Answer:
Length of pipe organ(L) = 0647 m (Approx)
Explanation:
Given:
Temperature (T) = 14°C
Fundamental frequency (F) = 262 Hz
Speed of sound (v) = 331 + 0.60(T) m/s
Find:
Length of pipe organ(L)
Computation:
Speed of sound (v) = 331 + 0.60(14) m/s
Speed of sound (v) = 339.4
Length of pipe organ(L) = Speed of sound (v) / 2(Fundamental frequency)
Length of pipe organ(L) = 339.4 / 2 (262)
Length of pipe organ(L) = 0647 m (Approx)
A convex spherical mirror has a radius of curvature of magnitude36.0cm.
(a) Determine the position of the virtual image and the magnification for object distances of25.0cm. Indicate the location of the image with the sign of your answer.
image location =cm
magnification =
(b) Determine the position of the virtual image and the magnification for object distances of47.0cm. Indicate the location of the image with the sign of your answer.
image location =cm
magnification =
(c) Are the images in parts (a) and (b) upright or inverted?
The image in part (a) is---Select---uprightinverted
The image in part (b) is---Select---uprightinverted
Answer:
Explanation:
a )
focal length of convex spherical mirror
f = 36/2 cm = 18 cm
object distance u = - 25 cm
mirror formula
[tex]\frac{1}{v} + \frac{1}{u} = \frac{1}{f}[/tex]
[tex]\frac{1}{v} + \frac{1}{- 25} = \frac{1}{18}[/tex]
[tex]\frac{1}{v} = \frac{1}{25} + \frac{1}{18}[/tex]
v = 6.28 cm .
It is positive hence the image will be erect / upright and formed on the back of the mirror.
For object distance of 47 cm
u = - 47 cm
Putting the values in the mirror formula
[tex]\frac{1}{v} + \frac{1}{- 47} = \frac{1}{18}[/tex]
[tex]\frac{1}{v} = \frac{1}{ 47} + \frac{1}{18}[/tex]
v = 13 cm
It is positive hence the image will be erect / upright and formed on the back of the mirror.
A light beam is traveling through an unknown substance. When it strikes a boundary between that substance and the air (nair≈1), the angle of reflection is 29.0∘ and the angle of refraction is 39.0∘. What is the index of refraction n of the substance?
Answer:
0.7707
Explanation:
From Snell's law,
n(1) * sin θ1 = n(2) * sinθ2
Where n(1) = refractive index of air = 1.0003
θ1 = angle of incidence
n(2) = refractive index of second substance
θ2 = angle of refraction
The angle of reflection through the unknown substance is the same as the angle of incidence of air. Thus this means that θ1 = 29°
=> 1.0003 * sin29 = n(2) * sin39
n(2) = (1.0003 * sin29) / sin39
n(2) = 0.7707
Explanation:
The index of refraction n of the substance is 0.7707
Snell law:Here we know that
n(1) * sin θ1 = n(2) * sinθ2
here
n(1) = refractive index of air = 1.0003
θ1 = angle of incidence
n(2) = refractive index of second substance
θ2 = angle of refraction
The angle of reflection should be via the unknown substance that represent the same as the angle of incidence of air.
So,
θ1 = 29°
1.0003 * sin29 = n(2) * sin39
n(2) = (1.0003 * sin29) / sin39
n(2) = 0.7707
learn more about refraction here: https://brainly.com/question/21312906
Six automobiles are initially traveling at the indicated velocities. The automobiles have different masses and velocities. The drivers step on the brakes and all automobiles are brought to rest.Automobile 1: 500kg, 10m/sAutomobile 2: 2000kg, 5m/sAutomobile 3: 500kg, 20m/sAutomobile 4: 1000kg, 20m/sAutomobile 5: 1000kg, 10m/sAutomobile 6: 4000kg, 5m/sRequired:a. Rank these automobiles based on the magnitude of their momentum before the brakes are applied, from largest to smallest.b. Rank these automobiles based on the magnitude of the impulse needed to stop them, from largest to smallest.c. Rank the automobiles based on the magnitude of the force needed to stop them, from largest to smallest.
