Answer:
Explanation:
compressibility = 1 / bulk modulus of elasticity ( B )
B = 1 / 4.8 x 10⁻⁵ = Δp / Δv /v ( Δp is change in pressure , Δv is change in volume )
1 / 4.8 x 10⁻⁵ = 429 / Δv /v
Δv /v = 429 x 4.8 x 10⁻⁵
= 2059.2 x 10⁻⁵
= .021
v = m / d ( d is density and m is mass of the water taken )
taking log and then differentiating
Δv /v = - Δd / d
- .021 = - Δd / d
Δd =.021 x d
= .021 x 998
= 20.9
new density
= 998 + 20.9
1018.9 kg/m3
The increase in the density of water when it is compressed from 1 atm to 430 atm pressure isothermally is 20.6 kg/m³.
Explanation:In this problem, we are asked to find the increase in the density of water when it is compressed from 1 atm to 430 atm pressure isothermally. The isothermal compressibility of water is given as 4.80 × 10−5 atm−1. The density of water at 1 atm pressure and 20°C is given as rho1 = 998 kg/m^3.
The formula that connects these quantities is Δρ = - ρ * β * ΔP, where Δρ is the change in density, β is the isothermal compressibility (in atm^−1), ΔP is the change in pressure (in atm), and ρ is the initial density (in kg/m^3).
Substituting the given values into the formula we get Δρ = - (998 kg/m³) * (4.80 × 10−5 atm−1) * (430 atm - 1 atm) = - ((998 kg/m³) * (4.80 × 10−5 atm−1) * (429 atm)) = -20.6 kg/m^3. The negative sign indicates an increase in density due to compression. So the increase in density of the water is 20.6 kg/m³.
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Match the following: (Statistical methods) Frequency counts Frequency distribution Measurements of central tendency Measurements of spread A. Applied to categorical values B. Applied to quantitative values C. Mean, median, and mode D. Max, min, percentiles, and standard deviation
Final answer:
Match the correct statistical methods to their definitions as follows: Frequency counts are A (applied to categorical values), Frequency distribution is B (applied to quantitative values), Measurements of central tendency are C (mean, median, and mode), and Measurements of spread are D (max, min, percentiles, and standard deviation).
Explanation:
When examining statistical methods, it is important to correctly match each concept with its definition to understand data analysis better. Here are the correct matches:
Frequency counts - Applied to categorical values (A)Frequency distribution - Applied to quantitative values (B)Measurements of central tendency - Mean, median, and mode (C)Measurements of spread - Max, min, percentiles, and standard deviation (D)These matches are crucial in the proper analysis and interpretation of data whether you are dealing with categorical or quantitative data. It's important to apply the correct statistical method for the level of measurement being dealt with, such as nominal, ordinal, interval, or ratio scales.
A student does 60. joules of work pushing a 3.0-kilogram box up the full length of a ramp that is 5.0 meters long. What is the magnitude of the force applied to the box to do this work?
Answer:
F= 12 N
Explanation:
Given that
Work done by student ,W= 60 J
The mass of the box ,m = 3 kg
Length ,x = 5 m
We know that ,The work done by a force a force F is given as
W= F .x
x=Displacement
F=Force
W=Work
Now by putting the values
60 = F x 5
[tex]F=\dfrac{60}{5}\ N[/tex]
F= 12 N
That is why the magnitude of the force will be 12 N.
The magnitude of the force applied to the box to do the work is 12 N
Definition of workWorkdone is defined as the product of force and distance moved in the direction of the force. Mathematically, it can be expressed as:
Workdone (Wd) = force (F) × distance (d)
Wd = Fd
With the above formula, we can obtain the force used to do the work.
Determination of the force•Workdone (Wd) = 60 J
•Distance (d) = 5 m
•Force (F) =?
Wd = Fd
60 = F × 5
Divide both side by 5
F = 60 / 5
F = 12 N
Thus, 12 N is required to do the work.
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Two identical small metal spheres are separated by 1.00 × 105 m and exert an 18.9-N repulsive force on each other. A wire is connected between the spheres and then removed. The spheres now repel each other, exerting a 22.5-N force. Assume both charges are positive.
1. Determine the total charge of this system.
q =______ (units)
2. Determine the charges of the spheres before the connection.
q1, q2 = ______ C
Answer:
(a). The total charge of this system is 10 C.
(b). The charges of the spheres before the connection is 5 C and 4.2 C.
