Answer:
final displacement = +24484.5 nm
Explanation:
The path difference when 158 bright spots were observed with red light (λ1 = 656.3 nm) is given as;
Δr = 2d2 - 2d1 = 150λ1
So, 2d2 - 2d1 = 150λ1
Dividing both sides by 2 to get;
d2 - d1 = 75λ1 - - - - eq1
Where;
d1 = distance between the fixed mirror and the beam splitter
d2 = position of moveable mirror from splitter when 158 bright spots are observed
Now, the path difference between the two waves when 114 bright spots were observed is;
Δr = 2d'2 - 2d1 = 114λ1
2d'2 - 2d1 = 114λ1
Divide both sides by 2 to get;
d'2 - d1 = 57λ1
Where;
d'2 is the new position of the movable mirror from the splitter
Now, the displacement of the moveable mirror is (d2 - d'2). To get this, we will subtract eq2 from eq1.
(d2 - d1) - (d'2 - d1) = 75λ1 - 57λ2
d2 - d1 - d'2 + d1 = 75λ1 - 57λ2
d2 - d'2 = 75λ1 - 57λ2
We are given;
(λ1 = 656.3 nm) and λ2 = 434.0 nm.
Thus;
d2 - d'2 = 75(656.3) - 57(434)
d2 - d'2 = +24484.5 nm
Two identical wooden barrels are fitted with long pipes extending out their tops. The pipe on the first barrel is 1 foot in diameter, and the pipe on the second barrel is only 1/2 inch in diameter. When the larger pipe is filled with water to a height of 20 feet, the barrel bursts. To burst the second barrel, will water have to be added to a height less than, equal to, or greater than 20 feet? Explain.
Answer:
The 1/2 inch barrel will burst at the same height of 20 ft
Explanation:
The pressure on a column of fluid increases with depth, and decreases with height. This means that if you increase the height of the fluid in the column, the pressure at the bottom will increase.
From the equation of fluid pressure,
P = ρgh
where
P is the pressure at the bottom of the fluid due to its height
ρ is the density of the fluid in question
h is the height to which the water stand.
You notice how apart from the height 'h' in the equation, all the other parts of the right hand side of the equation cannot be varied; they are a fixed property of the fluid and gravity. And there is no consideration for the horizontal diameter of the water's cross section area.
We can also think of the pressure at the bottom of the fluid to be as a result of an incremental weight of an infinitesimally small vertical section of the water down.
That been said, we can then say that if the barrel with the 1 ft diameter dimension bursts when filled with water up to 20 ft, then, the barrel with the reduced diameter will still burst at the same height as the former pipe.
NB: The only way to stop the pipe from bursting is to increase the thickness of the barrel wall to counteract the pressure forces due to the height.
In a polar coordinate system, the velocity vector can be written as . The term theta with dot on top is called _______________________ angular velocity transverse velocity radial velocity angular acceleration
Answer:
I believe it's called rapid growth
Explanation:
that is my answer no matter what
g A particle (charge = +40 mC) is located on the x axis at the point x = -20 cm, and a second particle (charge = -50 mC) is placed on the x axis at x = +30 cm. What is the magnitude of the total electrostatic force on a third particle (charge = -4.0 mC) placed at the origin (x = 0)? Group of answer choices
Answer:
Explanation:
We shall find electric field at origin due to two given charges sitting on the either side of origin .
Total field will add up due to their same direction .
Field due to a charge Q
= 9 x 10⁹ x Q / R² ; R is distance of point , Q is charge
Field due to first charge
= 9 x 10⁹ x 40 x 10⁻³ / 2² x 10⁻⁴
= 90 x 10¹⁰ N/C
Field due to second charge
= 9 x 10⁹ x 50 x 10⁻³ / 2² x 10⁻⁴
= 112.5 x 10¹⁰ N/C
Total field
= 202.5 x 10¹⁰ N/C
Force on given charge at origin
= charge x field
= 4 x 10⁻³ x 202.5 x 10¹⁰
= 810 x 10⁷ N .
