You probably already took this but if anyone wanted to know the answer it's D. "A system where matter and energy cannot enter or leave the system".
(I also took the quiz for it and got it correct)
A closed physical system is a system where matter and energy cannot enter or leave the system. The correct option is D.
What is a physical system?A physical system is any item or part of an object that can be analyzed using physics laws.
An atom and lake water are both examples of physical systems. Everything outside the system is referred to as the environment, and it is ignored in the analysis except for its effects on the system.
A physical system is a collection of parts or elements that, when combined, exhibit behavior that the constituents do not. Matter and energy make up physical systems.
A closed system is a natural physical system that does not allow the transfer of matter into or out of the system, though it does allow the transfer of energy in contexts such as physics, chemistry, or engineering.
Thus, the correct option is D.
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A diet decreases a person's mass by 6 %. Exercise creates muscle and reduces fat, thus increasing the person's density by 1 %. Determine the percent change in the person's volume.
Answer:
-5%
Explanation:
We know that volume is mass/density
So to find percentage
Let first mass be = M1
Second be 0.94m( 1-6%)
Second density be=0.99
First density be d
So finlq volume v2= 0.94m/0.99p=
V2= (0.94/0.99)V1
V2= 0.95v1
So volume decrease will be 1-0.95= 0.05
= -5%
An object has the acceleration graph shown in (Figure 1). Its velocity at t=0s is vx=2.0m/s. Draw the object's velocity graph for the values of t that are separated by the step Δt = 2s.
Answer:
Explanation:
We may notice that change in velocity can be obtained by calculating areas between acceleration lines and horizontal axis ("Time"). Mathematically, we know that:
[tex]v_{b}-v_{a} = \int\limits^{t_{b}}_{t_{a}} {a(t)} \, dt[/tex]
[tex]v_{b} = v_{a}+ \int\limits^{t_{b}}_{t_{a}} {a(t)} \, dt[/tex]
Where:
[tex]v_{a}[/tex], [tex]v_{b}[/tex] - Initial and final velocities, measured in meters per second.
[tex]t_{a}[/tex], [tex]t_{b}[/tex] - Initial and final times, measured in seconds.
[tex]a(t)[/tex] - Acceleration, measured in meters per square second.
Acceleration is the slope of velocity, as we know that each line is an horizontal one, then, velocity curves are lines with slopes different of zero. There are three region where velocities should be found:
Region I (t = 0 s to t = 4 s)
[tex]v_{4} = 2\,\frac{m}{s} +\int\limits^{4\,s}_{0\,s} {\left(-2\,\frac{m}{s^{2}} \right)} \, dt[/tex]
[tex]v_{4} = 2\,\frac{m}{s}+\left(-2\,\frac{m}{s^{2}} \right) \cdot (4\,s-0\,s)[/tex]
[tex]v_{4} = -6\,\frac{m}{s}[/tex]
Region II (t = 4 s to t = 6 s)
[tex]v_{6} = -6\,\frac{m}{s} +\int\limits^{6\,s}_{4\,s} {\left(1\,\frac{m}{s^{2}} \right)} \, dt[/tex]
[tex]v_{6} = -6\,\frac{m}{s}+\left(1\,\frac{m}{s^{2}} \right) \cdot (6\,s-4\,s)[/tex]
[tex]v_{6} = -4\,\frac{m}{s}[/tex]
Region III (t = 6 s to t = 10 s)
[tex]v_{10} = -4\,\frac{m}{s} +\int\limits^{10\,s}_{6\,s} {\left(2\,\frac{m}{s^{2}} \right)} \, dt[/tex]
[tex]v_{10} = -4\,\frac{m}{s}+\left(2\,\frac{m}{s^{2}} \right) \cdot (10\,s-6\,s)[/tex]
[tex]v_{10} = 4\,\frac{m}{s}[/tex]
Finally, we draw the object's velocity graph as follows. Graphic is attached below.