Explanation:
Given that,
Distance [tex]r= 1.00\times10^{5}\ m[/tex]
Force = 22.5 N
The two spheres are connected for a moment by a wire. After this, charge is re-distributed equally, and each sphere now has a charge
We need to calculate the charge
Using formula of electric force
[tex]F=\dfrac{kq_{1}q_{2}}{r^2}[/tex]
Here, charge is equal
Put the value into the formula
[tex]22.5=\dfrac{9\times10^{9}\times q^2}{(1.00\times10^{5})}[/tex]
[tex]q^2=\dfrac{22.5\times(1.00\times10^{5})^2}{9\times10^{9}}[/tex]
[tex]q=\sqrt{\dfrac{22.5\times(1.00\times10^{5})^2}{9\times10^{9}}}[/tex]
[tex]q=5\ C[/tex]
(a). We need to calculate the total charge of this system
We have the charge of each sphere after re-distribution.
We know that the charge is distributed equally.
Therefore, the total charge of the system is
[tex]q+Q=2q=2\times5=10\ C[/tex]
(b). We need to calculate the charge of the spheres before the connection
Using formula of electric force
[tex]F=\dfrac{kq_{1}q_{2}}{r^2}[/tex]
Put the value into the formula
[tex]18.9=\dfrac{9\times10^{9}\times q_{1}\times q_{2}}{(1.00\times10^{5})}[/tex]
[tex]q_{2}=\dfrac{18.9\times(1.00\times10^{5})^2}{9\times10^{9}\times5}[/tex]
[tex]q_{2}=4.2\ C[/tex]
Hence, (a). The total charge of this system is 10 C.
(b). The charges of the spheres before the connection is 5 C and 4.2 C.
In the circuit described by the diagram, which pair of resistors is connected in parallel?
*answer choices attached*
Option 4 ( R2 and R3 ) is the correct answer.
Explanation:
In the below given diagram, we can see a circuit diagram that has four resistors such as R1, R2, R3, and R4.The opening of the circuit is noted as "a" and the ending is noted as "b".By observing the above diagram, we can clearly see that R2 and R3 are the pair of resistors that are connected in a parallel manner.Where all the other resistors such as R1 and R4 are neither connected in parallel nor in series.Hence we can conclude that Resistor R2 and R3 are the ones that are connected in parallel.
Centripetal force Fc acts on a car going around a curve. If the speed of the car were twice as great, the magnitude of the centripetal force necessary to keep the car moving in the same path would be
Answer:
We need 4 times more force to keep the car in circular motion if the velocity gets double.
Explanation:
Lets take the mass of the car = m
The radius of the arc = r
[tex]F=\frac{m\times v^2}{r}[/tex]
Given that speed of the car gets double ,v' = 2 v
Then the force on the car = F'
[tex]F'=\frac{m\times v'^2}{r}[/tex] ( radius of the arc is constant)
[tex]F'=\frac{m\times (2v)^2}{r} [/tex]
[tex]F'=4\times \frac{m\times v^2}{r}[/tex]
We know that [tex]F=\frac{m\times v^2}{r}[/tex]
Therefore F' = 4 F
So we can say that we need 4 times more force to keep the car in circular motion if the velocity gets double.
The magnitude of centripetal force will become four times with the twice the greater speed of car.
Given data:
The magnitude of centripetal force on car is, Fc.
The force acting on any object undergoing the motion around the circular path, such that the motion is balanced is known as centripetal force. It is also known as the center-seeking force. And the expression for the centripetal force is given as,
Fc = mv²/r
Here,
m is the mass of car.
v is the speed of car.
r is the radius of circular track.
If the speed of car is twice as great, then the new centripetal force is given as,
F'c = m(2v)²/r
F'c = 4 × mv²/r
F'c = 4 × Fc
Thus, we can conclude that the magnitude of centripetal force will become four times with the twice the greater speed of car.
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Here is a simple example. The inhabitants of a small island begin exporting beautiful cloth made from a rare plant that grows only on their island. Seeing how popular the small quantity that they export has been, they steadily raise their prices. A clothing maker from New York, thinking that he can save money by "cutting out the middleman," decides to travel to the small island and buy the cloth himself. Ignorant of the local custom of offering strangers outrageous prices and then negotiating down, the clothing maker accepts (much to everyone's surpise) the initial price of 400 tepizes/m2. The price of this cloth in New York is 120 dollars/yard2.If the clothing maker bought 500 m2 of this fabric, how much money did he lose? Use 1tepiz=0.625dollar and 0.9144m=1yard.