Equal charges, one at rest, the other having a velocity of 104 m/s, are released in a uniform magnetic field. Which charge has the largest force exerted on it by the magnetic field
Answer:
case 1 of physics is the answer
The number of neutrons in the nucleus of zinc 65 Zn 30 is:
35
Need more data to answer
65
30
Explanation:
proton number + neutron number = atomic mass
30 + 35 = 65
Based on the graph below, what prediction can we make about the acceleration when the force is 0 newtons? A. It will be 0 meters per second per second. B. It will be 5 meters per second per second. C. It will be 10 meters per second per second. D. It will be 15 meters per second per second.
Answer:
Option A
Explanation:
From the graph, we came to know that Force and acceleration are in direct relationship.
Also,
Force = 0 when Acceleration = 0
Because Both are 0 at the origin.
Answer:
A. It will be 0 meters per second per second.
Explanation:
The force and acceleration is in a proportional relationship, that means the line goes through the origin.
On the graph, when the force is at 0, the acceleration is 0. The line passes through the origin.
A wheel rotating about a fixed axis has a constant angular acceleration of 4.0 rad/s2. In a 4.0-s interval the wheel turns through an angle of 80 radians. Assuming the wheel started from rest, how long had it been in motion at the start of the 4.0-s interval
Answer:
The time interval is [tex]t = 3 \ s[/tex]
Explanation:
From the question we are told that
The angular acceleration is [tex]\alpha = 4.0 \ rad/s^2[/tex]
The time taken is [tex]t = 4.0 \ s[/tex]
The angular displacement is [tex]\theta = 80 \ radians[/tex]
The angular displacement can be represented by the second equation of motion as shown below
[tex]\theta = w_i t + \frac{1}{2} \alpha t^2[/tex]
where [tex]w_i[/tex] is the initial velocity at the start of the 4 second interval
So substituting values
[tex]80 = w_i * 4 + 0.5 * 4.0 * (4^2)[/tex]
=> [tex]w_i = 12 \ rad/s[/tex]
Now considering this motion starting from the start point (that is rest ) we have
[tex]w__{4.0 }} = w__{0}} + \alpha * t[/tex]
Where [tex]w__{0}}[/tex] is the angular velocity at rest which is zero and [tex]w__{4}}[/tex] is the angular velocity after 4.0 second which is calculated as 12 rad/s s
[tex]12 = 0 + 4 t[/tex]
=> [tex]t = 3 \ s[/tex]
Following are the response to the given question:
Given:
[tex]\to \alpha = 4.0 \ \frac{rad}{s^2}\\\\[/tex]
[tex]\to \theta= 80\ radians\\\\\to t= 4.0 \ s\\\\ \to \theta_0=0\\[/tex]
To find:
[tex]\to \omega=?\\\\\to t=?\\\\[/tex]
Solution:
Using formula:
[tex]\to \theta- \theta_0 = w_{0} t+ \frac{1}{2} \alpha t^2\\\\ \to 80-0= \omega_{0}(4) + \frac{1}{2} (4)(4^2)\\\\ \to 80= \omega_{0}(4) + \frac{1}{2} (4)(16)\\\\\\to 80= \omega_{0}(4) + (4)(8)\\\\\to 80= \omega_{0}(4) + 32\\\\\to 80-32 = \omega_{0}(4) \\\\\to \omega_{0}(4)= 48 \\\\\to \omega_{0}= \frac{48}{4} \\\\ \to \omega_{0} = 12 \frac{rad}{ s} \\\\[/tex]
It would be the angle for rotation at the start of the 4-second interval.
This duration can be estimated by leveraging the fact that the wheel begins from rest.
[tex]\to \omega = \omega_{0} + \alpha t\\\\\to 12 = 0 +4(t) \\\\\to 12 = 4(t) \\\\ \to t=\frac{12}{4}\\\\\to t= 3\ s[/tex]
Therefore, the answer is "[tex]12\ \frac{rad}{s}[/tex] and [tex]3 \ s[/tex]".
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Two spherical objects at the same altitude move with identical velocities and experience the same drag force at a time t. If Object 1 has twice (2x) the diameter of Object 2, which object has the larger drag coefficient? Explain your answer using the drag equation.
Answer:
The object with the twice the area of the other object, will have the larger drag coefficient.