The velocity of a body under constant acceleration increases steadily with time
Please find attached the required velocity graph for values of t that are separated by Δt = 2s
The reasons the attached graph is correct are given as follows:
At t = 0 secondsThe initial velocity of the object at t = 0 is v = 2.0 m/s
The first point on the graph is (0, 2.0)
From t = 0 s, to t = 4 sThe acceleration from t = 0, to t = 4, a₁ = -2 m/s²
The velocity at t = 4 s, v₂ = 2.0 + (-2)×4 = -6
Therefore, the next point on the graph is (4, -6)
From t = 4 s to t = 6 sFrom t = 4 to t = 6, the acceleration, a₂ = 1 m/s²
Therefore, v₃ = -6 + 1 × 2 = -4
The third point on the velocity graph is (6, -4)
From t = 6 s to t = 10 sFrom t = 6 s to t = 10 s, we have, the acceleration, a₃ = 2 m/s²
The velocity, v₄ = -4 + 4 × 2 = 4
Therefore, the fourth point on the velocity graph is (10, 4)
With the above points, the velocity graph can be plotted using MS Excel
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A snowboarder on a slope starts from rest and reaches a speed of 8.9 m/s after 9.5 s.
How far does the snowboarder travel in this time?
Explanation:
Given that,
Initial speed, u = 0
Final speed, v = 8.9 m/s
Time, t = 9.5 s
Let a is the acceleration of the snowboarder. It can be calculated as follows :
[tex]a=\dfrac{v-u}{t}\\\\a=\dfrac{8.9-0}{9.5}\\\\a=0.936\ m/s^2[/tex]
The distance traveled by the snowboarder in this time is calculated using third equation of motion as follows :
[tex]v^2=u^2=2as\\\\s=\dfrac{v^2}{2a}\\\\s=\dfrac{(8.9)^2}{2\times 0.936}\\\\s=42.31\ m[/tex]
So, 42.31 m of distance is traveled by snowboarder in this time.
A cyclist traveling in a straight line has the velocity of +5 m/s for 10 sec. She then accelerates at a rate of 0.5 m/s2 over another 10 sec. How far does the cyclist travel during those 20 sec?
Answer: 125
Explanation: use the equation d=vt to find the first distance, 50. Then use the equation
X=Vot + 1/2at to find the second distance. Then add both together.
Vo=5
V=5
T=10
A=5
The train traveled 500 kilometers north to Odessa in 2 hours. What is the train’s speed? What is it’s velocity?
A girl is riding her bicycle. She rides 4 miles in 2.5 hours. What was her average speed?
Explanation:
4/2.5 = 1.6 miles/hour
What is the car's average velocity (in m/s) in the interval between t = 1.0 s
to t = 1.5 s?
Answer:
1.4 m/s
Explanation:
From the question given above, we obtained the following data:
Initial Displacement (d1) = 0.9 m
Final Displacement (d2) = 1.6 m
Initial time (t1) = 1.5 secs
Final time (t2) = 2 secs
Velocity (v) =..?
The velocity of an object can be defined as the rate of change of the displacement of the object with time. Mathematically, it can be expressed as follow:
Velocity = change of displacement /time
v = Δd / Δt
Thus, with the above formula, we can obtain the velocity of the car as follow:
Initial Displacement (d1) = 0.9 m
Final Displacement (d2) = 1.6 m
Change in displacement (Δd) = d2 – d1 = 1.6 – 0.9
= 0.7 m
Initial time (t1) = 1.5 secs
Final time (t2) = 2 secs
Change in time (Δt) = t2 – t1
= 2 – 1.5
= 0.5 s
Velocity (v) =..?
v = Δd / Δt
v = 0.7/0.5
v = 1.4 m/s
Therefore, the velocity of the car is 1.4 m/s
If an initially neutral insulator is charged on its left end, would you get an electrical shock by touching its right end?