Answer:
He lost $53,240.60
Explanation:
In order to solve this problem, we must start by determining how many dollars he spent for the [tex]500m^{2}[/tex] of cloth. We can determine this by doing the following conversion:
[tex]500m^{2}*\frac{400 tepizes}{m^{2}}*\frac{\$0.625}{1 tepiz}=\$125,000[/tex]
Next, we need to figure out what is the price of the [tex]500m^{2}[/tex] of coth is in New York:
[tex]500m^{2}*\frac{1yd^{2}}{(0.9144m)^{2}}*\frac{\$120}{1 yard^{2}}=\$71,759.40[/tex]
so now we subtract the two amounts so we get:
$125,000-$71,759.40=$53,240.60
According to the Revere and Black (2003) article, processes that result in an error probability of 0.000070 should be recognized as achieving the Six Sigma standard. Group of answer choices True False
Answer:
False
Explanation:
The ‘sigma’ refers to the Greek letter used to denote standard deviation, so ‘six sigma’ means that the error rate lies beyond six standard deviations from the mean.
every computer consists of physical components and non physical components the non physical component of a computer that understand how to work with the physical components are referred to as_________
Answer:
C. software
Explanation:
software, is a collection of data or computer instructions that tell the computer how to work. This is in contrast to physical hardware, from which the system is built and actually performs the work.
Answer:gbji
Explanation:
hbbju
All electric devices are required to have identifying plates that specify their electrical characteristics. The plate on a certain steam iron states that the iron carries a current of 6.00 A when connected to a source of 1.20 ✕ 102 V. What is the resistance of the steam iron?
The value of steam iron resistance is [tex]20ohm[/tex].
The relation between voltage, current and resistant shown below,
[tex]V=IR\\\\R=\frac{V}{I}[/tex]
Where V is voltage and I is current .
Given that, [tex]I=6A,V=1.20*10^{2}[/tex]
Substitute values in above relation.
[tex]R=\frac{1.2*10^{2} }{6}=20ohm[/tex]
Hence, the value of steam iron resistance is [tex]20ohm[/tex].
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The resistance of the steam iron can be calculated using Ohm's Law, expressed as the formula R = V/I. With a current of 6.00 A and a voltage of 1.20 ✕ 102 V, the steam iron has a resistance of 20.0 Ohms.
Explanation:The electrical characteristics of a device, including resistance, can usually be determined using Ohm's Law, which states that the current through a conductor between two points is directly proportional to the voltage across the two points. It can be expressed in the formula R = V/I, where R is resistance, V is voltage, and I is current.
In this case, our steam iron has a current (I) of 6.00 A and a voltage (V) of 1.20 ✕ 102 V. Substituting these values into the Ohm's Law formula gives us: R = (1.20 x 102 V) / 6.00 A.
This calculates to a resistance (R) of 20.0 Ohms for the steam iron.
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Why is it necessary to centrifuge out any precipitate formed in the unknown solution and continue testing the remaining unknown solution?
Answer:
Precipitation is the formation of a solid from a solution. It is necessary to centrifuge the precipitate to exert sufficient forces of gravity to bring the solid particles in the solution to come together and settle
Explanation:
When you centrifuge precipitate it enables the nucleation to form.
Centrifuging the precipitate helps in determining whether a certain element is present in a solution or not.
Final answer:
Centrifuging out the precipitate formed in an unknown solution prevents it from interfering with the analysis of remaining ions. The separated supernatant can then be further tested to identify other ions while avoiding unwanted reactions and ensuring precise characterizations of the substances involved.
Explanation:
It is necessary to centrifuge out any precipitate formed in the unknown solution during a chemical analysis for several reasons. The process of centrifugation uses inertia to separate particles in the fluid, so when the precipitate forms due to the reaction of different ions, it needs to be removed to isolate the remaining supernatant. The purpose is to ensure that subsequent testing only involves the dissolved substances, without interference from solids that have already reacted. Moreover, analysis of the residue is crucial after centrifugation and sometimes after supernatant evaporation to dryness, followed by reconstitution of the residue. This allows for a detailed study of the precipitate itself.
