Answer:
Create space
Watch your breath
Watch your thoughts
Be gentle with yourself
Affirm what you want
Explanation:
a. A light wave moves through glass (n = 1.5) at an angle of 15°. What angle will it have when it moves from the glass into water (n = 1.33)? (4 points)
b. Draw a ray diagram to locate the image of the arrow, as refracted through the lens shown. Write 2 - 3 sentences describing the type of image and its size relative to the object. What type of mirror could be used to form an image of the same type and size? (8 points)
c. An object is located 65 cm from a concave mirror with a focal length of 45 cm. What is the image distance? Is the image real or virtual? (6 points)
Answer in detail with the correct units and steps to solve. Will mark brainliest.
A wave is a phenomenon that does not cause a permanent displacement in the particles of the medium through which it passes. And it transfers energy from one end of the medium to the other. Examples of waves include light waves, sound waves, water waves, x-rays, radiowaves, etc. Thus the required answers for each part of the question are:
a. The angle that the light wave would have is [tex]7.5^{o}[/tex].
b. The type of mirror that can be used is a plane mirror.
c. The image distance is 146.3 cm.
ii. The image formed by the mirror is a real image.
a. When a ray of light passes from one medium to another, then refraction occurs. The refraction depends on the refractive index of the medium considered.
Thus from Snell's law, we have:
refractive index, n, = [tex]\frac{Sin i}{Sin r}[/tex]
where: i is the angle of incidence, and r is the refracted angle.
Now given that n = 1.5, and i = 15.
Then;
n = [tex]\frac{Sin i}{Sin r}[/tex]
1.5 = [tex]\frac{Sin 15}{Sin r}[/tex]
Sin r = [tex]\frac{0.2588}{1.5}[/tex]
= 0.17253
r = [tex]Sin^{-1}[/tex] 0.17253
= 9.936
r ≅ [tex]10^{o}[/tex]
Since the light wave now moves from the glass into water, the determined refracted angle now becomes its angle of incidence in water. So that;
n = [tex]\frac{Sin i}{Sin r}[/tex]
1.33 = [tex]\frac{Sin 10}{Sin r}[/tex]
Sin r = [tex]\frac{0.17365}{1.33}[/tex]
= 0.1306
r = [tex]Sin^{-1}[/tex] 0.1306
= 7.504
r = [tex]7.5^{o}[/tex]
Therefore, the angle that the light wave would have is [tex]7.5^{o}[/tex].
b. The image formed would be the same size as that of the object. And also the same distance as that of the object to the pole of the lens.
The type of mirror that can be used is a plane mirror.
The ray diagram is attached to this answer.
c. From the mirror formula;
[tex]\frac{1}{f}[/tex] = [tex]\frac{1}{u}[/tex] + [tex]\frac{1}{v}[/tex]
where; f is the focal length of the mirror, u is the object's distance to the mirror, and v is the image's distance to the mirror.
Given; u = 65 cm, and f = 45 cm, then:
[tex]\frac{1}{45}[/tex] = [tex]\frac{1}{65}[/tex] + [tex]\frac{1}{v}[/tex]
[tex]\frac{1}{v}[/tex] = [tex]\frac{1}{45}[/tex] - [tex]\frac{1}{65}[/tex]
[tex]\frac{1}{v}[/tex] = [tex]\frac{20}{2925}[/tex]
v = [tex]\frac{2925}{20}[/tex]
v = 146.25 cm
The image distance is 146.3 cm.
ii. The image formed by the mirror is a real image.
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Aster walks first 70 m in the direction 37° north of east, and then walks 82 m in the
direction 20° south of east, and finally walks 28 m in the direction 30° west of north.(2pt)
a) How far and at what angle is the Aster's final position from her initial position?
b) In what direction would she has to head to return to her initial position?
The Aster's final position from her initial position is 64 m
The angle is 300° and She has to head in West north direction to return to her initial position
What is Displacement ?The displacement is the distance travelled in a specific direction. Displacement is a vector quantity.
Given that Aster walks first 70 m in the direction 37° north of east, and then walks 82 m in the direction 20° south of east, and finally walks 28 m in the direction 30° west of north.
This can be solved by using bearing method. Cosine formula will be the best to solve for the distance D.