Explanation:
The equation for drag force is given as
[tex]F_{D} = \frac{1}{2}pu^{2} C_{D} A[/tex]
where [tex]F_{D}[/tex] IS the drag force on the object
p = density of the fluid through which the object moves
u = relative velocity of the object through the fluid
p = density of the fluid
[tex]C_{D}[/tex] = coefficient of drag
A = area of the object
Note that [tex]C_{D}[/tex] is a dimensionless coefficient related to the object's geometry and taking into account both skin friction and form drag. The most interesting things is that it is dependent on the linear dimension, which means that it will vary directly with the change in diameter of the fluid
The above equation can also be broken down as
[tex]F_{D}[/tex] ∝ [tex]P_{D}[/tex] A
where [tex]P_{D}[/tex] is the pressure exerted by the fluid on the area A
Also note that [tex]P_{D}[/tex] = [tex]\frac{1}{2}pu^{2}[/tex]
which also clarifies that the drag force is approximately proportional to the abject's area.
In this case, the object with the twice the area of the other object, will have the larger drag coefficient.
A thin film with an index of refraction of 1.60 is placed in one of the beams of a Michelson interferometer. If this causes a shift of 8 bright fringes in the pattern produced by light of wavelength 580 nm, what is the thickness of the film
Answer:
3.867 μm
Explanation:
The index of refraction, μ = 1.6
Wavelength of the light, λ = 580 nm
N2 - N1 = (2L / λ) (n2 - n1), Making L subject of formula, we have
(N2 - N1) λ = 2L (n2 - n1)
L = [(N2 - N1) * λ] / 2(n2 - n1)
L = (8 * 580) / 2(1.6 - 1.0)
L = 4640 nm / 1.2
L = 3867 nm or 3.867 μm
Therefore we can come to the conclusion that the thickness of the film is 3.867 nm
A charge is placed on a spherical conductor of radius r1. This sphere is then connected to a distant sphere of radius r2 (not equal to r1) by a conducting wire. After the charges on the spheres are in equilibrium:__________.
1. the electric fields at the surfaces of the two spheres are equal.
2. the amount of charge on each sphere is q/2.
3. both spheres are at the same potential. the potentials are in the ratio V2/V1 = q2/q1.
4. the potentials are in the ratio V2/V1 = r2/r1 .
Answer:
Option 3 = both spheres are at the same potential.
Explanation:
So, let us complete or fill the missing gap in the question above;
" A charge is placed on a spherical conductor of radius r1. This sphere is then connected to a distant sphere of radius r2 (not equal to r1) by a conducting wire. After the charges on the spheres are in equilibrium BOTH SPHERES ARE AT THE SAME POTENTIAL"
The reason both spheres are at the same potential after the charges on the spheres are in equilibrium is given below:
=> So, if we take a look at the Question again, the kind of connection described in the question above (that is a charged sphere, say X is connected another charged sphere, say Y by a conducting wire) will eventually cause the movement of charges(which initially are not of the same potential) from X to Y and from Y to X and this will continue until both spheres are at the same potential.
. A 24-V battery is attached to a 3.0-mF capacitor and a 100-ohm resistor. If the capacitor is initially uncharged, what is the voltage across the capacitor 0.16 seconds after the circuit is connected to the battery
Answer:
The voltage is [tex]V_c = 9.92 \ V[/tex]
Explanation:
From the question we are told that
The voltage of the battery is [tex]V_b = 24 \ V[/tex]
The capacitance of the capacitor is [tex]C = 3.0 mF = 3.0 *10^{-3} \ F[/tex]
The resistance of the resistor is [tex]R = 100\ \Omega[/tex]
The time taken is [tex]t = 0.16 \ s[/tex]
Generally the voltage of a charging charging capacitor after time t is mathematically represented as
[tex]V_c = V_o (1 - e^{- \frac{t}{RC} })[/tex]
Here [tex]V_o[/tex] is the voltage of the capacitor when it is fully charged which in the case of this question is equivalent to the voltage of the battery so
[tex]V_c = 24 (1 - e^{- \frac{0.16}{100 * 3.0 *10^{-1}} })[/tex]
[tex]V_c = 9.92 \ V[/tex]
The phenomenon of magnetism is best understood in terms ofA) the existence of magnetic poles.B)the magnetic fields associated with the movement of charged particles.C)gravitational forces between nuclei and orbital electrons.D) electrical fluid
Answer:
A) the existence of magnetic poles.Explanation:
Magnetism is defined as the ability of a magnet to attract magnetic substance to itself. Such magnet has the ability of being magnetized. A magnet is known to possess poles which are the north poles and south poles. The presence of this poles is what makes them possess the properties of a magnet. An ordinary steel bar doesn't have the properties of a magnet unless it is magnetized and when you are trying to magnetize a steel bar, you are invariably introducing the magnetic poles.