The remaining solution, after centrifugation, contains the supernate which may still contain ions or molecules of interest. If the precipitate is not removed, the solids could skew the results of further tests by hiding the presence of other ions or reacting further in unwanted ways. The separated liquid, or supernatant, can then be subjected to additional tests to identify other ions that may be present. For instance, if one is testing for the presence of barium sulfate in a mixture, tests like the precipitin ring test or radial immunodiffusion assay can be used, which demonstrates the presence of specific substances without the interference from the precipitate removed by centrifugation.
If the fundamental frequency of a violin string is 440 HzHz, what is the frequency of the second harmonic?
Answer:
880Hz
Explanation:
A violin string is an example of an open pipe. An open pipe is open at both ends.
The fundamental frequency is the first harmonic.In an open pipe the second harmonic is twice the fundamental frequency.
Hence, second harmonic = 2Fo
Fo=440Hz
Second harmonic = 2*440Hz
Second harmonic = 880Hz
. Assume you have a dot grid with 36 dots per sq.in. How many acres are represented by each dot using the following map scales: (a) 330 ft. per in., (b) 25 chains per in., (c) 1 mile per in.
Answer:
a) 0.069 acres
b) 0.114acres
c) 17.78acres
Explanation:
1 dot=1/36sqin
a) 1 sqin= 330×330=108900ft^2
1 dot=108900/36 =3025ft^2
Converting to acre,divide by43560
3025/43560 =0.069acre
b) 1 chain =22 yards
25×22=550yards
Converting to acre divide by 4840
550/4840 =0.114acre
c)1 sqmile =640acres
(1/36) ÷ 640
640/36
17.78acres
To calculate the area represented by each dot on the dot grid, you need to consider the scale of the map. With a scale of 330 ft. per inch, 1 dot represents (330^2 / 43,560) acres. With a scale of 25 chains per inch, 1 dot represents ((25 chains)^2 / 10) acres. And with a scale of 1 mile per inch, 1 dot represents ((1 mile)^2 / 640) acres.
Explanation:
The conversion from dots to acres varies depending on the scale:
A ball is thrown straight up from a bridge at a speed of 11.0 m/s. What will be its velocity (speed and direction) after 2.0 seconds?
Explanation:
Let upper direction be positive
We have equation of motion v = u + at
Initial velocity, u = 11 m/s
Final velocity, v = ?
Time, t = 2 s
Acceleration,a = -9.81 m/s²
Substituting
v = u + at
v = 11 + -9.81 x 2
v = -8.62 m/s
Speed of ball after 2 seconds is 8.62 m/s downward.
The forces exerted by earth and a skier become an action reaction force pair when the skier accelerates while earth doesnt seem to move at all. expalin why the skier accelerates whu,e tthe earth doesn't seem to move at all?
Explanation:
We know from newton's 2nd law that F= m/a. That is force directly proportional to mass and inversely proportional to acceleration. F is same for the skier and Earth but mass is different for both. m for earth is way larger than the m for skier. So, for constant Force , a of Earth must very small as compared with a of the skier.
Skiers and skateboarders accelerate due to action-reaction force pairs in accordance with Newton's third law, but the Earth's immense mass leads to negligible acceleration, rendering its movement imperceptible.
Explanation:Understanding Force and Acceleration
When discussing the action-reaction force pair between the Earth and a skier, Newton's third law of motion is essential. This law states that for every action, there is an equal and opposite reaction. When the skier accelerates, the force exerted on them by the Earth (gravity) is met with an equal and opposite force exerted by the skier back onto the Earth. The reason the Earth does not seem to move while the skier accelerates is due to the massive difference in mass between the Earth and the skier. Since acceleration is calculated as force divided by mass (F = ma), the much larger mass of the Earth compared to the skier results in a negligible acceleration of the Earth. This is similar to why stationary skater A does not move when pushing skater B; if skater A has more contact with the surface or better frictional grip, they may not move significantly despite exerting force.
Furthermore, when objects like a skateboarder jump off a building, both the Earth and the skateboarder exert mutual forces on each other. Despite experiencing the same total change of momentum, the Earth's massive size means its acceleration is imperceptible, while the skateboarder moves considerably more.
What is the new pressure of 150 ml of a gas that is compressed to 50 ml when the original pressure was 3.0 atm and the temperature is held constant?
Answer: 9.0 atm
Explanation:
To calculate the new pressure, we use the equation given by Boyle's law. This law states that pressure is directly proportional to the volume of the gas at constant temperature.
