Answer:
17.54N in -x direction.
Explanation:
Amplitude (A) = 3.54m
Force constant (k) = 5N/m
Mass (m) = 2.13kg
Angular frequency ω = √(k/m)
ω = √(5/2.13)
ω = 1.53 rad/s
The force acting on the object F(t) = ?
F(t) = -mAω²cos(ωt)
F(t) = -2.13 * 3.54 * (1.53)² * cos (1.53 * 3.50)
F(t) = -17.65 * cos (5.355)
F(t) = -17.57N
The force is 17.57 in -x direction
You are given a long length of string and an oscillator that can shake one end of the string at any desired frequency. The oscillator has a display that indicates the frequency. You are asked to design an experiment to study how the velocity of waves on the string depends on the string's tension. You do not have any way to measure time with sufficient accuracy to help in your investigation.
A. Describe your experimental setup and procedure, including any additional pieces of equipment you would need and the kind of data you would record. Include enough detail that another student could follow and complete the experiment successfully.
B. Describe how you would analyze your data to obtain information about the wave velocity's dependence on tension.
Answer:
Explanation:
a.
AIM :
TO STUDY HOW VELOCITY OF WAVES ON THE STRING DEPENDS ON THE STRING'S TENSION.
APPARATUS:
Oscillator, long strings , some masses( to create tension in string) and the support ( rectangular wooden piece).
EXPERIMENTAL SETUP:
1. Measure the length of the string and mass of the weights used.
2. Connect one end of string to the oscillator.
3. Place the support below string on table such that the string is in same line without touching table.
4. After the support, the string should hang freely.
5. The other end of string is connected with some small measured masses which should be hanging.
PROCEDURE:
1. Note down the length of string and mass of weights.
2. Adjust the frequency in the oscillator which creates standing waves in the string.
3. Start from lower frequency and note down the lowest frequency at which mild sound is heard or when string forms one loop while oscillating.
4. Calculate the wavelength using of waves using length of string.
5. Calculate the velocity using frequency and wavelength.
6. Calculate linear mass density.
8. Repeat the procedure with different masses.
7. plot a graph with tension in y axis and linear mass density in x axis.
8. Find slope and compare with velocity.
Linear mass density
µ = m/l(kg-1)
tension
T = m x 9.8N
wave length
ƛ = 2L
b.
We can analyze the data by comparing slope of the graph, tension Vs linear mass density with velocity which is constant for constant length.
Write the slope value in terms of value of velocity and find the relationship between velocity and string's tension.
The expected result is
slope = v²
T ∝ V²
Like a transverse wave, a longitudinal wave has
-wavelength, speed, and frequency.
-amplitude, frequency, wavelength, and speed.
-amplitude, frequency, and speed.
-amplitude, wavelength, and speed.
-amplitude, frequency, and wavelength.
Final answer:
A longitudinal wave, like a transverse wave, has properties of amplitude, frequency, wavelength, and speed. These characteristics define the wave's physical behavior and are related by the equation v = fλ, where 'v' is wave speed, 'f' is frequency, and 'λ' is wavelength. Option 2 is correct.
Explanation:
Like a transverse wave, a longitudinal wave has amplitude, frequency, wavelength, and speed. Both types of waves have these fundamental properties, but the way they propagate through mediums is different. In longitudinal waves, the particles of the medium move parallel to the wave's direction of travel, while in transverse waves, particles move perpendicular to the direction of the wave's travel. An example of a transverse wave is a wave on a string, like when playing a guitar. In contrast, sound waves in air are longitudinal waves.
The wavelength (λ) is the distance between adjacent identical parts of the wave, which can be considered from one compression to the next in the case of longitudinal waves. Wave speed (v) is the rate at which the wave travels through the medium. The frequency (f) is the number of wave cycles that pass a given point per unit time, and amplitude refers to the maximum displacement from the equilibrium position within the wave cycle.
It's important to remember that all these properties are related: the speed of a wave is given by the product of its frequency and wavelength (v = fλ).
A girl on a bike is moving at a speed of 1.40 m/s at the start of a 2.45 m high and 12.4 m long incline. The total mass is 60.0 kg, air resistance and rolling resistance can be modeled as a constant friction force of 41.0 N, and the speed at the lower end of the incline is 6.70 m/s. Determine the work done (in J) by the girl as the bike travels down the incline.
Answer:
Explanation:
Given that,
Initial speed of the girl is
u = 1.4m/s
Height she is going is
H = 2.45m
Incline plane she will pass to that height
L = 12.4m
Mass of girl and bicycle is
M=60kg
Frictional force that oppose motion is
Fr = 41N
Speed at lower end of inclined plane
V2 = 6.7m/s
Work done by the girl when the car travel downward
Using conservation of energy
K.E(top) + P.E(top) + work = K.E(bottom) + P.E(bottom) + Wfr
Where Wfr is work done by friction
Wfr = Fr × d
P.E(bottom) is zero, sicne the height is zero at the ground
K.E is given as ½mv²
Then,
½M•u² + MgH + W = ½M•V2² + 0 + Fr×d
½ × 60 × 1.4² + 60×9.8 × 2.45 + W = ½ × 60 × 6.7² + 41 × 12.4
58.8 + 1440.5 + W = 1855.1
W = 1885.1 —58.8 —1440.5
W = 355.8 J
Each of the following statements is arguably true of thermometers. Which of them is most helpful to keep in mind if you are conducting an experiment to measure the specific heat of a material? Group of answer choices It may take a few minutes for a thermometer to come into equilibrium with its surroundings. The temperature reported by a thermometer is never precisely the same as its surroundings. A thermometer can only report information about its own temperature. By definition, the temperature of the surrounding environment is exactly what the thermometer reports.