According to the law of magnetism, like poles repel but unlike poles attract. From the above explanation, it can be concluded that the phenomenon of magnetism is best understood interns of existence of magnetic poles. This poles are called the north and the south poles.
A conventional current of 3 A runs clockwise in a circular loop of wire in the plane, with center at the origin and with radius 0.093 m. Another circular loop of wire lies in the same plane, with its center at the origin and with radius 0.03 m. How much conventional current must run counterclockwise in this smaller loop in order for the magnetic field at the origin to be zero
Answer:
The current in the small radius loop must be 0.9677 A
Explanation:
Recall that the formula for the magnetic field at the center of a loop of radius R which runs a current I, is given by:
[tex]B=\mu_0\,\frac{I}{2\,R}[/tex]
therefore for the first loop in the problem, that magnetic field strength is:
[tex]B=\mu_0\,\frac{I}{2\,R} =\mu_0\,\frac{3}{2\,(0.093)} =16.129\,\mu_{0}\,[/tex]
with the direction of the magnetic field towards the plane.
For the second smaller loop of wire, since the current goes counterclockwise, the magnetic field will be pointing coming out of the plane, and will subtract from the othe field. In order to the addition of these two magnetic fields to be zero, the magnitudes of them have to be equal, that is:
[tex]16.129\,\,\mu_{0}=\mu_0\,\frac{I'}{2\,R'} =\mu_{0}\,\frac{I'}{2\,(0.03)} \\I'=16.129\,(2)\,(0.03)=0.9677\,\,Amps[/tex]
A flat loop of wire consisting of a single turn of cross-sectional area 8.20 cm2 is perpendicular to a magnetic field that increases uniformly in magnitude from 0.500 T to 2.60 T in 1.02 s. What is the resulting induced current if the loop has a resistance of 2.70
Answer:
The induced current is [tex]I = 6.25*10^{-4} \ A[/tex]
Explanation:
From the question we are told that
The number of turns is [tex]N = 1[/tex]
The cross-sectional area is [tex]A = 8.20 cm^2 = 8.20 * 10^{-4} \ m^2[/tex]
The initial magnetic field is [tex]B_i = 0.500 \ T[/tex]
The magnetic field at time = 1.02 s is [tex]B_t = 2.60 \ T[/tex]
The resistance is [tex]R = 2.70\ \Omega[/tex]
The induced emf is mathematically represented as
[tex]\epsilon = - N * \frac{ d\phi }{dt}[/tex]
The negative sign tells us that the induced emf is moving opposite to the change in magnetic flux
Here [tex]d\phi[/tex] is the change in magnetic flux which is mathematically represented as
[tex]d \phi = dB * A[/tex]
Where dB is the change in magnetic field which is mathematically represented as
[tex]dB = B_t - B_i[/tex]
substituting values
[tex]dB = 2.60 - 0.500[/tex]
[tex]dB = 2.1 \ T[/tex]
Thus
[tex]d \phi = 2.1 * 8.20 *10^{-4}[/tex]
[tex]d \phi = 1.722*10^{-3} \ weber[/tex]
So
[tex]|\epsilon| = 1 * \frac{ 1.722*10^{-3}}{1.02}[/tex]
[tex]|\epsilon| = 1.69 *10^{-3} \ V[/tex]
The induced current i mathematically represented as
[tex]I = \frac{\epsilon}{ R }[/tex]
substituting values
[tex]I = \frac{1.69*10^{-3}}{ 2.70 }[/tex]
[tex]I = 6.25*10^{-4} \ A[/tex]
A motorcycle travels up one side of a hill over the top and down the other side. The crest of the hill can be considered to be a circular arc with radius of 45.0 m. Determine the maximum speed that the cycle can have while moving over the crest without losing contact with the road.