Answer:
No there won't be.
Explanation:
An insulator does not conduct electricity readily. An insulator that is charged at one end will have the charges remain at that end, since it does not allow the free flow of charges through it. If you touch the other end of the insulator that is void of electric charges, there would be no charge transfer between you and the insulator, and hence no electric shock.
Which element is most likely to be shiny? sulfur (S) boron (B) calcium (Ca) carbon (C)
Answer:
C.
Explanation:
The element that is most likely to be shiny is calcium. The correct option is 3.
Calcium (Ca) is the most likely to be sparkly of the elements shown. Calcium is a metallic element with a glossy or shining look when newly cut or polished.
It belongs to the alkaline earth metals group, which have metallic characteristics and a distinctive lustre.
Sulphur (S), boron (B), and carbon (C), on the other hand, are non-metallic elements with a dull look.
They come in a variety of forms, including powders, crystals, and amorphous solids, although they are not noted for their metallic lustre.
Thus, the answer is 3. calcium.
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Your question seems incomplete, the probable complete question is:
Which element is most likely to be shiny?
sulfur (S) boron (B) calcium (Ca) carbon (C)The period of a simple pendulum, defined as the time necessary for one complete oscillation, is measured in time units and is given by T=2πlg where l is the length of the pendulum and g is the acceleration due to gravity, in units of length divided by time squared. Show that this equation is dimensionally consistent.
Answer:
Explanation:
T = 2π √l/g
The dimension for l = m
The dimension for g = m/s²
The dimension for 2π is nothing. Since it's a constant, it is dimensionless.
Now we proceed ahead. Since we are not using the 2π, for the sake of this proving, our formula will temporarily be written as
T = √l/g
Inputting the dimensions, we have
T = √(m) / (m/s²)
T = √(m * s²/m)
T = √s²
T = s
Since the unit of period itself is in s, we can adjudge that the equation is dimensionally constant.
A rigid rod of length L rotates about an axis perpendicular to the rod, with one end of the rod fixed to the axis. Which of the following must always be equal at all points on the rod?
1. Angular position.
2. Angular velocity.
3. Angular acceleration.
4. Tangential acceleration.
A. 1 and 2.
B. 1, 2 and 4.
C. 1 and 3.
D. 1, 2, 3 and 4.
Answer:
D
Explanation:
The Angular Position: The angular position is also called the angular displacement, and it is the angle through which a point revolves around the centre line that has been rotated in a particular manner about a given axis.
The Angular Velocity: The angular velocity of an object is the measure of the speed, or say, how fast an object rotates at one point with respect to a point.
The Angular Acceleration: This is very simple as it's a derivative of angular velocity. They both share the same relationship velocity and acceleration share. Angular acceleration is the rate of change of angular velocity.
The only options that are equal at all points on the rod are;
1. Angular position.
2. Angular velocity.
3. Angular acceleration.
Let us look at each of the options;
1) Angular Position: This is also referred to as angular displacement. In centripetal motion, it is defined as the angle through which a a body has been rotated from the initial position or reference position. This must always be equal at all points on the rod.
2) Angular Velocity: This is defined as the speed with which an object is moving in a circular motion. This must always be equal at all points on the rod.
3) Angular Acceleration: This is simply defined as the rate of change of angular velocity with time. Now, since angular velocity must be equal at all points on the rod, then angular acceleration must also be equal at all points on the rod.
4) Tangential acceleration; This is defined as the rate of change of tangential velocity with time. Tangential velocity is the linear component of the speed and it is not equal at all points on the rod and as such the tangential acceleration will also not be equal at all points on the rod.
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A boat is pulled into the dock by a rope attached to the bow of the boat and passing through a pulley on the dock that is 1 m higher than the bow of the boat.
If the rope is pulled in at a rate of 1.4 m/s, how fast is the boat approaching the dock when it is 9 m from the dock?