Answer:
A. largest: (4000 kg, 5 m/s; 1000 kg, 20 m/s)
medium: (2000 kg, 5 m/s; 500 kg, 20 m/s; 1000 kg, 10 m/s)
smallest: (500 kg, 10 m/s)
B. largest: (4000 kg, 5 m/s; 1000 kg, 20 m/s)
medium: (2000 kg, 5 m/s; 500 kg, 20 m/s; 1000 kg, 10 m/s)
smallest: (500 kg, 10 m/s)
C. You can't say anything about the forces required until we know about the time frames required for each one to stop. So If they all stopped in the same time interval, then the rankings are the same.
When a magnet moves above a conducting ladder, the currents induced in the ladder produces a magnetic field. This field interacts with the magnetic field of the magnet to produce a force that A) pulls the ladder in the direction opposite to the direction of the moving magnet. B) pulls the ladder in the same direction as the moving magnet. C) brakes the ladder. D) None of the above.
Answer:
) pulls the ladder in the direction opposite
Explanation:
This is in line with lenz law that states that the magnetic field induced in a conductor act to oppose the magnetic field that produced it
The maximum gauge pressure in a hydraulic system is 15 atm. What is the largest mass that could be lifted by this system if the diameter of the piston is 65 cm
Answer:
The maximum force that can be lifted by this system is 51,478.4 kg
Explanation:
Given;
maximum gauge pressure of the hydraulic system, Hp = 15 atm = 1.52 x 10⁶ N/m²
diameter of the piston, d = 65 cm = 0.65 m
The maximum gauge pressure of the piston is given as;
[tex]Hp = \frac{F}{A}[/tex]
Where;
F is the maximum force of the piston
A is the area of the piston
[tex]A = \pi (\frac{0.65}{2} )^2\\\\A = 0.3319 \ m^2[/tex]
F = Hp x A
F = 1.52 x 10⁶N/m² x 0.3319m²
F = 504488 N
Force is given as;
F = mg
m = F/g
m = 504488/9.8
m = 51,478.4 kg
Therefore, the maximum force that can be lifted by this system is 51,478.4 kg
White light is spread out into spectral hues by a diffraction grating. If the grating has 1000 lines per cm, at what angle will red light (λ = 640 nm) appear in first order?
Answer:
3.67°
Explanation: Given that λ=640nm , m = 1
Considering the slit separation
d = 1cm/1000
= 1.000×10^-3cm
= 1.000×10-5m
We then have
Sinθ = mλ/d
Sinθ= (1×640×10^-9)/1.000×10-5m
Sinθ = 0.064
θ= sin-1 0.064
θ= 3.669°
= 3.67°
Terms to describe the opposition by a material.to being magnetised is
Answer:
Repulsion
Explanation:
An organ pipe open at both ends has two successive harmonics with frequencies of 220 Hz and 240 Hz. What is the length of the pipe? The speed of sound is 343 m/s in air.
Answer:
The length is [tex]l = 8.6 \ m[/tex]
Explanation:
From the question we are told that
The frequencies of the two successive harmonics are [tex]f_1 = 220 \ Hz[/tex] , [tex]f_2 = 240 \ Hz[/tex]
The speed of sound in the air is [tex]v_s = 343 \ m/s[/tex]
Generally the frequency of a given harmonic is mathematically represented as
[tex]f_n = \frac{n v }{2l}[/tex]
Here n defines the position of the harmonics
Now since the position of both harmonic is not know but we know that they successive then we can represented them mathematically as
[tex]220 = \frac{n v}{2l}[/tex]
and
[tex]240 = \frac{(n+1) v}{2l}[/tex]
So
[tex]\frac{(n + 1 ) v}{2l} - \frac{n v}{2l} = 240-220[/tex]
=> [tex]\frac{v}{2l} = 20[/tex]
=> [tex]l = 8.6 \ m[/tex]
In which type of indicating valve is the valve stem housed in a hollow metal post that contains a movable plate with a small glass window
Answer:
Post indicator valve
Explanation:
Post Indicator Valves are commonly used to control the water flow of sprinkler systems used in public and private buildings, warehouses, and factories for fire suppression. PIVs control water flow from the public system into the building's fire suppression system.
