Answer:
Explanation:
The whole wave front may be divided into two halves , the upper half and the lower half . Waves coming from top of the slit or top of upper half and top of lower half or from the mid point of slit can form minima at given point only when there is phase difference of π radian or path difference of λ or one wavelength. Every other point in upper half and corresponding point in lower half will interfere destructively at that point and will form dark spot at the given point . In this way minima will be formed at that point
Hence the phase difference between the Huygens wavelet from the top of the slit and the wavelet from the midpoint of the slit at first minima is π radian .
If your metal car moves over a wide, closed loop of wire embedded in a road surface, is the magnetic field of the Earth within the loop altered? Does this produce a current pulse?
Answer:
Yes it produces current
Explanation:
Because If this enclosed field is somehow changed, then following the law of electromagnetic induction, a pulse of current will be produced in the loop. Dues to a change is produced in the electric field when the iron parts of a car pass over it, momentarily increasing the strength of the field.
Two stars of masses M and 6M are separated by a distance D. Determine the distance (measured from M) to a point at which the net gravitational force on a third mass would be zero.
Answer:
0.29D
Explanation:
Given that
F = G M m / r2
F = GM(6m) / (D-r)2
G Mm/r2 = GM(6m) / (D-r)2
1/r2 = 6 / (D-r)2
r = D / (Ö6 + 1)
r = 0.29 D
See diagram in attached file
At which temperature do the lattice and conduction electron contributions to the specific heat of Copper become equal.
Answer:
At 3.86K
Explanation:
The following data are obtained from a straight line graph of C/T plotted against T2, where C is the measured heat capacity and T is the temperature:
gradient = 0.0469 mJ mol−1 K−4 vertical intercept = 0.7 mJ mol−1 K−2
Since the graph of C/T against T2 is a straight line, the are related by the straight line equation: C /T =γ+AT². Multiplying by T, we get C =γT +AT³ The electronic contribution is linear in T, so it would be given by the first term: Ce =γT. The lattice (phonon) contribution is proportional to T³, so it would be the second term: Cph =AT³. When they become equal, we can solve these 2 equations for T. This gives: T = √γ A .
We can find γ and A from the graph. Returning to the straight line equation C /T =γ+AT². we can see that γ would be the vertical intercept, and A would be the gradient. These 2 values are given. Substituting, we f ind: T =
√0.7/ 0.0469 = 3.86K.
A 0.10 kg point mass moves in a circular path with a radius of 0.36 m with a net force of 10.0 N toward the center of the circle. Select all of the following that are true statements.
a. The velocity of the object is 6 m/s toward the center of the circle.
b. The speed of the object is 6 m/s and decreasing.
c. The speed of the object is 6 m/s and increasing.
d. The velocity of the object is a constant 6 m/s.
e. The speed of the object is a constant 6 m/s.
Answer:
e. The speed of the object is a constant 6 m/s
Explanation:
Since the net force is towards the centre , hence there is no tangential acceleration . Only centripetal acceleration is there . Hence point mass is moving with uniform speed . Let it be u .
Centripetal force = m v² / r , r is radius of circular path .
Putting the given values
.10 x v² / .36 = 10
v = 6 m /s
A conducting sphere with radius R is charged until the magnitude of the electric field just outside its surface is E. The electric potential of the sphere, relative to the potential for away, is: Group of answer choices 0 E/R E/R2 ER ER2
Answer:
he correct answer is V = ER
Explanation:
In this exercise they give us the electric field on the surface of the sphere and ask us about the electric potential, the two quantities are related
ΔV = ∫ E.ds
where E is the elective field and normal displacement vector.
Since E is radial in a spray the displacement vector is also radial, the dot product e reduces to the algebraic product.
ΔV = ∫ E ds
ΔV = E s
since s is in the direction of the radii its value on the surface of the spheres s = R
ΔV = E R
checking the correct answer is V = ER
The principles of magnetism apply everywhere on earth. What does this tell us about God and His character?
Answer:
God is omnipresent.