Answer:
The temperature reported by a thermometer is never precisely the same as its surroundings
Explanation:
In this experiment to determine the specific heat of a material the theory explains that when a heat interchange takes place between two bodies that were having different temperatures at the start, the quantity of heat the warmer body looses is equal to that gained by the cooler body to reach the equilibrium temperature. This is true only if no heat is lost or gained from the surrounding. If heat is gained or lost from the surrounding environment, the temperature readings by the thermometer will be incorrect. The experimenter should therefore keep in mind that for accurate results, the temperature recorded by the thermometer is similar to that of the surrounding at the start of the experiment and if it differs then note that there is either heat gained or lost to the environment.
When measuring the specific heat of a material, it is crucial to remember that thermometers require time to come into equilibrium with their surroundings for accurate temperature measurements. This ensures that the temperature readings accurately represent the material's temperature, which is essential for precise specific heat calculations.
The statement, "It may take a few minutes for a thermometer to come into equilibrium with its surroundings," is arguably the most helpful to keep in mind when conducting an experiment to measure the specific heat of a material. This principle is essential because it underlines the importance of allowing time for a thermometer to accurately reflect the temperature of the material it is measuring. Thermal equilibrium is a critical concept in thermodynamics, emphasizing that for accurate temperature measurement, both the thermometer and the material being tested must reach a state where no net heat flow occurs between them. This ensures that the temperature reading is actually representative of the material's temperature, rather than being influenced by initial differences in temperature between the thermometer and the material.
Understanding this concept is vital when measuring specific heat because specific heat calculations rely on accurate temperature measurements before and after a heat transfer occurs. If the thermometer does not accurately reflect the material's temperature due to inadequate time for equilibrium, the calculated specific heat could be significantly off.
A 54 kg man holding a 0.65 kg ball stands on a frozen pond next to a wall. He throws the ball at the wall with a speed of 12.1 m/s (relative to the ground) and then catches the ball after it rebounds from the wall. How fast is he moving after he catches the ball? Ignore the projectile motion of the ball, and assume that it loses no energy in its collision with the wall. Answer in units of m/s.
Answer:
The velocity of the man is 0.144 m/s
Explanation:
This is a case of conservation of momentum.
The momentum of the moving ball before it was caught must equal the momentum of the man and the ball after he catches the ball.
Mass of ball = 0.65 kg
Mass of the man = 54 kg
Velocity of the ball = 12.1 m/s
Before collision, momentum of the ball = mass x velocity
= 0.65 x 12.1 = 7.865 kg-m/s
After collision the momentum of the man and ball system is
(0.65 + 54)Vf = 54.65Vf
Where Vf is their final common velocity.
Equating the initial and final momentum,
7.865 = 54.65Vf
Vf = 7.865/54.65 = 0.144 m/s
Which one of the following statements concerning the proper time interval between two events is true? a) The proper time interval is the longest time interval that any inertial observer can measure for the event. b) The proper time interval is the shortest time interval that any inertial observer can measure for the event. c) The proper time interval is the time measured by an observer who is in motion with respect to the event. d) The proper time interval depends upon the speed of the observer. e) The proper time interval depends upon the choice of reference frame.
Answer:
d) The proper time interval depends upon the speed of the observer
Explanation:
Proper time is the time as measured by a clock moving with the body in motion.
The proper time interval between two events on a world line is the change in proper time, and it depends on the events and the world line connecting them, and also on the motion of the clock between the events.
A vertical wire carries current in the upward direction. An electron is traveling parallel to the wire. What is the angle ααalpha between the velocity of the electron and the magnetic field of the wire? Express your answer in degrees. View Available Hint(s) ααalpha = nothing degreesdegrees
Answer:
First of all note that The magnetic field produced by the vertical wire will be into on the right hand side and it will be out of the page on the left hand side
Assuming that the electron beam is coming from the right hand side of the page parallel to the wire, The direction of the velocity vector(V) is left and the direction of magnetic field due the wire(B) is into the page. If u use right hand rule, you will get the direction downwards but as the formula also depends on q , the charge on electron is negative .Therefore the direction will be inverted i.e Upwards.
If you assume the electron beam coming from left hand side.Then also u will get the same answer.
So, the angle α between the velocity of the electron and the magnetic field of the wire is 90°.
Explanation:
For an electron traveling parallel to a wire carrying an upward direction current, the angle between its velocity and the magnetic field of the wire, according to the right-hand rule, is 90 degrees.
Explanation:The setting of this problem involves a vertical wire carrying an upward direction current and an electron traveling parallel to it. According to the right-hand rule in magnetism, which states that if your thumb points in the direction of the current, then your fingers will curl in the direction of the magnetic field, the magnetic field of the wire would form concentric circles around the wire.
For an electron traveling parallel to the wire, according to this rule, it would be always at a right angle or 90 degrees to the magnetic field as it moves along the circumference of these imaginary concentric circles. Therefore, the angle alpha (ααalpha) between the velocity of the electron and the magnetic field of the wire is 90 degrees.
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A straight wire carries a current of 10 A at an angle of 30° with respect to the direction of a uniform 0.30-T magnetic field. Find the magnitude of the magnetic force on a 0.50-m length of the wire.