Answer:
√82/9 m/s
Explanation:
Let's say x²+1²= y²(y length of rope)
Using Pythagoras
With repect to t diff.
So 2x dx / dt= 2y dy / dt
But dy/ dt = 1.4m/s
X = 9 so from Pythagoras y = √82
So 9dx/dt = √82
So the boat will approved de dock at √82/9 m/s
Which of the following is a negatively charged particle that is found in "clouds" around the nucleus?
Group of answer choices
proton
quark
electron
neutron
what will happen to the fieldof view for each resultant magnification as you change objectives from 4 to 10 to 43
Answer:
The field of view is reduced.
Explanation:
Given that,
The field of view for every resultant magnification like you change objectives from 4 to 10 to 43.
We know that,
Field of view :
When the view is observed at a point in a defined field then these field called field of view.
The normal angle of field of view is 90°.
The formula of field of view is define as,
[tex]field\ of\ view = \dfrac{field\ number}{magnification}[/tex]
We can say that,
The field of view is inversely proportional to the magnification.
When magnification is low then field of view will be large.
When magnification is higher then field of view will be small .
According to question,
When the magnification adjust from 4 to 10 to 43, the field of view is reduced.
Hence, The field of view is reduced.
A typical atom has a diameter of about 1.32 x 10-10 m. (a) what is this in inches (b) approximately how many atoms arethere along a 1.32 cm line.
Answer:
A ) [tex]5.197 \times 10 ^{-9} inches[/tex]
b) There are approximately 100,000,000 atoms on the 1.32 cm lines
Explanation:
To ensure the accuracy of our answer, we will have to make sure we work in the appropriate units.
A
To convert [tex]1.32 \times 10^{-10} m[/tex] to inches, we can use this factor.
1 m = 39.3701 inches
Therefore [tex]1.32 \times 10^{-10} m[/tex] =
B
To find the number of atoms on a 1.32 cm line, we will first of all need to convert 1.32 cm to meters.
1.32 cm = 0.0132 metres (divided 1.32 by 100 to convert to metres)
Assuming all the atoms are arranged side by side, with their edges touching on the 0.0132m line, the number of atoms present will be
[tex]0.0132/1.32\times 10^{-10}[/tex] = 100,000,000 atoms
The quantity represented by vi is a function of time (i.e., is not constant).A. TrueB. False
Answer:
The answer is false
A student throws a heavy red ball horizontally from a balcony of a tall building with an initial speed v0. At the same time, a second student drops a lighter blue ball from the same balcony. Neglecting air resistance, which statement is true?
Answer:
The correct statement must be: both balls hit the floor at the same time
Explanation:
This is a kinematics exercise. The ball thrown horizontally does not have vertical speed and the ball that is released does not have vertical speed, therefore both take the same time to reach the ground, if we neglect the air resistance
The correct statement must be: both balls hit the floor at the same time
You are driving along the New York State Thruway in a line of cars all travelling at a constant speed of 108.5 km/hr. The car in front of you applies its brakes for maximum acceleration. You then apply your brakes to achieve the same maximum acceleration after only a 1 s delay due to reaction time. What distance behind the car in front of you must you be to avoid a collision?
Answer:
Car first should be 30.13 m behind the second car.
Explanation:
Given that,
Constant speed = 108.5 Km/hr
Time = 1 sec
Let the distance covered by second car S and by first car S'
We need to calculate the distance covered by second car
Using equation of motion
[tex]S=ut-\dfrac{1}{2}at^2[/tex].....(I)
The distance covered by first car to avoid collision
[tex]S''=S'+S[/tex]
Put the value into the formula
[tex]S'+S=ut+ut-\dfrac{1}{2}at^2[/tex]...(II)
We need to calculate the distance covered by first car
Using equation (I) and (II)
[tex]S'=ut[/tex]
Put the value into the formula
[tex]S'=108.5\times\dfrac{5}{18}\times1[/tex]
[tex]S'=30.13\ m[/tex]
Hence, Car first should be 30.13 m behind the second car.