What is the pathway of sound through fluids starting at the oval window through to dissipation of the sound waves at the round window
When a mercury thermometer is heated, the mercury expands and rises in the thin tube of glass. What does this indicate about the relative rates of expansion for mercury and glass
Answer:
This means that mercury has a higher or faster expansion rate than glass
Explanation:
This is because When a container expands, the reservoir in the glass expands at the same rate as the glass. Thus, if there is something in a glass and both expand at the same rate, they have no change - but if the contents expand faster, they will fill the container to a higher level, and if the contents expand slower, they will fill the container to a lower level (relative to the new size of the container).
What will be the nature of the image formed from both a convex lens and a concave
lens of 20 centimeter focus distance, when the object is placed at a distance of
10 centimeters?
Answer:
Explanation:
Using the lens formula
1//f = 1/u+1/v
f is the focal length of the lens
u is the object distance
v is the image distance
For convex lens
The focal length of a convex lens is positive and the image distance can either be negative or positive.
Given f = 20cm and u = 10cm
1/v = 1/f - 1/u
1/v = 1/20-1/10
1/v = (1-2)/20
1/V = -1/20
v = -20/1
v = -20 cm
Since the image distance is negative, this shows that the nature of the image formed by the convex lens is a virtual image
For concave lens
The focal length of a concave lens is negative and the image distance is negative.
Given f = -20cm and u = 10cm
1/v = 1/f - 1/u
1/v = -1/20-1/10
1/v = (-1-2)/20
1/V = -3/20
v = -20/3
v = -6.67 cm
Since the image distance is negative, this shows that the nature of the image formed by the concave lens is a virtual image
A pump is to deliver 10, 000 kg/h of toluene at 1140C and 1.1 atm absolute pressure from the Reboiler of a distillation tower to the second distillation unit without cooling the toluene before it enters the pump. If the friction loss in the line between the Reboiler and the pump is 7 kN/m2. The density of toluene is 886 kg/m3. How far above the pump must the liquid be maintained to avoid cavitation
Answer:
3.4093
Explanation:
NPSHa = hatm + hel + hf +hva
the elevation head is the hel
friction loss head is hf
NPSHa is the head of vapour pressure of fluid
atmospheric pressure head is hatm
log₁₀P* = [tex]A -\frac{B}{C+T}[/tex]
[tex]A, B, C are fixed[/tex]
log₁₀Pv = [tex]4.07827-\frac{1343.943}{387.15-53.773}[/tex]
= 4.07827 - 1343.943/333.377
=4.07827 - 4.0313009
= 0.0469691
we take the log
p* = 1.114218
we convert this value to get 111421.8
hvap = 111421.8 * 1/776.14 * 1/9.81
= 14.63
hatm = 1.1 *101325/1 * 1/9.81 *1/776.14
=14.64
hf = 7000/1 * 1/776.14 * 1/9.81
= 0.9193
NPSHa = 2.5
hel = 0.9193 + 2.5 + 14.63 - 14.64
hel = 3.4093
The NSPH values are used to calculate cavitation. The vapor pressure of the liquid is 1.114 atm.
The vapor pressure can be calculated by,
[tex]\mathrm {NPSH_A}= ( \frac {p_i}{\rho g} + \frac {V_i^2}{2g})- \frac {p_v}{\rho g}[/tex]
Where,
[tex]\mathrm {NPSH_A}[/tex] = available NPSH
[tex]p_i[/tex] = absolute pressure at the inlet = 1.1 atm
[tex]V_i[/tex] = average velocity at the inlet = 10, 000 kg/h
[tex]\rho[/tex] = fluid density = 886 kg/m3.
g = acceleration of gravity = 9.8 m/s²
[tex]p_v[/tex] = vapor pressure of the fluid = ?
Put the values in the equation, we get
[tex]p_v = 1.114\ atm[/tex]
Therefore, the vapor pressure of the liquid is 1.114 atm.
To know more about vapor pressure:
https://brainly.com/question/25356241
The momentum of an electron is 1.75 times larger than the value computed non-relativistically. What is the speed of the electron
Answer:
Speed of the electron is 2.46 x 10^8 m/s
Explanation:
momentum of the electron before relativistic effect = [tex]M_{0} V[/tex]
where [tex]M_{0}[/tex] is the rest mass of the electron
V is the velocity of the electron.
under relativistic effect, the mass increases.
under relativistic effect, the new mass M will be
M = [tex]M_{0}/ \sqrt{1 - \beta ^{2} }[/tex]
where
[tex]\beta = V/c[/tex]
c is the speed of light = 3 x 10^8 m/s
V is the speed with which the electron travels.