Answer:
Explanation:
Given that,
Current in wire is
I = 10A
And the current makes an angle of 30° with respect to the magnetic field
Then, θ = 30°
And the magnetic field is
B = 0.3 T
Length of the wire is
L = 0.5m
Force on the wire F?
The force on the wire in calculated using
F = iL × B
Where
The magnitude of the cross produce of L and B is
L × B = LB•Sinθ
Then, force becomes
F = iLB•Sinθ
F = 10 × 0.5 × 0.3 × Sin30
F = 0.75 N
The force on the wire is 0.75 Newton
Answer:
The magnitude of the magnetic force on the wire is 0.75 N
Explanation:
Given;
current in the wire, I = 10 A
angle of inclination to magnetic field, θ = 30°
magnetic field strength, B = 0.30-T
length of the wire, L = 0.50-m
The magnitude of magnetic force acting on the wire is given as;
F = BILSinθ
where;
B is magnetic field strength
I is the current in the wire
L is length of the wire
θ is the angle of inclination
F = 0.3 x 10 x 0.5 x sin(30)
F = 0.3 x 10 x 0.5 x 0.5
F = 0.75 N
Therefore, the magnitude of the magnetic force on the wire is 0.75 N
The tub of a washing machine goes into its spin cycle, starting from rest and gaining angular speed steadily for 9.00 s, at which time it is turning at 7.00 rev/s. At this point, the lid of the washing machine is opened, and a safety switch turns it off. The tub then smoothly slows to rest in 13.0 s. Through how many revolutions does the tub rotate while it is in motion
Answer:
Explanation:
7 rev /s = 7 x 2π rad /s
angular velocity = 14π rad /s
Angular acceleration α = increase in angular velocity / time
= 14π - 0 / 9
α = 4.8844 rad / s
θ = 1/2 α x t² θ is angle of rotation , t is time
= 1/2 x 4.8844 x 9²
= 197.8182 rad
2π n = 197.8182
n = 31.5 rotation
During acceleration , no of rotation made = 31.5
During deceleration : -----
deceleration =
Angular deceleration α = decrease in angular velocity / time
= 14π - 0 / 13
α = 3.3815 rad / s
θ = 1/2 α x t² θ is angle of rotation , t is time
= 1/2 x 3.3815 x 13²
= 285.736 rad
2π n = 285.736
n = 45.5 rotation
total rotation
= 45.5 + 31.5
= 77 .
A proton and a deuteron are moving with equal velocities perpendicular to a uniform magnetic field. A deuteron has the same charge as the proton but has twice its mass. The ratio of the magnetic force on the proton to that on the deuteron is:
a. 0.5.
b. 1.
c. 2.
d. There is no magnetic force in this case.
Answer:
option (b)
Explanation:
mass of proton, mp = m
mass of deuteron, md = 2m
charge on proton, qp = q
charge on deuteron, qd = q
The magnetic force on the charged particle when it is moving is given by
F = q v B Sinθ
where, θ is the angle between the velocity and magnetic field.
Here, θ = 90°
Let v is the velocity of both the particle when they enters in the magnetic field.
The force on proton is given by
Fp = q x v x B ...... (1)
The force on deuteron is
Fd = q x v x B .... (2)
Divide equation (1) by equation (2)
Fp / Fd = 1
Thus, the ratio of force on proton to the force on deuteron is 1 : 1.
Thus, option (b) is correct.
A visitor to a lighthouse wishes to determine the height of the tower. The visitor ties a spool of thread to a small rock to make a simple pendulum, then hangs the pendulum down a spiral staircase in the center of the tower. The period of oscillation is 9.49 s. What is the height of the tower?
Answer:
22.38 m
Explanation:
Using,
T = 2π√(L/g)................... Equation 1
Where T = period of the oscillation, L = Length of the pendulum,/Height of the tower. g = acceleration due to gravity .
Make L the subject of the equation
L = gT²/(4π²)..................... Equation 2
Given: T = 9.49 s, g = 9.8 m/s², π = 3.14
Substitute into equation 2
L = 9.8(9.49²)/(4×3.14²)
L = 22.38 m
Hence the height of the tower = 22.38 m
The pressure exerted by a phonograph needle on a record is surprisingly large, due to the very small width of the needle. show answer Incorrect Answer If the equivalent of 0.95 g is supported by a needle, the tip of which is a circle 0.205 mm in radius, what pressure is exerted on the record, in pascals
Answer:
Explanation:
Given that,
Mass support is
M = 0.95g= 0.95/1000 = 0.00095kg
Radius of circle R = 0.205mm
r = 0.205/1000 = 0.000205m
Then, area of the circle can be determined using
A = πr²
A = π × 0.000205²
A = 1.32 × 10^-7 m²
From pressure definitions
Pressure = Force / Area
The force is perpendicular to the area
Force = weight = mg
F = mg = 0.00095 × 9.8
F = 9.31 × 10^-3 N
Then,
Pressure = Force / Area
P = F/A
P = 9.31 × 10^-3 / 1.32 × 10^-7
P = 70,530.30 N/m²
Since 1 pascal = 1 N/m²
Then,
P = 70,530.30 Pascals
To find the pressure exerted by the phonograph needle, multiply the weight of the needle by the gravitational constant to get the force, calculate the area of the needle tip using the given radius, and divide the force by the area using the pressure formula P = F / A.
Explanation:The pressure exerted by the needle on the record can be calculated using the formula for pressure P = F / A, where F is the force (in this case the weight of the needle) and A is the area over which the force is applied (in this case the area of the needle tip).