[tex]D^{2}[/tex] = [tex]70^{2}[/tex] + [tex]82^{2}[/tex] - 70 x 82 x Cos (37 + 20)
[tex]D^{2}[/tex] = 4900 + 6724 - 5740Cos57
[tex]D^{2}[/tex] = 11624 - 3126.23
[tex]D^{2}[/tex] = 8497.8
D = [tex]\sqrt{8497.8}[/tex]
D = 92.2 m
a) The Aster's final position from her initial position is 92.2 - 28 = 64 m
The angle = 270° + 30° = 300°
b) She has to head in West north direction to return to her initial position
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By how much does the gravitational potential energy of a 55- kg pole vaulter change if her center of mass rises about 4.0 m during the jump?
Express your answer to two significant figures and include the appropriate units.
The change in the gravitational potential energy of the pole vaulter is 2,156 J.
Change in the gravitational potential energy
The change in the gravitational potential energy of the pole vaulter is calculated as follows;
ΔP.E = mg(Δh)
where;
m is massΔh is change in heightΔP.E = (55)(9.8)(4)
ΔP.E = 2,156 J
Thus, the change in the gravitational potential energy of the pole vaulter is 2,156 J.
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Derive the following equations of motion
1. v = at
2. s = ut + at²
3. v² = u² + 2as
according to definition of acceleration
a=v-u/t
t=v-u/a(equation 1)
according to the formula of average velocity
v+u/2*s/t
s=v+u/2*t(equation 2)
now putting the value of t in equation 2
s=v+u/2*v-u/a
s=v^2-u^2/2a
v^2=u^2+2as
please make me brainlist
deriving projectile motion formulas
Answer:
Projectile motion formula or equations derived (In Tabular format)Motion Path equation:
y = (tanθ) x – (1/2) g . x2/(V0 cosθ)2
Explanation:
A 1500 kg aircraft going 35 m/s collides with a 2000 kg aircraft that is parked and they stick together after the collision and are going 15 m/s after the collision. If they skid for 112.1 m before stopping, how long (in seconds) did they skid? Hint: Are the aircraft moving at a constant velocity after the collision or do they experience an acceleration?
The two aircrafts are skidded for 14.9 s.
To find the answer, we need to know about the Newton's equation of motion.
What is the Newton's equation that relates velocity, distance, acceleration and time?As per Newton's equation of motion
V²-U²= 2aSV= U+atV= final velocity, U = initial velocity, S = distance, a= acceleration, t= timeWhat's the acceleration of the aircrafts that skidded from 35 m/s after the collision for 112.1 m before achieving 15 m/s?Here, U = 35 m/s, V = 15m/s, S= 112.1 mSo, 15²- 35²= 2a×112.1= 224.2a=> a= -1000/224.2= -4.5 m/s²
What's the time taken to stop if these aircrafts are de-accelerated by 4.5 m/s² from 35m/s?Here, U = 35 m/s, V=15 m/s a= -4.5 m/s²35= 15+(-4.5)t=> t= 20/4.5 = 4.4 s
Thus, we can conclude that the two aircrafts are skidded for 4.4 s.
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When this surgical procedure is used to reduce the risk of stroke it will correct stenosis in the artery the most common cause of this condition is the buildup of plaque that forms in the artery name this procedure
Endarterectomy is used to reduce the risk of stroke and correct the stenosis in the artery.
What is carotid artery stenosis?The primary blood vessels that supply the brain with blood and oxygen are the carotid arteries.
The narrowing of these arteries is referred to as carotid artery disease. Carotid artery stenosis is another name for it. The main factor causing constriction is atherosclerosis.
This fat deposit reduces the blood flow to the brain which cause a stroke.
Carotid endarterectomy is a surgical treatment to remove plaque, an accumulation of fatty deposits that causes a carotid artery to become narrowed.
Hence, an Endarterectomy is used to reduce the risk of stroke, it will correct stenosis in the artery.
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Define a dipole. Hence, write the expression for calculating the electric moment of a dipole
Answer:
Explanation:
An electric dipole is formed by two point charges +q and −q connected by a vector a. The electric dipole moment is defined as p = qa
A man pushing a crate of mass
m = 92.0 kg
at a speed of
v = 0.880 m/s
encounters a rough horizontal surface of length
ℓ = 0.65 m
as in the figure below. If the coefficient of kinetic friction between the crate and rough surface is 0.359 and he exerts a constant horizontal force of 299 N on the crate.
