/s
Explanation:
Given
Original frequency of the horn [tex]f_o=228\ Hz[/tex]
Apparent frequency [tex]f'=246\ Hz[/tex]
Speed of sound is [tex]V=340\ m/s[/tex]
Doppler frequency is
[tex]\Rightarrow f'=f_o\left(\dfrac{v+v_o}{v-v_s}\right)[/tex]
Where,
[tex]v_o=\text{Velocity of the observer}\\v_s=\text{Velocity of the source}[/tex]
Insert values
[tex]\Rightarrow 246=228\left[\dfrac{340+v_o}{340-0}\right]\\\\\Rightarrow 366.84=340+v_o\\\Rightarrow v_o=26.8\ m/s[/tex]
Thus, the speed of the car is [tex]26.84\ m/s[/tex]
What type of wave is a microwave?
O heat
O longitudinal
sound
transverse
Answer:
Microwave is a types of a electromagnetic radiation
Answer:
Transvers
Explanation:
Because microwave is electromagnetic waves and all electromagnetic waves are transvers.
1:
Forces and Motion:Question 2
A car is travelling east, when suddenly a more massive car travelling
north hits it with a greater force. What is likely to happen to the car
that was originally travelling east?
Explanation:
the car will be brought back
Explain how newton's first law of motion follows from second law?
Answer:
Newton's First Law of Motion states that a body will stay at rest or continue its path with constant velocity unless an external force acts upon it. Newton's Second Law of Motion states that the net force that acts upon a body is equal to the mass of the body multiplied by the acceleration due to the net force.
A lightning bolt has a current of 56,000 A and lasts for 80 x 10-6 seconds (80 μs). How much charge (in Coulombs) has flowed in this bolt?
Answer:
A cloud can discharge as much as 20 coulombs in a lightning bolt.
A proton traveling at 17.6° with respect to the direction of a magnetic field of strength 3.28 mT experiences a magnetic force of 9.14 × 10-17 N. Calculate (a) the proton's speed and (b) its kinetic energy in electron-volts.
Answer:
a) The proton's speed is 5.75x10⁵ m/s.
b) The kinetic energy of the proton is 1723 eV.
Explanation:
a) The proton's speed can be calculated with the Lorentz force equation:
[tex] F = qv \times B = qvBsin(\theta) [/tex] (1)
Where:
F: is the force = 9.14x10⁻¹⁷ N
q: is the charge of the particle (proton) = 1.602x10⁻¹⁹ C
v: is the proton's speed =?
B: is the magnetic field = 3.28 mT
θ: is the angle between the proton's speed and the magnetic field = 17.6°
By solving equation (1) for v we have:
[tex]v = \frac{F}{qBsin(\theta)} = \frac{9.14 \cdot 10^{-17} N}{1.602\cdot 10^{-19} C*3.28 \cdot 10^{-3} T*sin(17.6)} = 5.75 \cdot 10^{5} m/s[/tex]
Hence, the proton's speed is 5.75x10⁵ m/s.
b) Its kinetic energy (K) is given by:
[tex] K = \frac{1}{2}mv^{2} [/tex]
Where:
m: is the mass of the proton = 1.67x10⁻²⁷ kg
[tex] K = \frac{1}{2}mv^{2} = \frac{1}{2}1.67 \cdot 10^{-27} kg*(5.75 \cdot 10^{5} m/s)^{2} = 2.76 \cdot 10^{-16} J*\frac{1 eV}{1.602 \cdot 10^{-19} J} = 1723 eV [/tex]
Therefore, the kinetic energy of the proton is 1723 eV.
I hope it helps you!
a stone is thrown vertically upwards with a velocity of 20 m per second determine the total time of flight of stone in air
Answer:
Explanation:
The best way to do this is to remember the rule about the halfway mark in a parabolic path. At a trajectory's half way point in its travels, it will be at its max height. To get the total time in the air, we take that time at half way and double it. Here's what we know that we are told:
initial velocity is 20 m/s
Here's what we know that we are NOT told:
a = -9.8 m/s/s and
final velocity is 0 at an object's max height in parabolic motion.
We will use the equation:
[tex]v=v_0+at[/tex] where v is final velocity and v0 is initial velocity. Filling in:
0 = 20 + (-9.8)t and
-20 = -9.8t so
t = 2 seconds. The stone reaches its max height 2 seconds after it is thrown; that means that after another 2 seconds it will be on the ground. Total air time is 4 seconds.