First, you need to convert the weight of the needle into a force. Since weight is a force caused by gravity acting on a mass, you can find it by multiplying the mass of the needle by the acceleration due to gravity (g = 9.8 m/s²). So, F = 0.95 g * 9.8 m/s². However, you need to convert grams to kilograms (as 1g = 0.001kg). Hence, F = 0.95 * 0.001 kg * 9.8 m/s².
Next, you find the area of the very tip of the needle. Since the tip is circular, we use the formula for the area of a circle, A = πr², where r is the radius of the needle tip. Substituting r = 0.205 mm = 0.205 * 10^-3 m (since 1mm = 10^-3m) into the formula will give you the area.
Finally, substitute F and A into the formula P = F / A to find the pressure.
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A uranium nucleus is traveling at 0.94 c in the positive direction relative to the laboratory when it suddenly splits into two pieces. Piece A is propelled in the forward direction with a speed of 0.43 c relative to the original nucleus. Piece B is sent backward at 0.34 c relative to the original nucleus. Find the velocity of piece A as measured by an observer in the laboratory. Do the same for piece B.
Answer with Explanation:
We are given that
Velocity of uranium=v=0.94 c
Speed of piece A relative to the original nucleus=[tex]u'_A=0.43c[/tex]
Speed of piece B relative to the original nucleus=[tex]u'_B=0.34c[/tex]
Velocity of piece A observed by observer
[tex]u_A=\frac{u'_A+v}{1+\frac{u'_A v}{c^2}}[/tex]
Substitute the values
[tex]u_A=\frac{0.43c+0.94c}{1+\frac{0.43c\times 0.94c}{c^2}}[/tex]
[tex]u_A=\frac{1.37c}{1+0.4042}=0.98c[/tex]
Velocity of piece B observed by observer
[tex]u_B=\frac{0.34c+0.94c}{1+\frac{0.34c\times 0.94c}{c^2}}[/tex]
[tex]u_B=\frac{1.28c}{1+0.3196}[/tex]
[tex]u_B=0.97 c[/tex]
The velocity of piece A and piece B as measured by an observer in the laboratory are not same.
Which of the following devices can be used to measure force?
A. Bathroom scale
(b) spring scale
c. Force sensor
d. All of the above.
Answer:
Explanation:
d.All of the above
Final answer:
All the listed devices -- bathroom scale, spring scale, and force sensor -- can be used to measure force. Bathroom and spring scales measure the weight of an object, which is a force, and display it typically in kilograms after calibration while force sensors provide direct measurement in newtons.
Explanation:
Devices that can be used to measure force include a bathroom scale, a spring scale, and a force sensor. All of these devices measure force, which can be the weight of an object (the force due to gravity acting on the object). A bathroom scale, for example, measures the normal force exerted by a person standing on it and gives a reading in kilograms by dividing the force in newtons by the acceleration due to gravity (9.80 m/s2). However, it is calibrated to display mass. A spring scale uses the extension of a spring under load to measure force which is typically indicated in newtons or pounds. Lastly, a force sensor is a more general device that can directly measure the force exerted on it in newtons and is often used for more precise scientific measurements.
If you stood on a bathroom scale in an elevator, the reading would change depending on the elevator's motion. The scale would display a higher value when the elevator starts moving upward (accelerating) due to the increase in the normal force. However, when the elevator moves at a constant speed, the scale would read your normal weight as no additional normal force is required once you're moving at constant velocity.
During a test, a NATO surveillance radar system, operating at 23 GHz at 197 kW of power, attempts to detect an incoming stealth aircraft at 101 km. Assume that the radar beam is emitted uniformly over a hemisphere. (a) What is the intensity of the beam when the beam reaches the aircraft's location
Answer:
Intensity will be equal to [tex]3.07\times 10^{-6}W/m^2[/tex]
Explanation:
We have given power P = 197 kW = 197000 watt
Distance R = 101 km
Area of the hemisphere will be [tex]A=2\pi R^2[/tex]
[tex]A=2\times 3.14\times 101000^2=6.4\times 10^{10}m^2[/tex]
We have to find the intensity
Intensity is equal to [tex]I=\frac{P}{A}[/tex]
[tex]I=\frac{1.97\times 10^5}{6.4\times 10^{10}}=3.07\times 10^{-6}W/m^2[/tex]
So intensity will be equal to [tex]3.07\times 10^{-6}W/m^2[/tex]
Sketch both, the time domain AM signal and its frequency spectrum and explain what you see in terms of themodulation property of the Fourier transform.(Hint: How is the frequency spectrum of the message signal(co-sinusoid of 880 Hz) plus a DC component in base band, i.e. before modulation?)
Answer:
see the attachment
Explanation:
In frequency spectrum there is no change in value of frequency of signal. However, The amplitude of signal after modulation increases. The fourier transform of sinwct is
1/2(F(w+wc) - F(w-wc))
For 2sinwct, the fourier transform is,
(F(w+wc) - F(w-wc))
In frequency domain, in AM only amplitude changes. Frequency remains same
One end of a string is attached to a rigid wall on a tabletop. The string is run over a frictionless pulley and the other end of the string is attached to a stationary hanging mass. The distance between the wall and the pulley is 0.405 meters, When the mass on the hook is 25.4 kg, the horizontal portion of the string oscillates with a fundamental frequency of 261.6 Hz (the same frequency as the middle C note on a piano). Calculate the linear mass density of the string.