Explanation:
This means God is everywhere and He works where ever we are in the world
The velocity selector in in a mass spectrometer consists of a uniform magnetic field oriented at 90 degrees to a uniform electric field so that a charge particle entering the region perpendicular to both fields will experience an electric force and a magnetic force that are oppositely directed. If the uniform magnetic field has a magnitude of 37.8 ~\text{mT}37.8 mT, then calculate the magnitude of the electric field that will cause a proton entering the velocity selector at 40.640.6 km/s to be undeflected. Give your answer in units of kV/m.
Answer:
50k/h is the answer to iy
The electric field of a sinusoidal electromagnetic wave obeys the equation E = (375V /m) cos[(1.99× 107rad/m)x + (5.97 × 1015rad/s)t].
(a) What is the speed of the wave?
(b) What are the amplitudes of the electric and magnetic fields of this wave?
(c) What are the frequency, wavelength, and period of the wave? Is this light visible to humans?
Answer:
a) v = 2,9992 10⁸ m / s , b) Eo = 375 V / m , B = 1.25 10⁻⁶ T,
c) λ = 3,157 10⁻⁷ m, f = 9.50 10¹⁴ Hz , T = 1.05 10⁻¹⁵ s , UV
Explanation:
In this problem they give us the equation of the traveling wave
E = 375 cos [1.99 10⁷ x + 5.97 10¹⁵ t]
a) what the wave velocity
all waves must meet
v = λ f
In this case, because of an electromagnetic wave, the speed must be the speed of light.
k = 2π / λ
λ = 2π / k
λ = 2π / 1.99 10⁷
λ = 3,157 10⁻⁷ m
w = 2π f
f = w / 2 π
f = 5.97 10¹⁵ / 2π
f = 9.50 10¹⁴ Hz
the wave speed is
v = 3,157 10⁻⁷ 9.50 10¹⁴
v = 2,9992 10⁸ m / s
b) The electric field is
Eo = 375 V / m
to find the magnetic field we use
E / B = c
B = E / c
B = 375 / 2,9992 10⁸
B = 1.25 10⁻⁶ T
c) The period is
T = 1 / f
T = 1 / 9.50 10¹⁴
T = 1.05 10⁻¹⁵ s
the wavelength value is
λ = 3,157 10-7 m (109 nm / 1m) = 315.7 nm
this wavelength corresponds to the ultraviolet
A beach ball filled with air is pushed about 1 m below the surface of a swimming pool and released from rest. Which of the following statements are valid, assuming the size of the ball remains the same?
a) The buoyant force on the ball decreases as the ball approaches the surface of the pool.
b) As the ball rises in the pool, the buoyant force on it increases.
c) The buoyant force on the ball equals its weight and remains constant as the ball rises.
d) The buoyant force on the ball while it is submerged is approximately equal to the weight of the volume of water that could fill the ball.
e) When the ball is released, the buoyant force exceeds the gravitational force, and the ball accelerates upward.
Answer:
e is correct
Explanation:
When a ball is pushed below the surface of a pool, it is submerged when the buoyant force is approximately equal to the water's weight of the volume that could fill the ball. When the ball is released, the buoyant force becomes greater than the gravitational force so that the ball accelerates upward.
What is buoyant force?The buoyant force can be described as the upward force exerted on an object wholly or partially immersed in a fluid and is also called Upthrust. A body submerged partially or fully in a fluid due to the buoyant force appears to lose its weight.
The following factors affect buoyant force the density of the fluid, the volume of the fluid displaced, and the local acceleration due to gravity.
When an object immerses in water, the object experiences a force from the downward direction opposite to the gravitational pull, which causes a decrease in its weight. The difference in this pressure gives the upward force on the object, as buoyancy.
Therefore, options (d), (e) are correct.
Learn more about buoyant force, here:
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Consider an electromagnetic wave where the electric field of an electromagnetic wave is oscillating along the z-axis and the magnetic field is oscillating along the x-axis.
In what directions is it possible that the wave is traveling?
A. The-z direction.
B. The ty direction
C. The +x direction.
D. The -y direction
E. The -x direction.
F. The +z direction.
Answer:
The wave will be travelling in the y-axis
Explanation:
An e-m wave has a spatially varying electric field that is always associated with a magnetic field that changes over time and vice versa. The electric field and the magnetic field oscillates perpendicularly to each other, and together form a wave that travels in a perpendicular direction to the magnetic and the electric field in space. The movement of the e-m wave through space is usually away from the source where it is generated. So, if the electric field travels in the z-axis, and the magnetic field travels through along the x-axis, then the e-m wave generated will travel in the y-axis direction.