The new momentum will therefore be
==> [tex]M_{0}V/ \sqrt{1 - \beta ^{2} }[/tex]
It is stated that the relativistic momentum is 1.75 times the non-relativistic momentum. Equating, we have
1.75[tex]M_{0} V[/tex] = [tex]M_{0}V/ \sqrt{1 - \beta ^{2} }[/tex]
the equation reduces to
1.75 = [tex]1/ \sqrt{1 - \beta ^{2} }[/tex]
square both sides of the equation, we have
3.0625 = 1/[tex](1 - \beta ^{2} )[/tex]
3.0625 - 3.0625[tex]\beta ^{2}[/tex] = 1
2.0625 = 3.0625[tex]\beta ^{2}[/tex]
[tex]\beta ^{2}[/tex] = 0.67
β = 0.819
substitute for [tex]\beta = V/c[/tex]
V/c = 0.819
V = c x 0.819
V = 3 x 10^8 x 0.819 = 2.46 x 10^8 m/s
An L-R-C series circuit is connected to a 120 Hz ac source that has Vrms = 82.0 V. The circuit has a resistance of 71.0 Ω and an impedance at this frequency of 107 Ω and an impedance at this frequency of 105Ω. What average power is delivered to the circuit by the source?
Explanation:
Given that,
Frequency of LCR circuit is 120 Hz
RMS voltage, [tex]V_{rms}=82\ V[/tex]
Resistance of circuit, R = 71 Ω
Impedance, Z = 107 Ω
We need to find the average power is delivered to the circuit by the source. Firstly, finding the rms value of current,
[tex]I_{rms}=\dfrac{V_{rms}}{Z}\\\\I_{rms}=\dfrac{82}{105}\\\\I_{rms}=0.78\ A[/tex]
Power is given by :
[tex]P=I_{rms}V_{rms}\cos\phi[/tex]
[tex]\cos\phi = \dfrac{R}{Z}\\\\\cos\phi = \dfrac{71}{105}\\\\\cos\phi =0.676[/tex]
Now, power,
[tex]P=0.78\times 82\times 0.676\\\\P=43.23\ W[/tex]
So, the average power of 43.23 watts is delivered to the circuit by the source.
A magnetic field near the floor points down and is increasing. Looking down at the floor, does the non-Coulomb electric field curl clockwise or counter-clockwise?
a. clockwiseb. counter-clockwise c. no curly E
Answer:
when a magnetic field near the floors points down and is increasing then the electric field curl (a) clockwise.
Explanation:
The magnetic field this is the area that is around a magnet which there is presence of magnetic force. The Moving electric charges can create magnetic fields. we say In physics, that the magnetic field is a field that passes through space and which makes a magnetic force move electric charges.
The Non-coulomb electric field curls ; ( B ) counterclockwise
Non-coulomb electric field also known as induced EMF is the Negative time rate of change of a magnetic flux in a closed loop through the loop. Non-coulomb electric field is expressed as ; Fnc = qEnc
Given that the magnetic field points downwards and the value of the electric field ( ε ) is increasing ( i.e. ε > 0 ) The direction of the non-coulomb electric field will curl in a counter-clockwise direction.
Hence we can conclude that The Non-coulomb electric field curls in a counterclockwise direction.
Learn more : https://brainly.com/question/12975267
The atomic number of a nucleus increases during which nuclear reactions?
Answer:
Option (A) : Nuclear Fusion and Beta Decay (electron emission)
Answer:
A : Fusion followed by beta decay (electron emission)
Explanation:
Ap3x
Difference between matter and energy
Answer:
Energy is the strength and vitality required for sustained physical or mental activity.
Matter occupies space and possesses rest mass, especially as distinct from energy.
Hope this helps! （づ￣3￣）づ╭❤～
A baseball (m=145g) traveling 35 m/s moves a fielder's glove backward 23 cm when the ball is caught. What was the average force exerted by the ball on the glove?
Answer:
386.13 N
Explanation:
The kinetic energy of the baseball is converted into workdone in moving the glove backward( work energy theorem).
Therefore, KE of the ball
[tex]\frac{1}{2} mv^2 =\frac{1}{2}(0.145)35^2\\ = 88.81 \text{J}[/tex]
Now, workdone in moving the glove
W= Fd
where F = Force applied, d = displacement of the glove= 0.23 cm.