Answer:
The linear mass density is of the string [tex]\mu= 5.51*10^{-3} kg / m[/tex]
Explanation:
From the question we are told that
The distance between wall and pulley is [tex]d = 0.405m[/tex]
The mass on the hook is [tex]m = 25.4\ kg[/tex]
The frequency of oscillation is [tex]f = 261.6 Hz[/tex]
Generally, the frequency of oscillation is mathematically represented as
[tex]f = \frac{1}{2d} \sqrt{\frac{T}{\mu} }[/tex]
Where T is the tension mathematically represented as
T = mg
Substituting values
[tex]T = 25.4 *9.8[/tex]
[tex]=248.92N[/tex]
[tex]\mu[/tex] is the mass linear density
Making [tex]\mu[/tex] the subject of the formula above
[tex]\mu = \frac{T}{(2df)^2}[/tex]
Substituting values
[tex]\mu = \frac{248.92}{(2 * 0.405 * 261.6)^2}[/tex]
[tex]\mu= 5.51*10^{-3} kg / m[/tex]
Answer:
0.005550 Kg/m
Explanation:
The picture attached below shows the full explanation
Electric potential is associated with both electric fields due to static charges and induced electric fields. Electric potential is associated with magnetic fields but not with electric fields due to static charges. Electric potential is associated with both electric fields due to static charges and magnetic fields. Electric potential is associated with electric fields due to static charges but not with induced electric fields. Electric potential is associated with induced electric fields but not with electric fields due to static charges.
Answer:
Electric potential is associated with electric fields due to static charges but not with induced electric fields.
Explanation:
An electric potential is the amount of work needed to move a unit charge from a reference point to a specific point inside the field without producing an acceleration. Typically, the reference point is the Earth or a point at infinity. Induced electricity is as a result of changing magnetic flux linkage (no charge is involved)
Electric potential is related to the electrostatic field created by static charges and is given by V = kQ/r; it is not related to induced electric fields or magnetic fields because they are nonconservative.
Electric potential is associated with electric fields due to static charges but not with induced electric fields.
This is because electric potential is defined for conservative fields, such as those produced by static charges, where the potential difference is related to the work done in moving a charge between two points in the field.
For point charges, the electric potential is given by the equation V = kQ/r, which clearly illustrates that electric potential depends on the charge and the distance from it.
On the other hand, induced electric fields are produced by changing magnetic fields and are nonconservative.
Nonconservative fields do not have a well-defined electric potential because the work done to move a charge can vary depending on the path taken, unlike the work done in a conservative electrostatic field.
Therefore, electric potential cannot be associated with induced electric fields or magnetic fields.
Two loudspeakers, 4.5 m apart and facing each other, play identical sounds of the same frequency. You stand halfway between them, where there is a maximum of sound intensity. Moving from this point toward one of the speakers, you encounter a minimum of sound intensity when you have moved 0.21 m . Assume the speed of sound is 340 m/s.
Part A
What is the frequency of the sound?
Part B
If the frequency is then increased while you remain 0.35m from the center, what is the first frequency for which that location will be a maximum of sound intensity?
Answer:
Explanation:
Given that,
Distance between speaker
L = 4.5m
Minimum intensity at L1 = 0.21m
Speed of sound is
V = 340m/s
A. Frequency of sound f?
The path difference Pd
Distance from the first speaker when you are 0.21m away
d1 = 2.25 + 0.21 = 2.46m
Distance from the second speaker when you move 0.21m closer
d2 = 2.25—0.21 = 2.04m
So, path difference is
Pd = ∆d = d1 — d2
Pd = 2.46—2.04 = 0.42m
Using the destructive interference condition
∆d = (m + ½)λ
m = 0,1,2,3....
When m= 0
∆d = ½λ
0.42 = ½λ
λ = 0.84
Then, using wave equation
v = fλ
Then, f = v / λ
f = 340 / 0.84
f = 404.76Hz
B. Incorrect question
If he is to remain at his initial positions then it is 0.21m from the center.
Then,
Constructive interference is given as
∆d = mλ
Where m = 0,1,2,3
So when m= 1
∆d = λ
And we already got the path difference to be 0.42m
So, ∆d = λ
λ = 0.42
So, applying wave equation
V = fλ
F = v/λ
F = 340/0.42
F = 809.52 Hz
But if we are to use the data given in part B
0.35m from the center..
Following the same principle as part A, the path difference will be 0.35
Therefore, since ∆d = λ
Then, λ = 0.35
So, f, = v/λ
F = 340 /0.35
F = 971.43 Hz
How did Kepler’s discoveries contribute to astronomy?
Answer:They established the laws of planetary motion.(option B)
Answer:
They established the laws of planetary motion
Explanation:
i did it on edge
You are driving along a highway at 35.0 m/s when you hear the siren of a police car approaching you from behind at constant speed and you perceive the frequency as . You are relieved that he is in pursuit of a different driver when he continues past you, but now you perceive the frequency as What is the speed of the police car? The speed of sound in a
The complete question is:
You are driving along a highway at 35.0 m/s when you hear the siren of a police car approaching you from behind at constant speed and you perceive the frequency as 1340 Hz. You are relieved that he is in pursuit of a different driver when he continues past you, but now you perceive the frequency as 1300 Hz. What is the speed of the police car? The speed of sound in air is 343m/s.
Answer:
V_s = 30 m/s
Explanation:
The change in frequency observation occur due to doppler effect is given by the equation;
f_o = [(V ± V_o)/(V ∓ V_s)]f_s
Where;
f_o is observed frequency
f_source is frequency of the source
V is speed of sound
V_o is velocity of the observer
V_s is velocity of the source
Now, When the police is coming to you , you hear a higher frequency and thus, we'll use the positive sign on the numerator and negative sign on denominator.