What is the car's average velocity (in m/s) in the interval between t = 0.5 s
to t = 2 s?
Answer:
[tex]1.0\; \rm m \cdot s^{-1}[/tex].
Explanation:
Consider a time period of duration [tex]\Delta t[/tex]. Let change in the position of an object during that period of time be denoted as [tex]\Delta x[/tex]. The average velocity of that object during that period would be:
[tex]\displaystyle \bar{v} = \frac{\Delta x}{\Delta t}[/tex].
For the toy car in this question, the time interval has a duration of [tex]2.0 \; \rm s - 0.5\; \rm s = 1.5\; \rm s[/tex]. That is: [tex]\Delta t = 1.5\; \rm s[/tex]. (One decimal place, two significant figures.)
On the other hand, what would be the change in the position of this toy car during that [tex]1.5\; \rm s[/tex]?
Note, that from readings on the snapshot in the diagram:
The position of the toy car was [tex]0.1\; \rm m[/tex] at [tex]t = 0.5\; \rm s[/tex] (the beginning of this [tex]1.5[/tex]-second time period.)The position of the toy car was [tex]1.6\; \rm m[/tex] at [tex]t = 2.0\; \rm s[/tex] (the end of this [tex]1.5[/tex]-second time period.)Therefore, the change to the position of this toy car over that time period would be [tex]\Delta x = 1.6\; \rm m - 0.1\; \rm m = 1.5\; \rm m[/tex]. (One decimal place, two significant figures.)
The average velocity of this car over this period of time would thus be:
[tex]\displaystyle \bar{v} = \frac{\Delta x}{\Delta t} = \frac{1.5\; \rm m}{1.5\; \rm s} = 1.0\; \rm m \cdot s^{-1}[/tex]. (Two significant figures.)
If you dropped a rock 4.9 m in 1 s how far would it fall in 3 s? express your answer in meters to three significant figures
Answer:
14.7 m
Explanation:
It would fall 14.7 m in 3 s . You multiply 4.9 by 3.
We can reasonably model a 75 W incandescent light bulb as a sphere 6.0 cm in diameter. Typically, only about 5 % of the energy goes to visible light: the rest foes largely to nonvisible infrared radiation.(a) What is the visible light intensity (in W/m^2) at the surface of the bulb?(b) What are the amplitudes of the electric and magnetic fields at this surface, for a sinusoidal wave with this intensity?
Answer:
a
[tex]I = 6637 \ W/m^2[/tex]
b
[tex]E_{max} = 500 \ N/m[/tex]
And
[tex]B_{max} = 1.67*10^{-6} \ T[/tex]
Explanation:
From the question we are told that
The power is [tex]P = 75 \ W[/tex]
The diameter is [tex]d = 6.0 \ cm = 0.06 \ m [/tex]
Generally the radius is mathematically represented as
[tex]r = \frac{d}{2}[/tex]
=> [tex]r = \frac{ 0.06}{2}[/tex]
=> [tex]r = 0.03 \ m[/tex]
Generally the area of the sphere is mathematically evaluated as
[tex]A = 4 \pi r^2[/tex]
=> [tex]A = 4 * 3.142 * (0.03)^2[/tex]
=> [tex]A = 0.0113 \ m^2[/tex]
Generally the total Intensity of the incandescent light bulb is mathematically represented as
[tex]I= \frac{P}{A}[/tex]
=> [tex]I = \frac{75}{ 0.0113}[/tex]
=> [tex]I = 6637 \ W/m^2[/tex]
Given that 5% of the energy goes to visible light
Then the intensity that goes visible light is
[tex]I_v = 0.05 * 6637[/tex]
[tex]I_v = 332 \ W/m^2[/tex]
The amplitude of the electric field at the surface is mathematically represented as
[tex]E_{max} = \sqrt{\frac{2 * I_v}{\epsilon_o * c } }[/tex]
=> [tex]E_{max} = \sqrt{\frac{2 * 332}{ 8.85*10^{-12} * 3.0*10^8} }[/tex]
=> [tex]E_{max} = 500 \ N/m[/tex]
The amplitude of the magnetic field at the surface is mathematically represented as
[tex]B_{max} = \frac{E_{max}}{c}[/tex]
=> [tex]B_{max} = \frac{ 500}{3.0*10^8}[/tex]
=> [tex]B_{max} = 1.67*10^{-6} \ T[/tex]
If a truck travels 100 miles in 2 hours, what is its speed?