The equation given by this law is:
[tex]P_1V_1=P_2V_2[/tex]
where,
[tex]P_1\text{ and }V_1[/tex] are initial pressure and volume.
[tex]P_2\text{ and }V_2[/tex] are final pressure and volume.
We are given:
[tex]P_1=3.0atm\\V_1=150mL\\P_2=?mmHg\\V_2=50mL[/tex]
Putting values in above equation, we get:
[tex]3.0\times 150mL=P_2\times 50mL\\\\P_2=9.0atm[/tex]
Thus new pressure of 150 ml of a gas that is compressed to 50 ml is 9.0 atm
A proton with kinetic energy of 1.16×105 eV is fired perpendicular to the face of a large plate that has a uniform charge density of σ = +6.60 μC/m2. What is the magnitude of the force on the proton?
Answer:
force on the proton = 5.96 × [tex]10^{-14}[/tex] N
Explanation:
given data
kinetic energy = 1.16 ×[tex]10^{5}[/tex] eV
charge density of σ = +6.60 μC/m²
solution
we get here force on the proton that is express as
F = qE ....................1
here q is charge on proton i.e = 1.6 × [tex]10^{-19}[/tex] C
and E is electric field due to charge i.e E = [tex]\frac{\sigma }{2*\epsilon_o }[/tex]
so put the value in equation 1 we get
force on the proton = 1.6 × [tex]10^{-19}[/tex] × [tex]\frac{6.60*10^{-6}}{2*8.85*10^{-12}}[/tex]
force on the proton = 5.96 × [tex]10^{-14}[/tex] N
Final answer:
The magnitude of the force on the proton is 1.19 × 10-13 N.
Explanation:
In order to find the magnitude of the force on the proton, we can use the formula F = qE, where F is the force, q is the charge of the proton, and E is the electric field. First, we need to convert the charge density from μC/m² to C/m², which gives us σ = 6.60 × 10-6 C/m². Since the electric field is uniform, we can use the equation E = σ/ε₀, where ε₀ is the electric constant.
Plugging in the values, we have E = (6.60 × 10-6 C/m²) / (8.85 × 10-12 C²/N·m²), which gives us E = 7.46 × 10^5 N/C. Now, we can find the force using F = qE.
F = (1.60 × 10-19 C)(7.46 × 10^5 N/C) = 1.19 × 10-13 N.
Four point charges each having charge Q are located at the corners of a square having sides of length a.
(a) Find a symbolic expression for the total electric potential at the center of the square due to the four charges (Use any variable or symbol stated above along with the following as necessary: ke.)
Vtotal = ______
(b) Find a symbolic expression for the work required to bring a fifth charge p from infinity to the center of the square (Use any variable or symbol stated above along with the following as necessary: ke.)
W = _____
Answer:
Explanation:
Check attachment
The total electric potential in the center of the square due to 4 chargers at its corners is given by V_total= (4 * ke * Q) / sqrt( a^2/2 ), and the work needed to bring a fifth charge p from infinity to that point is W = p * (4 * ke * Q) / sqrt( a^2/2 ).
Explanation:The electric potential at a particular point (center in this case) because of a point charge follows the formula V = k*Q / r where:
V is the electric potentialk is Coulomb's constantQ is the charger is the distance from the point to the chargeFor four charges Q at corners of a square of side length a, the distance from the charge to the center of square would be sqrt(a^2/2), so:
V_total= (4 * ke * Q) / sqrt( a^2/2 )
The work done to bring a charge from infinity to a point is given by W = q * V_total where q is the charge being moved. So, under given circumstances:
W = p * (4 * ke * Q) / sqrt( a^2/2 )
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Converging circuits with excitation and inhibition are associated most closely with which step of the perceptual process?
Answer:
Neural processing
Explanation:
Neural processing- it is referred to as the unit that is responsible for the implementation of arithmetic logic which is crucial for the execution of machine learning algorithms.
it is used to operate the artificial neural network in designing and can be helpful to enhance the efficiency of machine learning applications like artificial language.
Two workers push on a wooden crate. One worker push with a force of 543 N and the other with a force of 333 N. The mass of the wooden crate is 206 kg. What is the acceleration of the crate?
Answer:
[tex]a=4.2524\ m.s^{-2}[/tex]
Explanation:
Given:
two workers push a crate.
Force by first worker, [tex]F_1=543\ N[/tex]force by second worker, [tex]F_2=333\ N[/tex]mass of crate, [tex]m=206\ kg[/tex](Assuming that both the workers push in the same direction.)