88.81 = F×0.23
F= 88.81/0.23 = 386.13 N
Sonar is used to determine the speed of an object. A 38.0-kHz signal is sent out, and a 40.0-kHz signal is returned. If the speed of sound is 341 m/s, how fast is the object moving?
Answer:
The velocity is [tex]v = 8.743 \ m/s[/tex]
Explanation:
From the question we are told that
The frequency of the signal sent out is [tex]f_s = 38.0 \ kHz = 38.0 *10^{3} \ Hz[/tex]
The frequency of the signal received is [tex]f_r = 40.0 \ kHz = 40.0 *10^{3} \ Hz[/tex]
The speed of sound is [tex]v_s = 341 \ m/s[/tex]
Generally the frequency of the sound received is mathematically represented as
[tex]f_r = f_s [\frac{v_s + v}{v_s - v} ][/tex]
where v is the velocity of the object
=> [tex]40 *10^{3} = 38 *10^{3} * [\frac{341 + v}{341 - v} ][/tex]
=> [tex]1.05263 = \frac{341+v }{341-v}[/tex]
=> [tex]358.94 - 1.05263v = 341 + v[/tex]
=> [tex]17.947 = 2.05263 v[/tex]
=> [tex]v = 8.743 \ m/s[/tex]
Red and orange stars are found evenly spread throughout the galactic disk, but blue stars are typically found
Answer:
only in or near star-forming clouds
Explanation:
When in the galactic disk, Red and orange stars are found evenly spread so here Blue stars are hot and therefore massive and therefore short-lived, that is means they never have time to venture far from the places, where they were born. so correct answer is blue stars are typically found only in or near star-forming clouds
The speed of sound through air is 340 m/s. If a person hears the clap of thunder 9.6 s after seeing the bolt of lightning, how far away is the lightning?
Explanation:
Distance = speed × time
d = (340 m/s) (9.6 s)
d = 3264 m
Figure (3) shows a car travelling along the route PQRST in 30 minutes. What is the average speed of the car in km/hour?
Answer:
60 km/hour.
Explanation:
We'll begin by calculating the total distance traveled by the car. This is illustrated below:
Total distance traveled = sum of distance between PQRST
Total distance = 10 + 5 + 10 + 5
Total distance = 30 km
Next, we shall convert 30 mins to hour. This can obtained as follow:
Recall:
60 mins = 1 hour
Therefore,
30 mins = 30/60 = 0.5 hour.
Finally, we shall determine the average speed of the car as follow:
Distance = 30 km
Time = 0.5 hour
Speed =?
Speed = distance /time
Speed = 30/0.5
Speed = 60 km/hour
Therefore, the speed of the car is 60 km/hour.
You're conducting an experiment on another planet. You drop a rock from a height of 1 m and it hits the ground 0.4 seconds later. What is acceleration due to gravity on the planet ?
Answer:
Here,
v (final velocity) = 0
u (initial velocity) = u
a = ?
s = 1m
t = 0.4s
using the first equation of motion,
0 = u + 0.4a
= -0.4a = u
using the second equation of motion:
1 = 0.4u + 0.08a
from the bold equation
1 = 0.4(-0.4a) + 0.08a
1 = -0.16a + 0.08a
1 = -0.08a
a = -1/0.08
a = -100/8
a = -12.5 m/s/s
please make me brainly, i am 1 brainly away from the next rank
: A spaceship is traveling at the speed 2t 2 1 km/s (t is time in seconds). It is pointing directly away from earth and at time t 0 it is 1000 kilometers from earth. How far from earth is it at one minute from time t 0
Answer:
145060km
Explanation: Given that
speed = dx/dt = 2t^2 +1
integrate
x = 2/3t^3 + t + c (c is constant, x is in km, t is in second)
given that at t=0, x = 1000
so 1000 = 2/3 X (0)^3 + 0 + c
or c = 1000
So x = 2/3t^3 + t + 1000
for t = 1 min = 60s
x = 2/3 X 60^3 + 60 + 1000
x = 2/3×216000+ 1060
x = 144000+1060
= 145060km
At one minute, it will be 145060km far from the earth
Ultraviolet light having a wavelength of 97 nm strikes a metallic surface. Electrons leave the surface with speeds up to 3.48 × 105 m/s. What is the work function, in eV of the metal?