A circular loop in the plane of a paper lies in a 0.75 T magnetic field pointing into the paper. The loop's diameter changes from 18.0 cm to 6.8 cm in 0.46 s.
A) Determine the direction of the induced current and justify your answer.
B) Determine the magnitude of the average induced emf.
C) If the coil resistance is 2.5 Ω, what is the average induced current?
Answer:
Explanation:
A.the direction of induced current will be clockwise
B: Changing 18cm and 6.8cm into 0.18m and 0.68
2.5
Divide them both by 2 to find the radius . Now we have 0.09 and .034m.
Now use Φ=(π*0.09^2)(.75 T)cos0 and the 0.019wb
(π*0.034^2)(.75 T)cos0 and the 0.00272wb
ow use ε=-N(ΔΦ/Δt)
For ΔΦ, 0.091-0.0027=0.0883
C.
To find the current, use I=ε/R
0.0883/2.5= 0.035A
Astronauts increased in height by an average of approximately 40 mm (about an inch and a half) during the Apollo-Soyuz missions, due to the absence of gravity compressing their spines during their time in space. Does something similar happen here on Earth
Answer:
Yes. Something similar occurs here on Earth.
Explanation:
Gravity tends to pull objects perpendicularly to the ground. In space, the absence of this force means there is no compression on the spine due to gravity trying to pull it down. This means that astronauts undergo an increase in height in space.
Here on Earth, we experience gravity pull on our spine during the day. At night when we sleep, we lie down with our spine parallel to the ground, which means that our spine is no longer under compression from gravity force. The result is that we are a few centimetres taller in the morning when we wake up, than we are before going to bed at night. The increase is not much pronounced here on Earth because there is a repeated cycle of compression and decompression of our spine due to gravity, unlike when compared to that of astronauts that spend long duration in space, all the while without gravity forces on their spine
When a nerve cell depolarizes, charge is transferred across the cell membrane, changing the potential difference. For a typical nerve cell, 9.0 pC of charge flows in a time of 0.50 ms. Part A What is the average current
Answer:
1.8 x 10^-8A
Explanation:
Using the equation for current:
I= Q/t
current = (9X10^-12)/ (0.50X 10^-3)
= 1.8^-8A
A cylindrical shell of radius 7.00 cm and length 2.59 m has its charge uniformly distributed on its curved surface. The magnitude of the electric field at a point 20.1 cm radially outward from its axis (measured from the midpoint of the shell) is 36.0 kN/C.
A) Use approximate relationships to find the net charge on the shell.
B) Use approximate relationships to find the electric field at a point 4.00 cm from the axis, measured radially outward from the midpoint of the shell.
A bullet with a mass of 20 g and a speed of 960 m/s strikes a block of wood of mass 4.5 kg resting on a horizontal surface. The bullet gets embedded in the block. The speed of the block immediately after the collision is:________.
A) cannot be found because we don't know whether the surface is frictionless.
B) is 0.21 km/s.
C) is 65 m/s.
D) is 9.3 m/s.
E) None of these is correct
Answer:
4.25m/sE. None of the option is correctExplanation:
Using the law of conservation of momentum to solve the problem. According to the law, the sum of momentum of the bodies before collision is equal to the sum of the bodies after collision. The bodies move with the same velocity after collision.
Mathematically.
mu + MU = (m+M)v
m and M are the masses of the bullet and the block respectively
u and U are their respective velocities
v is their common velocity
from the question, the following parameters are given;
m = 20g = 0.02kg
u = 960m/s
M = 4.5kg
U =0m/s (block is at rest)
Substituting this values into the formula above to get v;
0.02(960)+4.5(0) = (0.02+4.5)v
19.2+0 = 4.52v
4.52v = 19.2
Dividing both sides by 4.52
4.52v/4.52 = 19.2/4.52
v = 4.25m/s
Since they have the same velocity after collision, then the speed of the block immediately after the collision is also 4.25m/s
Consider the uniform electric field E = (8.0ĵ + 2.0 ) ✕ 103 N/C. What is its electric flux (in N · m2/C) through a circular area of radius 9.0 m that lies in the xy-plane? (Enter the magnitude.)