Answer:50 miles per hour 50/1hr
Explanation:100 divided by 2 is 50, divide 2 by 2 thats 1
Answer:
50 miles/hr
Explanation:
[tex]Distance = 100\: miles\\Time = 2\:hours\\\\Speed = \frac{Distance}{Time} \\\\Speed = \frac{100\:miles}{2\:hours}\\\\ Speed = 50\:miles/hour[/tex]
An unmarked police car traveling a constant 85 km/h is passed by a speeder. Precisely 2.00 s after the speeder passes, the police officer steps on the acceleratorIf the police car accelerates uniformly at 2.00 m/s2 and overtakes the speeder after accelerating for 9.00 s , what was the speeder's speed?.
Answer:
observe o gráfico abaixo que mostra o número de internação em um hospital no município no período de 1960
A system of 1470 particles, each of which is either an electron or a proton has a net charge of -5.456 x 10^-17 C. (a) How many electrons are in this system? (b) What is the mass of this system?
Answer:
a
[tex]N_e = 906 \ electrons[/tex]
b
[tex]M = 9.43 *10^{-25} \ kg[/tex]
Explanation:
From the question we are told that
The total number of particles is n = 1470 particles
The the total amount of charge on the particles is [tex]Q = - 5.456*10^{-17} \ C[/tex]
Generally this total number of particles can be mathematically represented as
[tex]n = N_p + N_e[/tex]
Where [tex]N_p \ and \ N_e[/tex] represent the number of proton and electron respectively
=> [tex]N_p = 1470 - N_e[/tex]
Also the total charge of these particles can be mathematically represented as
[tex]Q = (N_p - N_e ) e[/tex]
Here e is the charge on a single electron or proton with value
The negative sign is due to the fact the electrons are negative signed
[tex]e = 1.60 *10^{-19} \ C[/tex]
[tex]-5.456*10^{-17} = ( 1470 -2N_e )* 1.60 *10^{-19} [/tex]
[tex]N_e = 906[/tex]
Thus
[tex]N_p = 1470 - 906[/tex]
[tex]N_p = 564[/tex]
Generally the mass of the system is mathematically represented as
[tex]M = N_e * M_e + N_p * M_p[/tex]
Here [tex]M_e \ and \ M_p \ are \ mass \ of \ electron \ and \ proton \ with \ values \\ M_e = 9.1 *10^{-31} \ kg \ and \ M_p = 1.67 *10^{-27} \ kg \ respectively[/tex]
So
[tex]M = 906 * 9.1 *10^{-31} + 564 * 1.67 *10^{-27}[/tex]
[tex]M = 9.43 *10^{-25} \ kg[/tex]
Name a statistical quantity that can be used to indicate an accuracy level.
Answer:
100%
Explanation:
The accuracy level can be measured by the percent of error which is defined as the average accepted 100%.
Accuracy is related to the proximity of the measurement. It is called the true value of the accuracy where as the precision is referred to the ability at the same time to be reproduced in measurement .
The instrument should be proper and produce the reliable and accurate measurement tools.
Thus here the statistical quantity that is used by the to indicate the accuracy level is purely 100%.
Refer to the data. How many males are taller than 175 cm and approximately what percentage of the total is that?