Answer:
12.45eVExplanation:
Before calculating the work function, we must know the formula for calculating the kinetic energy of an electron. The kinetic energy of an electron is the taken as the difference between incident photon energy and work function of a metal.
Mathematically, KE = hf - Ф where;
h is the Planck constant
f is the frequency = c/λ
c is the speed of light
λ is the wavelength
Ф is the work function
The formula will become KE = hc/λ - Ф. Making the work function the subject of the formula we have;
Ф = hc/λ - KE
Ф = hc/λ - 1/2mv²
Given parameters
c = 3*10⁸m/s
λ = 97*10⁻⁹m
velocity of the electron v = 3.48*10⁵m/s
h = 6.62607015 × 10⁻³⁴
m is the mass of the electron = 9.10938356 × 10⁻³¹kg
Substituting the given parameters into the formula Ф = hc/λ - 1/2mv²
Ф = 6.63 × 10⁻³⁴*3*10⁸/97*10⁻⁹ - 1/2*9.11*10⁻³¹(3.48*10⁵)²
Ф = 0.205*10⁻¹⁷ - 4.555*10⁻³¹*12.1104*10¹⁰
Ф = 0.205*10⁻¹⁷ - 55.163*10⁻²¹
Ф = 0.205*10⁻¹⁷ - 0.0055.163*10⁻¹⁷
Ф = 0.1995*10⁻¹⁷Joules
Since 1eV = 1.60218*10⁻¹⁹J
x = 0.1995*10⁻¹⁷Joules
cross multiply
x = 0.1995*10⁻¹⁷/1.60218*10⁻¹⁹
x = 0.1245*10²
x = 12.45eV
Hence the work function of the metal in eV is 12.45eV
In _____ research, a group of people of one age is compared to a group of people who are another age.
Answer:
cross-sectional
Explanation:
The full definition of this is ''a research design in which several different age-groups of participants are studied at one particular point in time.''
550 J of heat is added to the gas in an isothermal process. As the gas expands, pushing against the piston, how much work does it do
Answer:
The work done by the system is 550 J
Explanation:
Given;
heat added to the system, Q = 550 J
Apply the first law of thermodynamics;
ΔU = Q - W
Where;
ΔU is change in internal energy
Q is the heat added to the system
W is the work done by the system
During an isothermal process, the temperature of the system is constant for the entire process. During this process, the change in the internal energy is zero.
0 = Q - W
W = Q
W = 550 J
Therefore, the work done by the system is 550 J
What is the maximum speed with which a 1200-kg car can round a turn of radius 94.0 m on a flat road if the coefficient of static friction between tires and road is 0.50?
Answer:
v= 21.47m/sExplanation:
For the car to turn at the about the centripetal force must not be greater than the static friction between the tires and the road
we will use the expression relating centripetal force and static friction below
let U represent the coefficient of static friction
Given that
U= 0.50
mass m= 1200-kg
radius r= 94.0 m
Assuming g= 9.81 m/s^2
[tex]U*m*g=\frac{mv^2}{r}[/tex]
[tex]U*g=\frac{v^2}{r}[/tex]
substituting our given data in to expression we can solve for the speed V
[tex]0.5*9.81=\frac{v^2}{94}[/tex]
making v the subject of formula we have
[tex]0.5*9.81=\frac{v^2}{94}\\\v= \sqrt{0.5*9.81*94} \\\\v= \sqrt{461.07} \\\\v= 21.47[/tex]
v= 21.47m/s
hence the maximum velocity of the car is 21.47m/s
A disk between vertebrae in the spine is subjected to a shearing force of 375 N. Find its shear deformation, taking it to have a shear modulus of 1.60×109 N/m2. The disk is equivalent to a solid cylinder 0.750 cm high and 6.50 cm in diameter.
Answer:
5.29×10^-7
Explanation:
shear stress τ = F/ A
shear deformation δ = (VL)/ (AG)
= (τL)/ G
V=shear force
L=height of disk=6.50×10^-2
A=cross sectional area
G= shear modulus= (1.60x10^9N/m^2)
A=πd^2/4
Then substitute the values we have
4×(375N)(0.00750m)
________________ = δ
(π*0.00650^2)(1.60x10^9N/m^2)
= 5.29×10^-7