Answer:
The maximum speed of the motorcycle should be 21 m/s
Explanation:
Since the hill is considered to be a circular arc, the motorcycle will experience centripetal force that tends to flip it away from the center of the hill.
Since the motorcycle does not lose contact with the ground, it means that the weight of the motorcycle downwards just balances the centripetal force on the motorcycle.
we know that the centripetal force on the motorcycle is equal to
centripetal force = [tex]\frac{mv^{2} }{r}[/tex]
where m is the mass of the motorcycle,
v is the velocity of the motorcycle,
and r is the radius of the hill = 45.0 m
Also we now that the weight of the motorcycle is equal to
weight = mg
where m is still the mass of the motorcycle,
and g is the acceleration due to gravity = 9.81 m/s
Equating the both forces since they are equal, we'll have
[tex]\frac{mv^{2} }{r}[/tex] = mg
the mass of the motorcycle will cancel out, and we'll be left with
[tex]v^{2} = gr[/tex]
[tex]v = \sqrt{gr}[/tex]
[tex]v = \sqrt{9.81*45}[/tex]
[tex]v = \sqrt{441.45}[/tex]
[tex]v[/tex] = 21 m/s
The rotor of a gas turbine is rotating at a speed of 7000 rpm when the turbine is shut down. It is observed that 3.5 minutes is required for the rotor to coast to rest. Assuming uniformly accelerated motion, determine the number of revolutions that the rotor executes before coming to rest. Hint: there will be a large number of rotations.
Answer:
The rotor of the gas turbine rotates 12250 revolutions before coming to rest.
Explanation:
Given that rotor of gas turbine is decelerating at constant rate, it is required to obtained the value of angular acceleration as a function of time, as well as initial and final angular speeds. That is:
[tex]\dot n = \dot n_{o} + \ddot n \cdot t[/tex]
Where:
[tex]\dot n_{o}[/tex] - Initial angular speed, measured in revolutions per minute.
[tex]\dot n[/tex] - Final angular speed, measured in revolutions per minute.
[tex]t[/tex] - Time, measured in minutes.
[tex]\ddot n[/tex] - Angular acceleration, measured in revoiutions per square minute.
The angular acceleration is now cleared:
[tex]\ddot n = \frac{\dot n - \dot n_{o}}{t}[/tex]
If [tex]\dot n_{o} = 7000\,\frac{rev}{min}[/tex], [tex]\dot n = 0\,\frac{rev}{min}[/tex] and [tex]t = 3.5\,min[/tex], the angular acceleration is:
[tex]\ddot n = \frac{0\,\frac{rev}{min}-7000\,\frac{rev}{min} }{3.5\,min}[/tex]
[tex]\ddot n = -2000\,\frac{rev}{min^{2}}[/tex]
Now, the final angular speed as a function of initial angular speed, angular acceleration and the change in angular position is represented by this kinematic equation:
[tex]\dot n^{2} = \dot n_{o}^{2} + 2\cdot \ddot n \cdot (n-n_{o})[/tex]
Where [tex]n[/tex] and [tex]n_{o}[/tex] are the initial and final angular position, respectively.
The change in angular position is cleared herein:
[tex]n-n_{o} = \frac{\dot n^{2}-\dot n_{o}^{2}}{2\cdot \ddot n}[/tex]
If [tex]\dot n_{o} = 7000\,\frac{rev}{min}[/tex], [tex]\dot n = 0\,\frac{rev}{min}[/tex] and [tex]\ddot n = -2000\,\frac{rev}{min^{2}}[/tex], the change in angular position is:
[tex]n-n_{o} = \frac{\left(0\,\frac{rev}{min} \right)^{2}-\left(7000\,\frac{rev}{min} \right)^{2}}{2\cdot \left(-2000\,\frac{rev}{min^{2}} \right)}[/tex]
[tex]n-n_{o} = 12250\,rev[/tex]
The rotor of the gas turbine rotates 12250 revolutions before coming to rest.