Thus,
f_o = [(V + V_o)/(V - V_s)]f_s
Plugging in relevant values, we have;
1340 = [(343 + 35)/(343 - V_s)]f_s
1340 = [(378)/(343 - V_s)]f_s - - (eq1)
when the police is passing you , you hear a lesser frequency, and thus, we'll use the negative sign on the numerator and positive sign on denominator. thus;
f_o = [(V - V_o)/(V + V_s)]f_s
Plugging in the relevant values to get;
1300 = [(343 - 35)/(343 + V_s)]f_s
1300 = [(308)/(343 + V_s)]f_s - - eq2
Divide eq2 by eq1 with f_s canceling out to give
1340/1300 = [(378)/(343 - V_s)]/[(308)/(343 + V_s)]
V_s = 30 m/s
A steel ball of mass 0.500 kg is fastened to a cord that is 70.0 cm long and fixed at the far end. The ball is then released when the cord is horizontal. At the bottom of its path, the ball strikes a 2.50 kg steel block initially at rest on a frictionless surface. The collision is elastic. Find (a) the speed of the ball and (b) the speed of the block, both just after the collision.
Answer:
a) The speed of the ball is 2.47 m/s (in -x direction)
b) The speed of the block, both just after the collision is 1.236 m/s (in +x direction)
Explanation:
Please look at the solution in the attached Word file.
An object of mass M is dropped near the surface of Earth such that the gravitational field provides a constant downward force on the object. Which of the following describes what happens to the center of mass of the object-Earth system as the object falls downward toward Earth? a. It moves toward the center of Earth. b. It moves toward the object.c. It does not move. d. The answer cannot be determined without knowing the mass of Earth and the distance between the object and Earth’s center.
Answer:
The answer is: c. It does not move
Explanation:
Because the gravitational force is characterized by being an internal force within the Earth-particle system, in this case, the object of mass M. And since in this system there is no external force in the system, it can be concluded that the center of mass of the system will not move.
As the object is dropped at a constant downward force, the center mass of the object-Earth system does not move.
In the absence of an external force, the center mass of the object-Earth system will remain constant as the object fall to the ground.
The force of attraction on the object above the surface of the Earth is given as;
[tex]F = \frac{Gm_1 m_2}{R^2}[/tex]
where;
G is gravitational constantm is massR is the distance of the object from the center of the earthThus, as the object is dropped at a constant downward force, the center mass of the object-Earth system does not move.
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Light from an LED with a wavelength of 4.90 ✕ 102 nm is incident on (and perpendicular to) a pair of slits separated by 0.310 mm. An interference pattern is formed on a screen 2.20 m from the slits. Find the distance (in mm) between the first and second dark fringes of the interference pattern.
Answer:
Δx = 3.477 x 10⁻³ m = 3.477 mm
Explanation:
The distance between two consecutive dark fringes is given by the following formula, in Young's Double Slit experiment:
Δx = λL/d
where,
Δx = distance between two consecutive dark fringes = ?
λ = wavelength of light = 4.9 x 10² nm = 4.9 x 10⁻⁷ m
L = Distance between slits and screen = 2.2 m
d = slit separation = 0.31 mm = 0.31 x 10⁻³ m
Therefore,
Δx = (4.9 x 10⁻⁷ m)(2.2 m)/(0.31 x 10⁻³ m)
Δx = 3.477 x 10⁻³ m = 3.477 mm
The distance (in mm) between the first and second dark fringes of the
interference pattern is 3.477 mm
This is calculated by using the formula in Young's Double Slit experiment:
Δx = λL/d
where,
Δx = distance between two consecutive dark fringes which is unknown
λ which is wavelength of light = 4.9 x 10² nm = 4.9 x 10⁻⁷ m
L which is distance between slits and screen = 2.2 m
d which is slit separation = 0.31 mm = 0.31 x 10⁻³ m
We then substitute them into the equation
Δx = (4.9 x 10⁻⁷ m ×2.2 m)/(0.31 × 10⁻³ m)
Δx = 3.477 x 10⁻³ m = 3.477 mm
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Shrinking Loop. A circular loop of flexible iron wire has an initial circumference of 168 cm , but its circumference is decreasing at a constant rate of 15.0 cm/s due to a tangential pull on the wire. The loop is in a constant uniform magnetic field of magnitude 0.900 T , which is oriented perpendicular to the plane of the loop. Assume that you are facing the loop and that the magnetic field points into the loop.
Required:
Find the magnitude of the emf EMF induced in the loop after exactly time 8.00s has passed since the circumference of the loop started to decrease.
Final answer:
To find the magnitude of the emf induced in the loop after 8.00 seconds has passed since the circumference started to decrease, we can use Faraday's law of electromagnetic induction. We calculate the rate of change of magnetic flux through the loop based on the changing area of the loop, and then determine the magnitude of the emf induced in the loop. The emf induced is -253.30 V, indicating that the induced current flows in a direction that opposes the change in magnetic flux.
Explanation:
To find the magnitude of the emf induced in the loop after 8.00 seconds has passed since the circumference started to decrease, we can use Faraday's law of electromagnetic induction. According to Faraday's law, the emf induced in a loop is equal to the rate of change of magnetic flux through the loop. In this case, as the loop shrinks, its area decreases, resulting in a decrease in magnetic flux.