(a) Find the magnitude and direction of the net force on the crate while it is on the rough surface.
magnitude _______N
Direction?:
1. Same as the motion of the crate
2. opposite as the motion of the crate
(b) Find the net work done on the crate while it is on the rough surface.
___________J
(c) Find the speed of the crate when it reaches the end of the rough surface.
_________m/s
(a) The magnitude and direction of the net force on the crate while it is on the rough surface is 24.67 N, opposite as the motion of the crate.
(b) The net work done on the crate while it is on the rough surface is -16.04 J.
(c) The speed of the crate when it reaches the end of the rough surface is 0.65 m/s.
Magnitude of net force on the crateF(net) = F - μFf
F(net) = 299 - 0.351(92 x 9.8)
F(net) = -24.67 N
Net work done on the crateW = F(net) x L
W = -24.67 x 0.65
W = - 16.04 J
Acceleration of the cratea = F(net)/m
a = -24.67/92
a = - 0.268 m/s²
Speed of the cratev² = u² + 2as
v² = 0.88² + 2(-0.268)(0.65)
v² = 0.426
v = √0.426
v = 0.65 m/s
Thus, the magnitude and direction of the net force on the crate while it is on the rough surface is 24.67 N, opposite as the motion of the crate.
The net work done on the crate while it is on the rough surface is -16.04 J.
The speed of the crate when it reaches the end of the rough surface is 0.65 m/s.
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Three strings, attached to the sides of a rectangular frame, are tied together by a knot as shown in the figure. The magnitude of the tension in the string labeled C is 56.3 N.
Calculate the magnitude of the tension in the string marked A. (You'll need to get the various positions from the graph. The ends of the strings are exactly on one of the tic marks.)
The magnitude of the tension in the string marked A is 52.5N
Generally, the equation for is mathematically given as
Let's take θ be an angle at A
So, tanθ = 3/8
Let's take α be an angle at B (Below X)
tanα = 5/4
Let's take β be an angle at C (Below x)
tanβ = 1/6
First we take the Horizontal Components
74.9cos(9.46°) = Acos(20.6°) + Bcos(51.3°)
By solving the equation, we get
A = 78.9 - 0.668B … (1)
Now, we take the vertical components
74.9sin(9.46°) + Asin(20.6°) = Bsin(51.3°)
By solving the equation, we get
40.07 = 1.015B
B = 39.5N
By substituting the value of B in equation (1)
A = 78.9 - 0.6668× 39.5
A = 52.5N
Hence, the magnitude of the tension in the string marked A is 52.5N
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8. The pulley is assumed massless and frictionless and rotates freely about its axle. The blocks have masses m = 40 g and m₂ = 20 g, and block mi is pulled to the right by a horizontal force of magnitude F = 0.03 N. Find the magnitude of the acceleration of block m2 and the tension in the cord if the surface is frictionless. (2pt) m₂ a₂ T₂ T₂ T₁ m₁
The magnitude of the acceleration of block m2 is 0.09 m/s² and the tension in the cord is 1.8 x 10⁻³ N.
Acceleration of the blocksThe acceleration of the blocks is calculated from the net force on the blocks.
∑F = ma
a = ∑F/m
a = (F) / (m₁ + m₂)
where;
F is the horizontal force appliedm₁ is mass of first block = 40 g = 0.04 kgm₂ is mass of the second block = 20 g = 0.02 kga = (0.03)/(0.04 + 0.02)
a = 0.09 m/s²
Tension due to block m₂T = m₂a
T = (0.02 x 0.09) = 1.8 x 10⁻³ N
Thus, the magnitude of the acceleration of block m2 is 0.09 m/s² and the tension in the cord is 1.8 x 10⁻³ N.
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The motor of a weed trimmer spins
at 9,000 rpm. The amount of time
required for the motor to reach this
speed would NOT be affected by...
A. the distribution of
trimmer line inside the
spool
B. the mass of the spool
containing the trimmer
line
C. the direction in which
the torque is applied
The speed of the trimmer is not affected by the the distribution of trimmer line inside the spool. Option A
What is a weed trimmer?The weed trimmer is a device that is used to trim the grasses on a lawn or a field. This device has a rotating shaft that does the actual trimming of the grasses.