An electron traveling with a speed v enters a uniform magnetic field directed perpendicular to its path. The electron travels for a time t0 along a half-circle of radius R before leaving the magnetic field traveling opposite the direction it initially entered the field. Which of the following quantities would change if the electron had entered the field with a speed 2v? (There may be more than one correct answer.)
a. The radius of the circular path the electron travels
b. The magnitude of the electron's acceleration inside the field
c. The time the electron is in the magnetic field
d. The magnitude of the net force acting on the electron inside the field
Answer:
Explanation:
For circular path in magnetic field
mv² / R = Bqv ,
m is mass , v is velocity , R is radius of circular path , B is magnetic field , q is charge on the particle .
a )
R = mv / Bq
If v is changed to 2v , keeping other factors unchanged , R will be doubled
b )
magnitude of acceleration inside field
= v² / R
= Bqv / m
As v is doubled , acceleration will also be doubled
c )
If T be the time inside the magnetic field
T = π R / v
= π / v x mv / Bq
= π m / Bq
As is does not contain v that means T remains unchanged .
d )
Net force acting on electron
= m v² / R = Bqv
Net force = Bqv
As v becomes twice force too becomes twice .
So a . b , d are correct answer.
a figures skater rotating at 5 rads with arms extended has a moment of inertia of 2.25 kg. if the arms are pulled in so the moment of inertia decrease to 1.8 what is the final angular speed
Answer:
The final angular speed is 6.25 rad/s
Explanation:
Given;
initial angular speed, ω₁ = 5 rad/s
initial moment of inertia, I₁ = 2.25 kg.m²
Final moment of inertia, I₂ = 1.8 kg.m²
final angular speed, ω₂ = ?
Based on conservation of angular momentum, we will have the following expression;
ω₁I₁ = ω₂I₂
ω₂ = (ω₁I₁ ) / I₂
ω₂ = (5 x 2.25) / 1.8
ω₂ = 6.25 rad/s
Therefore, the final angular speed is 6.25 rad/s
An electron moves to the left along the plane of the page, while a uniform magnetic field points into the page. What direction does the force act on the moving electron
Answer:
acting force is the answer
The direction of the magnetic force on the moving electron is upward.
The direction of the magnetic force on the electron can be determined by applying right hand rule.
This rule states that when the thumb is held perpendicular to the fingers, the thumb will point in the direction of the speed while the fingers will point in the direction of the field and the magnetic force will be perpendicular to the field.
Thus, we can conclude that, the direction of the magnetic force on the moving electron is upward.
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A department store expects to have 225 customers and 20 employees at peak times in summer. Determine the contribution of people to the total cooling load of the store. The average rate of heat generation from people doing light work is 115 W, and 70% of it is in sensible form.
Answer:
The contribution of people to the cooling load of the store is 19722.5 W
Explanation:
Total amount of customers = 225
Total amount of employees = 20
Total amount of people in the store at that instant n = 245 people
Average rate of heat generation Q = 115 W
percentage of these heat generated that is sensible heat = 70%
Sensible heat raises the surrounding temperature. Latent heat only causes a change of state.
The total heat generated by all the people in the store = n x Q
==> 245 x 115 = 28175 W
but only 70% of this heat is sensible heat that raises the temperature of the store, therefore, the contribution of people to the cooling load of the store = 70% of 28175 W
==> 0.7 x 28175 = 19722.5 W
A 1.8 kg microphone is connected to a spring and is oscillating in simple harmonic motion up and down with a period of 3s. Below the microphone is 1.8 hz, calculate the spring constant
Answer:
230N/m
Explanation:
Pls see attached file
A physics professor wants to perform a lecture demonstration of Young's double-slit experiment for her class using the 633-nm light from a He-Ne laser. Because the lecture hall is very large, the interference pattern will be projected on a wall that is 4.0 m from the slits. For easy viewing by all students in the class, the professor wants the distance between the m
What slit separation is required in order to produce the desired interference pattern?