We know that the circumference of the loop is decreasing at a constant rate of 15.0 cm/s. Using the formula for the circumference of a circle, we can determine the radius of the circle at the given time: r = C / (2*pi), where C is the circumference and pi is a mathematical constant approximately equal to 3.14159. Substituting the given values, we get r = 168 cm / (2*3.14159) = 26.79 cm.
Next, we can calculate the area of the loop as a function of time using the equation A = pi*r^2. Substituting the value of the radius at 8.00 seconds, we get A = 3.14159 * (26.79 cm)^2 = 2252.68 cm^2.
Since the magnetic field is perpendicular to the loop and uniform in magnitude, we can calculate the rate of change of magnetic flux as: dPhi/dt = B*dA/dt, where B is the magnitude of the magnetic field and dA/dt is the rate of change of the area.
Finally, we can calculate the magnitude of the emf induced in the loop as: EMF = -dPhi/dt. The negative sign indicates that the induced current flows in a direction that opposes the change in magnetic flux. Substituting the given values, we get EMF = -0.900 T * (2252.68 cm^2) / 8.00 s = -253.30 V.
In class we learned that you can detect The Big C (molecular weight of 303 grams/mol) by mass spectrometry. The mass spec has a magnetic field of 0.3 T and the singly charged ions in the sample are moving at 50,000 m/s. How far from the entrance in m do the Florida Snow ions strike the spectrometer detector after travelling through a semicircle trajectory?
Answer:
1.04m
Explanation:
the distance is determined by the diameter of the trajectory.
you can find the radius of the trajectory by using the following formula:
[tex]r=\frac{m_cv}{qB}[/tex]
mc : mass of the Big C = 0.303kg/mol/(6.02*10^{23}/mol)=5.03*10^{-25}kg
v: velocity of the ion = 50000m/s
B: magnetic constant = 0.3T
q: 1.6*10^{-19}C
By replacing you obtain:
[tex]r=\frac{(5.03*10^{-25}kg)(50000m/s)}{(1.6*10^{-19}C)(0.3T)}=0.52m[/tex]
the diameter wiil be:
d=2r=1.04m
the ion strikes the detector from 1.04m to the entrance of the spectrometer
The Florida Snow ions strike the spectrometer detector after travelling through a semicircle trajectory will be "1.04 m" far.
SpectrometerAccording to the question,
Big C's mass, [tex]m_c[/tex] = [tex]\frac{0.303}{6.02\times 10^{23}}[/tex]
= 5.03 × 10⁻²⁵ kg
Ion's velocity, v = 50000 ms
Magnetic constant, B = 0.3 T
Charge, q = 1.6 × 10⁻¹⁹ C
We know that,
→ r = [tex]\frac{m_c v}{qB}[/tex]
or,
The radius, r = [tex]\frac{5.03\times 10^{-25}\times 50000}{1.6\times 10^{-19}\times 0.3 }[/tex]
= 0.52 m or,
Diameter, d = 2 × r
= 2 × 0.52
= 1.04 m
Thus the above approach is correct.
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Susan’s 10 kg baby brother Paul sits on a mat. Susan pulls the mat across the floor using a rope that is angled 30° above the floor. The tension is a constant 30 N and the coefficient of friction is 0.20. Use work and energy to find Paul’s speed after being pulled 3.0 m.
To determine Paul's speed, we must calculate the net work done on him using the work-energy theorem. This includes the work done by Susan and the work done against friction. Paul’s speed after being pulled 3.0 m is approximately 1.96 m/s.
Explanation:Solving this problem involves understanding the work-energy theorem and forces. First, let's calculate the work done. The work done by the force Susan applies (W1) is the product of the tension (T), the distance (d), and the cosine of the angle (θ). W1 = T * d * cos(θ) = 30N * 3.0m * cos(30) = 77.94J.
Next, the work done against friction (W2) is the product of the frictional force and the distance, which is µmgd. Here, µ is the coefficient of friction (0.20), m (10kg) is the mass of the baby, g (9.8m/s2) is the acceleration due to gravity, and d is the distance (3.0 m). W2 = µmgd = 0.20 * 10kg * 9.8m/s2 * 3.0m = 58.8J.
According to the work-energy theorem, the net work done on an object is equal to the change in its kinetic energy. Therefore, the final kinetic energy (and thus the final speed) of Paul will be the initial kinetic energy plus the net work done on him. His initial speed is assumed to be zero, hence the initial kinetic energy is zero. The net work done on him is W = W1 - W2= 77.94J - 58.8J = 19.14J. Setting this equal to the final kinetic energy, (1/2)mv2, allows us to solve for the final speed, v = sqrt((2 * W)/m) = sqrt((2 * 19.14J)/10kg) = 1.96 m/s approximately.
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An electric generator contains a coil of 99 turns of wire, each forming a rectangular loop 73.9 cm by 34.9 cm. The coil is placed entirely in a uniform magnetic field with magnitude B = 2.96 T and initially perpendicular to the coil's plane. What is in volts the maximum value of the emf produced when the loop is spun at 1200 rev/min about an axis perpendicular to the magnetic field?