If at a particular instant and at a certain point in space the electric field is in the x-direction and has a magnitude of 3.70 V/m, what is the magnitude of the magnetic field of the wave at this same point in space and instant in time
Answer:
the magnitude of the magnetic field is 1.23 x 10⁸ T.
Explanation:
Given;
magnitude of the electric field, E = 3.7 V/m
The magnitude of the magnetic field is calculated as;
E = cB
where;
B is the magnitude of the magnetic field
c is the speed of light = 3 x 10⁸ m/s
From the above equation, the magnetic field, B, is calculated as;
[tex]B = \frac{E}{c} \\\\B = \frac{3.7 }{3\times 10^8 } \\\\B = 1.23 \times 10^{-8 } \ T[/tex]
Therefore, the magnitude of the magnetic field is 1.23 x 10⁸ T.
A bar of steel has the minimum properties Se = 40 kpsi, Sy = 60 kpsi, and Sut = 80 ksi. The bar is subjected to a steady torsional stress of 15 kpsi and an alternating bending stress of 25 ksi. Find the factor of safety guarding against a static failure, and either the factor of safety guarding against a fatigue failure or the expected life of the part. For the fatigue analysis use Modified Goodman criterion.
Answer:
The correct solution is:
(a) 1.66
(b) 1.05
Explanation:
Given:
Bending stress,
[tex]\sigma_b = 25 \ kpsi[/tex]
Torsional stress,
[tex]\tau= 15 \ kpsi[/tex]
Yield stress of steel bar,
[tex]\delta_y = 60 \ kpsi[/tex]
As we know,
⇒ [tex]\sigma_{max}^' \ = \sqrt{\sigma_b^2 + 3 \gamma^2}[/tex]
[tex]= \sqrt{(25)^2+3(15)^2}[/tex]
[tex]=36.055 \ kpsi[/tex]
(a)
The factor of safety against static failure will be:
⇒ [tex]\eta_y = \frac{\delta_y}{\sigma_{max}^'}[/tex]
By putting the values, we get
[tex]=\frac{60}{36.055}[/tex]
[tex]=1.66[/tex]
(b)
According to the Goodman line failure,
[tex]\sigma_a = \sigma_b = 25 \ kpsi[/tex]
[tex]S_e = 40 \ kpsi[/tex]
[tex]\sigma_m = \sqrt{3} \tau[/tex]
[tex]=\sqrt{3}\times 15[/tex]
[tex]=26 \ kpsi[/tex]
[tex]Sut = 80 \ kpsi[/tex]
⇒ [tex]\frac{\sigma_a}{S_e} +\frac{\sigma_m}{Sut} =\frac{1}{\eta_y}[/tex]
[tex]\frac{25}{40}+\frac{26}{80}=\frac{1}{\eta_y}[/tex]
[tex]\eta_y = 1.05[/tex]
g Light that is incident upon the eye is refracted several times before it reaches the retina. As light passes through the eye, at which boundary does most of the overall refraction occur?
Answer
Explanation
:giác mạc
You attach a 2.30 kg weight to a horizontal spring that is fixed at one end. You pull the weight until the spring is stretched by 0.500 m and release it from rest. Assume the weight slides on a horizontal surface with negligible friction. The weight reaches a speed of zero again 0.400 s after release (for the first time after release). What is the maximum speed of the weight (in m/s)
Answer: [tex]3.92\ m/s[/tex]
Explanation:
Given
Mass of the attached object is [tex]m=2.3\ kg[/tex]
Spring is stretched by [tex]A=0.5\ m[/tex]
Speed reaches zero after [tex]t=0.4\ s[/tex]
Speed is zero at the extremities of the S.H.M motion that is
[tex]\Rightarrow \dfrac{T}{2}=0.4\\\\\Rightarrow T=0.8\ s[/tex]
Time period of motion is [tex]0.8\ s[/tex] which can also be given by
[tex]\Rightarrow \omega T=2\pi\\\\\Rightarrow \omega=\dfrac{2\pi }{T}\\\\\Rightarrow \omega =\dfrac{2\pi }{0.8}\\\\\Rightarrow \omega=\dfrac{5\pi }{2}[/tex]
Maximum speed for S.H.M. is [tex]v_{max}=A\omega[/tex]
[tex]\Rightarrow v_{max}=0.5\times 2.5\pi\\\Rightarrow v_{max}=3.92\ m/s[/tex]
Hey guys,I hope u r gonna answer this question fast,SI system is extended from of MKS system.Why? I will be waiting for the answer. Good luck thank u
Answer:
Because SI system has fundamental units of MKS System
Answer:
Explanation: the unit of length ,mass , and time are same in both the system , thus, the SI system is the extended from of MKS system.