Answer:
5.09 x 10⁵ Nm²/C
Explanation:
The electric flux φ through a planar area is defined as the electric field Ε times the component of the area Α perpendicular to the field. i.e
φ = E A
From the question;
E = (8.0j + 2.0k) ✕ 10³ N/C
r = radius of the circular area = 9.0m
A = area of a circle = π r² [Take π = 3.142]
A = 3.142 x 9² = 254.502m²
Now, since the area lies in the x-y plane, only the z-component of the electric field is responsible for the electric flux through the circular area.
Therefore;
φ = (2.0) x 10³ x 254.502
φ = 5.09 x 10⁵ Nm²/C
The electric flux is 5.09 x 10⁵ Nm²/C
A platinum wire is used to determine the melting point of indium. The resistance of the platinum wire is 2.000 Ω at 20°C and increases to 3.072 Ω as indium starts to melt. What is the melting point of indium? (The temperature coefficient of resistivity for platinum is 3.9 ×
Answer:
The melting point of indium is 157.436 degrees Celsius.
Explanation:
The resistance of the platinum wire, R1 = 2
The temperature at R1 is, T1 = 20 degrees Celsius.
The increased resistance, R2 = 3.072
Let the temperature at 3.072 = T2
Now find the temperature at which the indium starts melting.
We know that α = ( R2 - R1 ) / [ R1 × ( T2 - T1 ) ]
Given, α = 3.9 x 10^-3/ degrees Celsius.
T2- T1 = ( R2 - R1 ) / R1 α
T2 – T1 = (3.072 – 2) / (2 × 3.9 x 10^-3)
T2 – T1 = 137.436
T2 = T1 + 137.436
T2 = 20 + 137.436
T2 = 157.436 degree Celsius
Answer:
1,772 C
Explanation:
one arm of a u shaped tube contains water and the other alcohol. if the two fluids meet exactly at the bottom of the U and the alcohol is at a height of 18 cm at what height will water be
Complete Question
One arm of a U-shaped tube (open at both ends) contains water, and the other alcohol.
If the two fluids meet at exactly the bottom of the U, and the alcohol is at a height of 18 cm, at what height will the water be?Assume the density of alcohol is [tex]\rho_a = 790\ kg/m^3[/tex]
Answer:
The height of water is [tex]h_w = 0.142 \ m[/tex]
Explanation:
From the question we are told that
The height of the alcohol is [tex]h_a =18 \ cm = 0.18 \ m[/tex]
The density of the alcohol is [tex]\rho_a = 790\ kg/m^3[/tex]
Generally the pressure on both arm of the tube are equal given that they are both open
i,e [tex]P_a = P_w[/tex]
Where [tex]P_a[/tex] is pressure of alcohol and [tex]P_w[/tex] is pressure of water
So the pressure on the arm of the tube containing the alcohol is mathematically evaluated as
[tex]P_a = g * h * \rho[/tex]
substituting values
[tex]P_a =9.8 * 0.18 * 790[/tex]
[tex]P_a = 1394 \ Pa[/tex]
Generally the pressure on the arm of the tube containing the water is mathematically evaluated as
[tex]P_w = g * h_w * \rho_w[/tex]
where [tex]\rho_w[/tex] is the density of water which has a value [tex]\rho _w = 1000 \ kg/m^3[/tex]
So
[tex]1394 = 9.8 * h_w * 1000[/tex]
=> [tex]h_w = \frac{1394}{9800}[/tex]
=> [tex]h_w = 0.142 \ m[/tex]
A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse is sent down the string. How long does it take the pulse to travel the 3.00 m of the string
Answer:
The time interval is [tex]t = 5.48 *10^{-3} \ s[/tex]
Explanation:
From the question we are told that
The length of the string is [tex]l = 3.00 \ m[/tex]
The mass of the string is [tex]m = 5.00 \ g = 5.0 *10^{-3}\ kg[/tex]
The tension on the string is [tex]T = 500 \ N[/tex]
The velocity of the pulse is mathematically represented as
[tex]v = \sqrt{ \frac{T}{\mu } }[/tex]
Where [tex]\mu[/tex] is the linear density which is mathematically evaluated as
[tex]\mu = \frac{m}{l}[/tex]
substituting values
[tex]\mu = \frac{5.0 *10^{-3}}{3}[/tex]
[tex]\mu = 1.67 *10^{-3} \ kg /m[/tex]
Thus
[tex]v = \sqrt{\frac{500}{1.67 *10^{-3}} }[/tex]
[tex]v = 547.7 m/s[/tex]
The time taken is evaluated as
[tex]t = \frac{d}{v}[/tex]
substituting values
[tex]t = \frac{3}{547.7}[/tex]
[tex]t = 5.48 *10^{-3} \ s[/tex]