Answer:
12078.46 V
Explanation:
Applying,
E₀ = BANω.................... Equation 1
Where: E₀ = maximum emf, B = magnetic Field, A = Area of the coil, N = Number of turns of the coil. ω = angular velocity
Given: N = 99 turns, B = 2.96 T, ω = 1200 rev/min = (1200+0.10472) = 125.664 rad/s
A = L×W, where L= Length = 93.9 cm = 0.939 m, W = width = 34.9 cm = 0.349 m
A = (0.939×0.349) = 0.328 m²
Substitute into equation 1
E₀ = 99(2.96)(0.328)(125.664)
E₀ = 12078.46 V
Hence the maximum value of the emf produced = 12078.46 V
Answer:
The maximum emf induced in the loop is 9498.268 V
Explanation:
Given;
number of turns of coil, N = 99 turns
area of the rectangular loop, A = 0.739 m x 0.349 m = 0.2579 m²
magnetic field strength, B = 2.96 T
angular speed of the loop, ω = 1200 rev/min
angular speed of the loop, ω (rad/s) = (2π x 1200) / 60 = 125.68 rad/s
The maximum value of the emf produced is calculated using the formula below;
ξ = NABω
Substitute the given values and calculate the maximum emf induced;
ξ = (99)(0.2579)(2.96)(125.68)
ξ = 9498.268 volts
Therefore, the maximum emf induced in the loop is 9498.268 V
Two long, parallel wires are attracted to each other by a force per unit length of 305 µN/m. One wire carries a current of 25.0 A to the right and is located along the line y = 0.470 m. The second wire lies along the x axis. Determine the value of y for the line in the plane of the two wires along which the total magnetic field is zero.
To solve this problem we will use the concepts related to the electromagnetic force related to the bases founded by Coulumb, the mathematical expression is the following as a function of force per unit area:
[tex]\frac{F}{L} = \frac{kl_1l_2}{d}[/tex]
Here,
F = Force
L = Length
k = Coulomb constant
I =Each current
d = Distance
Force of the wire one which is located along the line y to 0.47m is [tex]305*10^{-6}N/m[/tex] then we have
[tex]l_2 = \frac{F}{L} (\frac{d}{kl_1})[/tex]
[tex]l_2 = (305*10^{-6}N/m)(\frac{0.470m}{(2*10^{-7})(25A))})[/tex]
[tex]l_2 = 28.67A[/tex]
Considering the B is zero at
[tex]y = y_1[/tex]
[tex]\frac{kI_2}{2\pi y} =\frac{kI_1}{2\pi y_1}[/tex]
[tex]\frac{(4\pi*10^{-7})(28.67)}{2\pi (y_1)} = \frac{(4\pi *10^{-7})(25)}{2\pi (0.47-y_1)}[/tex]
[tex]y_1 = 0.25m[/tex]
Therefore the value of y for the line in the plane of the two wires along which the total B is zero is 0.25m
You have been sent to a new home. The homeowner reports that sometimes the electric furnace trips the 240-V, 60-A circuit breaker connected to it. Upon examination, you find that the furnace contains three 5000-W heating elements designed to turn on in stages. For example, when the thermostat calls for heat, the first 5000-W element turns on. After some period of time, the second element will turn on, and then, after another time delay, the third element will turn on. What do you think the problem is, and what would be your recommendation for correcting it? Explain your answer.
The problem is likely an excessive combined power draw from the three heating elements, exceeding the circuit breaker's capacity; the recommendation is to upgrade the circuit breaker or adjust the furnace's operation to stay within the breaker's limit.
We have,
The problem seems to be related to the load on the circuit exceeding the capacity of the circuit breaker.
When the furnace is turned on and all three 5000-W heating elements turn on in stages, the combined power consumption becomes 3 * 5000 W = 15000 W.
This is a substantial load that exceeds the circuit breaker's capacity, which is likely 240 V * 60 A = 14400 W (due to the product of voltage and current rating).
As a result, the circuit breaker is tripping to protect the circuit from overloading, as the total power drawn from the furnace exceeds its rated capacity.
To correct this issue and prevent the circuit breaker from tripping, you could consider the following recommendations:
- Check the Circuit Breaker Rating:
Confirm the rating of the circuit breaker connected to the furnace. If it's indeed 60 A, you might need to upgrade the circuit breaker to a higher rating that can handle the combined power consumption of all three heating elements.
- Reduce Load:
Alternatively, you could reconfigure the furnace to operate only one or two heating elements at a time to reduce the load and stay within the circuit breaker's capacity.
This may involve adjusting the furnace's internal settings or installing additional controls to manage the heating elements' activation.
- Consider Energy Management:
Implement an energy management system that staggers the activation of the heating elements over time.
This would ensure that the power demand doesn't exceed the circuit breaker's capacity during startup.
- Professional Electrician:
Thus,
The problem is likely an excessive combined power draw from the three heating elements, exceeding the circuit breaker's capacity; the recommendation is to upgrade the circuit breaker or adjust the furnace's operation to stay within the breaker's limit.
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The issue is likely that the electric furnace surpasses the 60-A limit of the circuit breaker when all heating elements are active, causing it to trip. A potential solution could be to upgrade the circuit breaker to a higher amperage rating or adjust the furnace so that it never exceeds 60 A.
Explanation:The problem the homeowner is encountering with their electric furnace is likely due to the furnace exceeding the capacity of the 240-V, 60-A circuit breaker when all three heating elements activate. Each 5000-W heating element at 240 V consumes about 20.8 A of current. If all three elements are powered simultaneously, the total current draw is 62.4 A, which surpasses the 60-A circuit breaker limit and causes it to trip.
One potential solution would be to upgrade the circuit breaker to one with a higher amperage rating. However, any modifications should comply with local electricity standards and should be carried out by a qualified electrician. Alternatively, the furnace could be rewired or adjusted to ensure that the cumulative draw never surpasses 60 A at any given time.
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