The speed of the trimmer is not affected by the the distribution of trimmer line inside the spool thus it does not affect the amount of time required to reach 9,000 rpm speed.
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Calculate the speed of a satellite moving in a stable circular orbit about the Earth at a height of 4930 km .
The speed of the satellite moving in a stable circular orbit about the Earth is 5,916.36 m/s.
Speed of the satellite
v = √GM/r
where;
M is mass of EarthG is universal gravitation constantr is distance from center of Earth = Radius of earth + 4930 kmv = √[(6.626 x 10⁻¹¹ x 5.97 x 10²⁴) / ((6371 + 4930) x 10³)]
v = 5,916.36 m/s
Thus, the speed of the satellite moving in a stable circular orbit about the Earth is 5,916.36 m/s.
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9.80 1. The density of mercury is 13600 kg/m³. What is this value in g/cm³? 2. Find the mass of water which will fit in a large tank measuring 2 m x 1 m x 20 cm. Density of water is 1000 kg/m³ or 1.0 g/cm³. 3. Find the volume of a lump of softwood whose mass is 120 g. Density of softwood is 0.6 gcm-3 or 600 kgm-³. -3
Answer:
1. 13..6 grams per centimeters cubed.
2. Mass = 400kg
3. Volume = 200cm = 2m
Explanation:
1. The conversion for kg/m^3 to g/cm^3 is divide by 1000.
2. [tex]density=\frac{mass}{volume}[/tex]
[tex]1000=\frac{mass}{2 * 1 * 0.2}[/tex]
[tex]1000*0.4=mass[/tex]
[tex]400kg = mass[/tex]
3. [tex]density=\frac{mass}{volume}[/tex]
[tex]0.6=\frac{120}{volume}[/tex]
[tex]volume=\frac{120}{0.6}[/tex]
[tex]volume= 200cm[/tex]
Compare and contrast visible light, infrared light, and ultraviolet light.
Answer:
NE BİLİM
Explanation:
PUAN İÇİN YAZIYORUM SEN BENİ GÖRMEZDEN GEL ANLDIN MI DOSTUM bu arada Suriye >>>>>
The four tires of an automobile are inflated to a gauge pressure of 2.02×10^5 Pa. Each tire has an area of 217 cm^2 in contact with the ground. Determine the weight of the automobile.
The weight of the automobile is 17,533.6 N.
Weight of the automobile
The weight of the automobile is calculated as follows;
P = F / A
F = (W/4)
P = (W/4) / A
P = W/4A
W = 4AP
where;
P is pressure A is areaW = (4)(217 x 10⁻⁴ m²)(2.02 x 10⁵ Pa)
W = 17,533.6 N
Thus, the weight of the automobile is 17,533.6 N.
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b. Calculate the total resistance of the circuit below. (4 points)
c. In the circuit diagram above, the meters are labeled 1 and 2. Write 2 - 3 sentences identifying each type of meter and how it is connected with the 12 Ω resistor. (4 points)
d. In the circuit diagram above, predict which resistors (if any) will stop working when the switch is opened. Write 2 - 3 sentences explaining your reasoning. (4 points)
Please answer in complete sentences. Will mark brainliest.
The total resistance in the circuit, R is 4 Ω .
The meters connected to the 12 Ω resistance are:
Voltmeter - connected in parallelAmmeter - connected in seriesIf the switch is opened, only the 4 Ω and 2 Ω resistances will stop working.
What is the equivalent or total resistance in the circuit?The resistances in the circuit are connected both in series and in parallel
The resistances in series are the 4 Ω and the 2 Ω resistances.
Equivalent resistance = 4 + 2 = 6 Ω
The 4 Ω and the 2 Ω resistances are then connected in parallel with the 12 Ω resistance.
Total resistance, R is calculated as follows:
1/R = 1/12 + 1/6
1/R = 3/12
R = 12/3
R = 4Ω
The meters connected to the 12 Ω resistance are:
Voltmeter - connected in parallelAmmeter - connected in seriesIf the switch is opened, only the 4 Ω and 2 Ω resistances will stop working because the circuit connecting them to the cell is broken whereas the circuit to the 12 Ω resistance is continuous.
In conclusion, resistances can be connected in parallel or in series.
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For a projectile launched horizontally at 650 m/s and travels for 5.75 seconds, What is the range?
The range of horizontal projectile motion is 3737.5 m.