d=________m
Note: if the professor wants the distance between the m = 0 and m = 1 maxima to be 25 cm
Answer:
d = 1.0128×10⁻⁵m
Explanation:
given:
length L = 4.0m
maximum distance between m = 0 and m = 1 , y = 25cm = 0.25m
wavelength λ = 633nm = 633×10⁻⁹m
note:
dsinθ = mλ (constructive interference)
where d is slit seperation, θ is angle of seperation , m is order of interference , and λ is wavelength
for small angle
sinθ ≈ tanθ
[tex]d (\frac{y}{L}) =[/tex] mλ
[tex]d (\frac{y}{L}) = (1)(633nm)[/tex]
[tex]d(\frac{0.25}{4} ) = (1)(633nm)[/tex]
d = 1.0128×10⁻⁵m
. If you live in a region that has a particular TV station, you can sometimes pick up some of its audio portion on your FM radio receiver. Explain how this is possible. Does it imply that TV audio is broadcast as FM
Answer:
Please see below as the answer is self-explanatory.
Explanation:
The low band of the VHF TV Spectrum, spans channels 2-6, from 54 to 88 Mhz.
In the analog TV, in the Americas, the total bandwidth of any channel is 6 Mhz, with the visual carrier modulated in VSS (Vestigial Side Band) at 1.25 Mhz from the lowest frequency of the channel.
The aural carrier is located at 4.5 Mhz from the visual carrier, and is FM modulated.
For Channel 6, which spans between 82 and 88 Mhz, the visual carrier is at 83.25 Mhz, so the aural carrier is at 87.75 Mhz, which falls within the FM Band, so it is possible to listen the audio part of this channel in a FM radio receiver, even at a lower volume, due to the FM radio has a greater deviation than TV aural carrier.
The reason why it is possible for TV station to sometimes pick up some of the audio portion on your FM radio receiver is because; TV waves can sometimes deviate into the FM radio frequency range.
Let us start with explaining the waves of TV and radio.
The frequency range utilized by TV stations is either the range 54 MHz to 88 MHz or 174 MHz to 222 MHz. In contrast, the frequency range utilized by FM Radio band is between 88 MHz and 174 MHz.
Now, in some cases, it is possible that the TV signal may deviate into the range of the FM Radio and as such in that case, the TV signal will pick the audio portion of an FM Radio. These TV waves are very high frequency waves.
Finally, it does not imply that the TV wave is broadcasting as an FM because it only deviated a bit from the TV range and not like that is where it is made to operate.
Read more about TV waves at; https://brainly.com/question/9684913
g How many rpm would a 25 m diameter Ferris wheel need to travel if a 75 kg person were to experience an effective weight of 810 N at the lower-most point of the ride
Answer:
2.52 rpm
Explanation:
given that
diameter of the wheel, d = 25 m
Mass of the person, m = 75 kg
Weight experienced, N = 810 N
Since diameter is 25, radius then is 25/2 = 12.5 m
We all know that,
v = rw
Also, the passengers weight is equal to the centripetal acceleration, and thus
mg = mv²/r
Substitute for v, we have
mg = m/r * (rw)²
mg = mr²w²/r
g = rw²
If we make w the subject of formula, we have
w² = g/r
w = √(g/r)
mg = 810
75 * g = 810
g = 810 / 75
g = 1.08 m/s²
w = √(g/r)
w = √(1.08 / 12.5)
w = √0.0864
w = 0.294 rad/s
Since the question asked us in rpm, we convert to rpm
0.294 * (60 / 2π)
2.52 revolution per minute.
g Question 11 pts Consider two masses connected by a string hanging over a pulley. The pulley is a uniform cylinder of mass 3.0 kg. Initially m1 is on the ground and m2 rests 2.9 m above the ground. After the system is released, what is the speed of m2 just before it hits the ground? m1= 30 kg and m2= 35 kg Group of answer choices 2.1 m/s 1.4 m/s 9.8 m/s 4.3 m/s 1.9 m/s
Answer:
The speed of m2 just before it hits the ground is 2.1 m/s
Explanation:
mass on the ground m1 = 30 kg
mass oat rest at the above the ground m2 = 35 kg
height of m2 above the ground =2.9 m
Let the tension on the string be taken as T
for the mass m2 to reach the ground, its force equation is given as
[tex]m_{2} g - T = m_{2}a[/tex] ....equ 1
where g is acceleration due to gravity = 9.81 m/s^2
and a is the acceleration with which it moves down
For mass m1 to move up, its force equation is
[tex]T - m_{1} g = m_{1} a[/tex]
[tex]T = m_{1}a + m_{1}g[/tex]
[tex]T = m_{1}(a + g)[/tex] ....equ 2
substituting T in equ 1, we have
[tex]m_{2} g - m_{1}(a+g) = m_{2}a[/tex]
imputing values, we have
[tex](35*9.81) - 30(a+9.81) = 35a[/tex]
[tex]343.35 - 30a-294.3 = 35a[/tex]
[tex]343.35 -294.3 = 35a+ 30a[/tex]
[tex]49.05 = 65a[/tex]
a = 49.05/65 = 0.755 m/s^2
The initial velocity of mass m2 = u = 0
acceleration of mass m2 = a = 0.755 m/s^2
distance to the ground = d = 2.9 m
final velocity = v = ?