1 A diffraction grating has a spacing of 1.6 10-m.
A beam of light is incident normally on the
grating. The first order maximum makes an angle
of 20° with the undeviated beam.what is wavelength of the incident light
Two parallel wires run in a north-south direction. The eastern wire carries 15.0 A northward while the western wire carries 6.0 A northward. If the wires are separated by 30 cm, what is the magnetic field magnitude and direction at a point between the wires at a distance of 10 cm from the western wire?
Answer:
The magnitude and direction of the magnetic field is 2.7 x 10⁻⁵ T upwards
Explanation:
Given;
current in the eastern wire, [tex]I_e[/tex] = 15 A
current in the western wire, [tex]I_w[/tex] = 6 A
distance between the wires, d = 30 cm = 0.3 m
The magnetic field at a distance R from a line current I, is given as;
[tex]B = \frac{\mu_o I }{2 \pi R}[/tex]
The magnetic field between the wires, are in opposite directions, and since the currents are also in opposite directions, the magnetic fields of the wires will be added.
The total field = magnetic field (east) + magnetic field (west);
[tex]B = \frac{\mu_o I_e}{2 \pi R_e} + \frac{\mu_0 I_w}{2 \pi R_w} \\\\B = \frac{\mu_o}{2\pi} (\frac{I_e}{R_e} + \frac{I_w}{R_w})[/tex]
where;
[tex]R_w[/tex] is the distance of the field from west = 10cm = 0.1 m
[tex]R_e[/tex] is the distance of the field on east from west = d - 10cm = 30cm - 10cm = 20cm = 0.2 m
The total magnetic field is;
[tex]B = \frac{\mu_o}{2\pi} (\frac{I_e}{R_e} + \frac{I_w}{R_w})\\\\B = \frac{4\pi *10^{-7}}{2\pi} (\frac{15}{0.2} + \frac{6}{0.1})\\\\B = 2*10^{-7}(75 + 60)\\\\B = 2*10^{-7}(135)\\\\B = 2.7*10^{-5} \ T[/tex]
Since total magnetic field is positive, the direction of the field is upwards (positive y direction)
Therefore, the magnitude and direction of the magnetic field is 2.7 x 10⁻⁵ T upwards
Rank in order, from largest to smallest, the magnitudes of the electric field at the black dot. A. 2, 1, 3, 4 B. 1, 4, 2, 3 C. 3, 2, 1, 4 D. 3, 1, 2, 4
Given that,
Rank in order from largest to smallest the magnitude of the electric field at block dot.
Electric field :
Electric field is proportional to the charge divided by square of distance.
In mathematically,
[tex]E\propto\dfrac{q}{r^2}[/tex]
Where, q = charge
r = distance
If the charge is greater then electric field will be greater.
If the distance is greater then electric field will be smaller.
We need to find the electric field at black dot
According to figure,
(I). The electric field at black dot due to positive charge point q to left. the distance is r.
The electric field will be
[tex]E=\dfrac{kq}{r^2}[/tex]
The electric field will be largest.
(II). The electric field at black dot due to positive charge point 2q to left. The distance is 2r.
Then, the electric field will be
[tex]E=\dfrac{k2q}{(2r)^2}[/tex]
[tex]E=\dfrac{kq}{2r^2}[/tex]
The electric field will be smallest.