A projectile's horizontal velocity is constant (a never changing value), Gravity causes a downward vertical acceleration with a magnitude of 9.8 m/s/s. A projectile's vertical velocity fluctuates by 9.8 m/s per second, A projectile's vertical motion is unrelated to its horizontal motion. we know that sinθ is maximum at 90°. Therefore horizontal range will maximum at 45°.
To calculate the range of horizontal projectile motion we use;
Δx = vₓ t
where , Δx = Range
vₓ = Velocity
t = Time
Initial horizontal velocity, vₓ = 650 m/s
Time = 5.75 sec
Δx = 650 × 5.75
= 3737.5 m
Therefore, the range of horizontal projectile motion is 3737.5 m.
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200 g of water is heated and its temperature goes from 280 K to 300 K. What was the change in temperature for this process?
A. 280 K
B. 20 F
C. 20 K
D. 300 K
The change in temperature is 20 kelvin
What is Temperature?
Temperature can simply be described as how hot or how cold an object is at a particular period in time. The unit of temperature is Kelvin.
The formula for calculating change in temperature is
Final temperature - Initial temperature
Final temperature = 300 kelvin
Initial temperature= 280 kelvin
Change in temperature= 300-280
= 20 kelvin
Thus the change in temperature for this process is 20 kelvin
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What is the formula that relates current and voltage?
A. V = I/R
B. V = R/I
C. V = IR
D. V = I^2/R
"V = IR" is the equation that relates current to voltage.
Relationship between current and voltage:
Georg Simon Ohm, a German scientist, and mathematician carried out an experiment in 1827 using multiple circuits with variable wire lengths. He took measurements of both the voltage across the electrical component and the current flowing through the circuit.Ohm's law outlines the connection between voltage, current, and resistance. According to the equation,V = IR or,
I = V/R,
the amount of current (I) flowing through a circuit is inversely
related to the amount of resistance (r) and directly proportional to
the voltage (v).
Therefore the correct answer is option C i.e., "V = IR".
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A man exerts a horizontal force of 123 N on a crate with a mass of 40.2 kg.
(A) If the crate doesn't move, what's the magnitude of the static friction force (in N)?
______N
(B) What is the minimum possible value of the coefficient of static friction between the crate and the floor? (Assume the crate remains stationary.)?
The magnitude of the static friction force = 123 N
The coefficient of static friction = 0.31
What is static friction?Static friction is the frictional force that must be overcome in other for a body to that moving over another.
Since the crate does not move, the magnitude of the static frictional force is equal to the applied force.
The magnitude of the static frictional force = 123 N
The coefficient of static friction = frictional force/normal reaction
The coefficient of static friction = 123/(40* 9.8)
The coefficient of static friction = 0.31
In conclusion, the static frictional force on the on the crate is equal to the applied force on the crate.
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If the magnetic flux through a loop increases according to the relation; where , is in milliweber (mWb) and t is in seconds. Calculate the magnitude of e.m.f induced in the loop when t = 2s
The magnitude of e.m.f induced in the loop when t = 2 s is 31 Volts.
emf induced in the loopThe magnitude of e.m.f induced in the loop is calculated as follows;
emf = dФ/dt
Ф = 6t² + 7t
dФ/dt = 12t + 7
at t = 2 seconds
emf = dФ/dt = 12(2) + 7 = 31 V
Thus, the magnitude of e.m.f induced in the loop when t = 2 s is 31 Volts.
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Please show work if possible! Thank you!!
A 2.0 x 103 kg roller coaster travels around a vertical 24-m radius loop. If the coaster has a tangential speed of 18 m/s at the lowest point of the loop, what is the normal force that is exerted on the coaster by the track at this point?
a. 5.3 x 10^4 N
b. 4.7 x 10^4 N
c. 3.0 x 10^4 N
d. 2.7 x 10^4 N
B. The normal force that is exerted on the coaster by the track at the lowest point is 4.7 x 10⁴ N.
Normal force exerted on the coaster at the lowest point
Fₙ = mg + mv²/r
where;
m is mass of the coasterv is speed of the coasterr is radius of the pathFₙ = (2,000 x 9.8) + (2,000 x 18²)/24
Fₙ = 46,600 N
Fₙ = 4.7 x 10⁴ N
Thus, the normal force that is exerted on the coaster by the track at the lowest point is 4.7 x 10⁴ N.