using Newton's equation of motion
[tex]v^{2}= u^{2} + 2ad[/tex]
substituting values, we have
[tex]v^{2}= 0^{2} + 2*0.755*2.9[/tex]
[tex]v^{2}= 2*0.755*2.9 = 4.379\\v = \sqrt{4.379}[/tex]
v = 2.1 m/s
If a bicycle starts from rest and is pedaled normally until the bike is moving at 6 meters per second across level ground, what kinds of energy have its tires been given? (Select all that apply) g
Answer: Translational Kinetic Energy
Rotational Kinetic Energy
Explanation:
An object has translational kinetic energy when it is undergoing through a linear displacement.
Rotational energy is kinetic energy due to the rotation of an object .
Here the wheel of bicycle undergoes both translational and rotational kinetic energy has it moves with linear displacement with rotation in it.
Hence, the tires have been two kinds of energy : translational and rotational kinetic energy
If the current flowing through a circuit of constant resistance is doubled, the power dissipated by that circuit will
Two people, one of mass 85 kg and the other of mass 50 kg, sit in a rowboat of mass 90 kg. With the boat initially at rest, the two people, who have been sitting at opposite ends of the boat, 3.5 m apart from each other, now exchange seats. How far does the boat move?
Answer:
0.11m
Explanation:
let's assume the boat is of uniform construction
Ignoring friction losses
Also assume the origin is at the end of the boat originally with the heavier person
the center of mass of the whole system will not change relative to the water when the two swap ends
Originally, the center of mass is
85[0] + 90[3.5/2] + 50[3.5] / (85 + 90 + 50) = 1.14m from the origin
after the swap, the center of mass is
50[0] + 90[3.5/2] + 85[3.5] / (85 + 90+ 50) = 1.030m from the origin
The center of mass has shifted
1.14-1.030 = 0.11m
as no external force acted on the system, the center of mass relative to the water will not change. The boat will therefore shift towards the end where the heavier person originally sat
Sergio has made the hypothesis that "the more time that passes, the farther away a person riding a bike will be." Do the data in the table below support his hypothesis? A. Yes, the data support the hypothesis. B. No, the data support the opposite of the hypothesis. C. The data show no relationship between the time passed and the distance.
Answer:
Option A
Explanation:
Given that
Distance = Speed / Time
So, they are in inverse relation.
Such that when the time passes, the distance from the reacing point will become less and vice versa.
So, Yes! The more time that passes, the farther away a person riding a bike will be.
Two point charges of +2.0 μC and -6.0 μC are located on the x-axis at x = -1.0 cm and x 12) = +2.0 cm respectively. Where should a third charge of +3.0-μC be placed on the +x-axis so that the potential at the origin is equal to zero?
Answer:
x = 0.006 m
Explanation:
The potential at one point is given by
V = k ∑ [tex]q_{i} / r_{i}[/tex]
remember that the potential is to scale, let's apply to our case
V = k (q₁ / x₁ + q₂ / x₂ + q₃ / x)
in this case they indicate that the potential is zero
0 = k (2 10⁻⁶ / (- 1 10⁻²) + (-6 10⁻⁶) / 2 10⁻² + 3 10⁻⁶ / x)
3 / x = + 2 / 10⁻² + 3 / 10⁻²
3 / x = 500
x = 3/500
x = 0.006 m