(III). The electric field at black dot due to positive charge point 2q to left. The distance is r.
Then, the electric field will be
[tex]E=\dfrac{k2q}{(r)^2}[/tex]
The electric field will be very largest.
(IV). The electric field at black dot due to positive charge point q to left. The distance is 2r.
Then, the electric field will be
[tex]E=\dfrac{kq}{(2r)^2}[/tex]
[tex]E=\dfrac{kq}{4r^2}[/tex]
The electric field will be very smallest.
So, The electric field from largest to smallest will be
[tex]E_{3}>E_{1}>E_{2}>E_{4}[/tex]
Hence, The ranking will be 3, 1, 2, 4.
(D) is correct option.
The speed of a particle moving in a circle 2.0 m in radius increases at the constant rate of 4.4 m/s2. At an instant when the magnitude of the total acceleration is 6.0 m/s2, what is the speed of the particle? Group of answer choices
Answer:
The speed of the particle is 2.86 m/s
Explanation:
Given;
radius of the circular path, r = 2.0 m
tangential acceleration, [tex]a_t[/tex] = 4.4 m/s²
total magnitude of the acceleration, a = 6.0 m/s²
Total acceleration is the vector sum of tangential acceleration and radial acceleration
[tex]a = \sqrt{a_c^2 + a_t^2}\\\\[/tex]
where;
[tex]a_c[/tex] is the radial acceleration
[tex]a = \sqrt{a_c^2 + a_t^2}\\\\a^2 = a_c^2 + a_t^2\\\\a_c^2 = a^2 -a_t^2\\\\a_c = \sqrt{a^2 -a_t^2}\\\\a_c = \sqrt{6.0^2 -4.4^2}\\\\a_c = \sqrt{16.64}\\\\a_c = 4.08 \ m/s^2[/tex]
The radial acceleration relates to speed of particle in the following equations;
[tex]a_c = \frac{v^2}{r}[/tex]
where;
v is the speed of the particle
[tex]v^2 = a_c r\\\\v= \sqrt{a_c r} \\\\v = \sqrt{4.08 *2}\\\\v = 2.86 \ m/s[/tex]
Therefore, the speed of the particle is 2.86 m/s
To understand the meaning of the variables in Gauss's law, and the conditions under which the law is applicable. Gauss's law is usually written
ΦE=∫E.dA =qencl/ϵ0
, where ϵ0=8.85×10−12C2/(N⋅m2) is the permittivity of vacuum.
How should the integral in Gauss's law be evaluated?
a. around the perimeter of a closed loop
b. over the surface bounded by a closed loop
c. over a closed surface
Answer:
Explanation:
jjjjjjjjjjjjjjjj
Why does front side of spoon forms inverted image but the back side form opposite of inverted image?
Answer:
our face is outside the focal length of the concave side of the spoon. We see a virtual inverted image whereas in case of concave mirror we can see a virtual image which is erect.
The wave number (k) and the angular frequency of a wave are 6.2rad/m and 12rad/s respectively. Which of the followings could be the equation of the wave?
a. 8sin(12x-6.2t)
b. 8sin(6.2x-12t)
c. 12 sin (6.2x-8t)
d. all of the above
Answer:
B. [tex]8sin(6.2x-12t)[/tex]
Explanation:
The general equation of a wave is expressed as [tex]y = Asin(kx-\omega t)[/tex]
A is the amplitude of the wave
k is the wave number and it is expressed as [tex]k =\frac{2\pi}{\lambda}[/tex]
[tex]\omega[/tex] is the angular frequency expressed as [tex]\omega = 2\pi f[/tex]
[tex]\lambda[/tex] is the wavelength and f is the angular frequency
Given k = 6.2rad/m and [tex]\omega = 12rad/s[/tex]
On substituting this value into the general wave equation;
[tex]y = Asin(kx-\omega t)\\y = Asin(6.2x-12t)[/tex]
From the expression gotten, the only equation that could be the equation of the wave is [tex]y = 8sin(6.2x-12t)[/tex]
The speed of sound in air is 340 m/s, and the density of air is 1.2 kg/m3. If the displacement amplitude of a 330-Hz sound wave is 14 µm, what is its pressure-variation amplitude?