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A 80-kg base runner begins his slide into second base when he is moving at a speed of 3.0 m/s. The coefficient of friction between his clothes and Earth is 0.70. He slides so that his speed is zero just as he reaches the base.
(a) How much mechanical energy is lost due to friction acting on the runner?
______J
(b) How far does he slide?
________m
The mechanical energy lost due to friction is 360 J.
The distance the runner slides is, s = 0.655 m
What is the mechanical energy of the runner?
The mechanical energy lost due to friction acting on the runner is equal to the change in kinetic energy.
Change in Kinetic energy = 1/2m(v -u)²
Change in Kinetic energy = 80 * (0 - 3.0)²/2
Change in Kinetic energy = 360 J
Mechanical energy lost = 360 J
Distance he slide is determined using the formula below as follows:
Acceleration of runner = coefficient of friction * acceleration due to gravity
acceleration, a = 0.7 * 9.8 = 6.86
v² = u² - 2as
s = v² + u²/2a
s = 0 + 3³/2 * 6.86
s = 0.655 m
In conclusion, the mechanical energy lost due to friction is the loss in kinetic energy.
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This technique allowed multiple peoples DNA to be compared to look for
similarities and differences in order to solve crime
DNA Examination
DNA Elimination
ODNA Sequencing
ODNA Fingerprinting
*2
DNA sequencing is the technique which allows multiple peoples DNA to be compared to look for similarities and differences in order to solve crime and is denoted as option C.
What is DNA?This is referred to as the deoxyribonucleic acid and contains the genetic components of organisms and is also located in the nucleus. On the other hand, DNA sequencing refers to the process of determining the nucleic acid sequence.
Each organisms has a unique DNA sequence which is why it is used to identify individuals through the use of bodily fluids such as saliva, blood etc.
It is used to compare the similarities and differences in order to solve crime thereby making it the most appropriate choice.
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A long uniform board weighs 52.8 N (10.6 lbs) rests on a support at its mid point. Two children weighing 206.0 N (41.2 lbs) and 272.0 N (54.4 lbs) stand on the board so that the board is balanced.
What is the upward force exerted on the board by the support?
The upward force exerted on the board by the support is 530.8 N.
Upward force exerted on the board by the supportThe upward force exerted on the board by the support is calculated as follows;
F(up) = 52.8 N + 206.0 N + 272.0 N
F(up) = 530.8 N
Thus, the upward force exerted on the board by the support is 530.8 N.
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Help me <3 please
Thank you :)
Answer:
11,890
Explanation:
First we need to know what is considered a significant figure.
A significant figure is a value that is not a zero at the start OR end of a value.
Which means, the 0 in the value of 90 or 0.363 are not considered a significant figure.
The 0 in the value of 3056 is considered a significant figure.
So from the table, we can deduce:
0.275 has 3 significant figures
750 has 2 significant figures
[tex]10.4 \times {10}^{5} = 1040000[/tex]
has 3 significant figures.
11,890 has 4 significant figures.
320,050 has 5 significant figures.
So from the above, we can already see the answer.
Forces always act in ________. A. Solitude B. Pairs C. Unpredictable ways D. Interesting ways
Three strings, attached to the sides of a rectangular frame, are tied together by a knot as shown in the figure. The magnitude of the tension in the string labeled C is 56.3 N.
Calculate the magnitude of the tension in the string marked A.
The magnitude of the tension in the string marked A and B is mathematically given as
A = 52.5 N
What is the magnitude of the tension in the string marked A?Generally, the equation for is mathematically given as
tan=3/8
negative x
[tex]B= tan\phi \\tan \phi=5/4[/tex]
negative x
C= tan=1/6
Hence, considering the Horizontal components
74.9cos(9.46) = A*cos(20.6) + B*cos(51.3)
A = 78.9 - 0.668B
Vertical components
74.9*sin(9.46) + Asin(20.6) = Bsin(51.3)
40.07 = 1.015B
B = 39.5 N
In conclusion, Sub the value of B is the equation of A
A = 78.9 - 0.668B
Sub
A = 78.9 - 0.668( 39.5 N)
A = 52.5 N
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Answer:
overloading can be avoided if two many appliances are not connected to a single socket short circuiting is a name given to a situation in which they live and the natural voice accidentally coming contact