Answer:
Males over 175 cm = 10 Percentage of total = 50%Explanation:
Males measuring;
161 - 165 = 2
166 - 170 = 1
171 - 175 = 7
176 - 180 = 1
181 - 185 = 6
186 - 190 = 3
Males taller than 175 cm = 1 + 6 + 3 = 10
Total number of Males = 2 + 1 + 7 + 1 + 6 + 3 = 20
Percentage of total over 175 cm
= 10 / 20
= 50%
Which graph shows a negative acceleration
Answer:
It's d on edg
Explanation:
how to find average acceleration?
Answer:
Explanation:
A way to see this is that the definite integral of the acceleration is the change in velocity (i.e. the final velocity minus the initial velocity), and the change in velocity divided by the length of the time interval is the average acceleration on the interval.
The portion of a uniform violin string that vibrates is from the "nut" to the "bridge" at the end of the finger board, and has length LA and mass m. The string tension F can only be increased or decreased by tightening or loosening the tuning pegs above the nut. Say that a string is tuned to produce a note with fundamental frequency ft.
a) Then, to play a different note with fundamental frequency fp, the violinist uses her finger to push the string against the fingerboard, reducing the length of the vibrating part of the string to L2 (The string now vibrates from her finger to the bridge.)
i) When she makes this change, determine if each of the following quantities increase, decrease, or remain unchanged: (1) string tension, (2) mass density, (3) wave speed, and (4) wavelength. Explain each.
ii) Is fy greater or less than fx ? Explain.
iii) Find an equation for Lg in terms of all or some of the given parameters. Simplify.
b) Find an equation for F in terms of the given parameters. c) Calculate numerical values for L, and F if m= 2.00 g, LA = 60.0 cm, fa = 440 Hz (A note), fB = 494 Hz (B note). -Bridge
Answer:
A)i) 1. constant, 2. constant, 3. constant, 4. decrease
ii) frecuency increase
iii) L = n /2f √T/μ
B) L_b = 0.534 m
Explanation:
We can approximate the violin string as a system of a fixed string at its two ends, therefore we have a node at each end and a maximum in the central part for a fundamental vibration,
λ = 2L / n
where n is an integer
The wavelength and frequency are related
v = λ f
and the speed of the wave is given by
v = √T /μ
with these expressions we can analyze the questions
A)
i) In this case the woman decreases the length of the rope L = L₂
therefore the wavelength changes
λ₂ = 2 (L₂) / n
as L₂ <L₀ the wavelength is
λ₂ < λ₀
The tension of the string is given by the force of the plug as it has not moved, the tension must not change and the density of the string is a constant that does not depend on the length of the string, therefore the speed of the string wave in the string should not change.
ii) how we analyze if the speed of the wave does not change
v = λ f
as the wavelength decreases, the frequency must increase so that the speed remains constant
fy> fx
iii) It is asked to find the length of the chord
let's use the initial equations
λ = 2L / n
v = λ f
v = 2L / n f
v = √ T /μ
we substitute
2 L / n f = √ T /μ
L = n /2f √T/μ
this is the length the string should be for each resonance
b) in this part they ask to calculate the frequency
f = n / 2L √ T /μ
the linear density is
μ = m / L
μ = 2.00 10⁻³ / 60.0 10⁻²
μ = 3.33 10⁻³ kg / m
we assume that the length is adequate to produce a fundamental frequency in each case
f_{a} = 440Hz
λ = 2La / n
λ = 2 0.60 / 1
λ = 1.20 m
v = λ f
v = 1.20 440
v = 528 m / s
v² = T /μ
T = v² μ
T = 528² 3.33 10⁻³
T = 9.28 10² N
Let's find the length of the chord for fb
f_{b} = 494 hz
L_b = 1 /(2 494) √(9.28 10² / 3.33 10⁻³)
L_b = 0.534 m
Explain the Law of Conservation of Momentum