I bel.ieve the answer is 279Ghz
The required value of pressure-variation amplitude of the given sound wave is 11.84 Pa.
Given data:
The speed of sound in air is, v = 340 m/s.
The density of air is, [tex]\rho = 1.2 \;\rm kg/m^{3}[/tex].
The frequency of sound wave is, f = 330 Hz.
The displacement amplitude of sound wave is, [tex]A = 14 \;\rm \mu m= 14 \times 10^{-6} \;\rm m[/tex].
The standard expression for the pressure variation amplitude for the sound wave propagating in air medium is,
[tex]\Delta P= B \times A \times K[/tex]
Here,
B is the Bulk Modulus and its value is, [tex]B = \rho \times v^{2}[/tex].
K is the wave form constant and its value is, [tex]K = \dfrac{2 \pi f}{v}[/tex].
Solving as,
[tex]\Delta P= (\rho \times v^{2}) \times A \times \dfrac{2 \pi f}{v}\\\\\Delta P= (\rho \times v) \times A \times (2 \pi f)\\\\\Delta P= (1.2 \times 340) \times (14 \times 10^{-6}) \times (2 \pi \times 330)\\\\\Delta P= 11.84 \;\rm Pa[/tex]
Thus, we can conclude that the required value of pressure-variation amplitude of the given sound wave is 11.84 Pa.
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Do the math: How many seconds would it take an echo sounder’s ping to make the trip from a ship to the Challenger Deep (10,994 meters) and back? Recall that depth in meters = ½ (1500 m/sec × Echo travel time in seconds). Round your answer to two decimal places.
Answer:
14.66secsExplanation:
Given the formula for calculating the depth in metres expressed as
depth in meters = ½ (1500 m/sec × Echo travel time in seconds)
Given depth of the challenger = 10, 994 meters, we will substitute this given value into the formula given to calculate the time take for the echo to travel.
10, 994 = depth in meters = ½ * 1500 m/sec × Echo travel time in seconds
10,994 = 750 * Echo travel time in seconds
Dividing both sides by 750;
Echo travel time in seconds = 10,994 /750
Echo travel time in seconds ≈ 14.66secs (to two decimal places)
Therefore, it would take an echo sounder’s ping 14.66secs to make the trip from a ship to the Challenger Deep and back
To test the resiliency of its bumper during low-speed collisions, a 2 010-kg automobile is driven into a brick wall. The car's bumper behaves like a spring with a force constant 4.00 106 N/m and compresses 3.18 cm as the car is brought to rest. What was the speed of the car before impact, assuming no mechanical energy is transformed or transferred away during impact with the wall?
Answer:
Vi = 2 m/s
Explanation:
First we find the force applied to the car by wall to stop it. We use Hooke's Law:
F = kx
where,
F = Force = ?
k = spring constant = 4 x 10⁶ N/m
x = compression = 3.18 cm = 0.0318 m
Therefore,
F = (4 x 10⁶ N/m)(0.0318 m)
F = 127200 N
but, from Newton's Second Law:
F = ma
a = F/m
where,
m = mass of car = 2010 kg
a = deceleration = ?
Therefore,
a = 127200 N/2010 kg
a = 63.28 m/s²
a = - 63.28 m/s²
negative sign due to deceleration.
Now, we use 3rd equation of motion:
2as = Vf² - Vi²
where,
s = distance traveled = 3.18 cm = 0.0318 m
Vf = Final Speed = 0 m/s
Vi = Initial Speed = ?
Therefore,
2(- 63.28 m/s²)(0.0318 m) = (0 m/s)² - Vi²
Vi = √4.02 m²/s²
Vi = 2 m/s
For this study, the researcher is analyzing data by using __________ measures. A. experimental B. qualitative C. quantitative D. naturalistic
Answer:
B. qualitative
Explanation:
Since in the question it is mentioned that it takes the teacher opinion with respect to the student lunchroom that arranged the opinion in four classifications and according to that the report is also written
So this research represents the quality of the data collected as it does not specify the rating in terms of liking the student lunchroom
Therefore the correct option is B. qualitative