If a boy lifts a mass of 6kg to a height of 10m and travels horizontally with a constant velocity of 4.2m/s, calculate the work done? Explain your answer.
Answer:
W = 641.52 J
Explanation:
The work done here will be the sum of potential energy and the kinetic energy of the boy. Here potential energy accounts for vertical motion part while the kinetic energy accounts for the horizontal motion part:
[tex]Work\ Done = Kinetic\ Energy + Potential\ Energy\\\\W = K.E +P.E\\\\W = \frac{1}{2}mv^2+mgh\\\\[/tex]
where,
W = Work Done = ?
m = mass = 6 kg
v = speed = 4.2 m/s
g = acceleration dueto gravity = 9.81 m/s²
h = height = 10 m
Therefore,
[tex]W = \frac{1}{2}(6\ kg)(4.2\ m/s)^2+(6\ kg)(9.81\ m/s^2)(10\ m)[/tex]
W = 52.92 J + 588.6 J
W = 641.52 J
which planet composed entirely of hydrogen and helium?
Answer:
The composition of Jupiter is similar to that of the Sun—mostly hydrogen and helium. Deep in the atmosphere, pressure and temperature increase, compressing the hydrogen gas into a liquid. This gives Jupiter the largest ocean in the solar system—an ocean made of hydrogen instead of water.
Which of the following is not an example of approximate simple harmonic motion
Answer:
where are the options
it's not full question
A crude approximation is that the Earth travels in a circular orbit about the Sun at constant speed, at a distance of 150,000,000 km from the Sun. Which of the following is the closest for the acceleration of the Earth in this orbit?
A. exactly 0 m/s2.
B. 0.006 m/s2.
C. 0.6 m/s2.
D. 6 m/s2.
E. 10 m/s2.
Answer:
The answer is "Option B".
Explanation:
[tex]r=15\times 10^{7}\ km\ = 15\times 10^{10}\ m\\\\w=\frac{2\pi}{1\ year}\\\\=\frac{2\pi}{1\times 365.24 \times 24 \times 60 \times 60\ sec}\\\\a=w^2r\\\\=(\frac{2\pi}{1\times 365.24 \times 24 \times 60 \times 60\ sec})^2 \times 15 \times 10^{10}\ \frac{m}{s^2}\\\\[/tex]
[tex]=5.940 \times 10^{-3} \ \frac{m}{s^2}\\\\=6 \times 10^{-3} \ \frac{m}{s^2}\\\\=0.006\ \frac{m}{s^2}\\\\[/tex]
An object moving with initial velocity 10 m/s is subjected to a uniform acceleration of 8 m/s ^² . The displacement in the next 2 s is: (a) 0m (b) 36 m (c) 16 m (d) 4 m
I’m a photoelectric effect, which property of the incident light determines how much kinetic energy the ejected electrons have ?
A) brightness
B) frequency
C) size of the beam
D) none of the above
Answer:
b = frequency
1.03 Transformation of energy flvs science question
Explanation:
the process of conversion of energy from one form to another is called transformation of energy.
friction always opposes the _____
Answer:
Friction always opposes the motion
I HOPE ITS RIGHT IF NOT THEN SORRYHAVE A GREAT DAY :)
The primary purpose of a switch in a circuit is to ___________.
A)either open or close a conductive path
B)change a circuit from parallel to series
C)change a circuit from series to parallel
D)store a charge for later use
Answer:
store a charge for later use
what is threshold frequency?
Answer:
"the minimum frequency of radiation that will produce a photoelectric effect."
Explanation:
That answer was derived from gogle cuz my explanations was harder to explain but good luck
What star is known as the "cold planet"?
Explanation:
OGLE-2005-BLG-390Lb.
PSR B1620-26 b. Surface Temperature: 72 Kelvin. ...
Neptune. Surface Temperature: 72 Kelvin. ...
Uranus. Surface Temperature: 76 Kelvin. ...
Saturn. Surface Temperature: 134 Kelvin. ...
Jupiter. Image Courtesy: NASA. ...
OGLE-2016-BLG-1195Lb. Surface Temperature: Unknown
what is the dimensional formula of young modulas
Answer:
The dimensional formula of Young's modulus is [ML^-1T^-2]
Answer:
G.oogle : The dimensional formula for Young’s modulus is:
A. [ML−1T−2]A. [ML−1T−2]
B. [M0LT−2]B. [M0LT−2]
C. [MLT−2]C. [MLT−2]
D. [ML2T−2]