We know that,
Acceleration is given as:
[tex]a=\frac{F}{m}[/tex]
[tex]a=\frac{F_1+F_2}{m}[/tex]
[tex]a=\frac{543+333}{206}[/tex]
[tex]a=4.2524\ m.s^{-2}[/tex]
Your companion on a train ride through Illinois notices that telephone poles near the tracks appear to be passing by very quickly, while telephone poles in the distance are passing by much more slowly. This is an example of
Answer: Relative motion
Explanation: If two objects are moving either towards or away from each other with both having their velocities in a reference frame and someone is outside this reference frame seeing the motion of the two objects.
The observer ( in his own frame of reference) will measure a different velocity as opposed to the velocities of the two object in their own reference frame. p
Both the velocity measured by the observer in his own reference frame and the velocity of both object in their reference is correct.
Velocities of this nature that have varying values based on motion referenced to another body is known as relative velocity.
Motion of this nature is known as relative motion.
Note that the word reference frame is simply any where the motion is occurring and the specified laws of motion is valid
For this example of ours, the reference frame of the companion is the train and the telephone poles has their reference frame as the earth.
The companion will measure the velocity of the telephone poles relative to him and the velocity of the telephone pole relative to an observer outside the train will be of a different value.
Charge A is sitting in an electric field you know the following information:________
the magnitude of charge A and the magnitude of both the field and the potential at charge A position.
What would you do to get the potential energy of charge A?
Answer:
The equation that will relate all the given parameters, in other to calculate the potential energy of charge A is:
∆V = ∆U/q, ∆V is potential at charge A position, q is magnitude of charge A, ∆U will be made the subject of the relation, which is the Potential Energy of charge A. The notation "∆" show, the quantities have both in values and final values, in the electric field.(Change in Electric potential and potential energy, due to the effect of the field)
Explanation:
The potential energy of a charged particle (Charge A) in an electric field depends on the magnitude of the charge(Known as stated in the question). However, the potential energy per unit charge has a unique value at any point in the electric field.
A 0.60 Kilo gram basketball poses on the rim of a basketball hoop that is 3.0 m high find the potential energy of the ball there (use G=10 N/KG.)
Answer:
18Joules
Explanation:
Potential Energy = mass * gravity * height
PE = 0.6 * 10 * 3
PE = 18Joules
The wavelength of visible light is about 5500 times longer than the wavelength of x-rays. If you represent the wavelenth of an x-ray photon by a line 1 inch long, how long a line (in feet) must you draw to represent the wavelength of visible light?
Answer:
458.33 ft
Explanation:
We are given that
Wavelength of of an x- ray photon=1 in
Wavelength of of an x- ray photon==[tex]\frac{1}{12}[/tex]ft
1 in=[tex]\frac{1}{12}[/tex]feet
We have to find the length of line (in feet) drawn by you to represent the wavelength of visible light.
According to question
Wavelength of visible light=[tex]5500\times \frac{1}{12}[/tex] ft
Wavelength of visible light[tex]=\frac{5500}{12}[/tex]ft
Wavelength of visible light=458.33 ft
Hence, the 458.33 ft line must drawn by you to represent the wavelength of visible light.
Answer:
458.33 ft
Explanation:
I had the same question once
Two wires, each of length 1.3 m, are stretched between two fixed supports. On wire A there is a second-harmonic standing wave whose frequency is 640 Hz. However, the same frequency of 640 Hz is the third harmonic on wire B. Find the speed at which the individual waves travel on each wire.
Answer:
Explanation:
Given
Length of each wire [tex]L=1.3\ m[/tex]
On wire A second harmonic frequency is given by
[tex]f_2_{a}=2\times (\frac{v}{2L})[/tex]
where f=frequency
v=velocity of wave
L=length of wire
[tex]v_a=f_2\times L[/tex]
[tex]v_a=640\times 1.3=832\ m/s[/tex]
For wire B third harmonic is given by
[tex]f_3_{b}=3\times (\frac{v}{2L})[/tex]
[tex]v_b=\frac{2L}{3}\cdot f_3_{b}[/tex]
[tex]v_b=\frac{2\times 1.3}{3}\times 640=554.66\ m/s[/tex]
A trumpet player, using her ear, hears 5 beats per second when she plays a note on her trumpet and simultaneously sounds a 440 Hz tuning fork. After pulling her tuning valve out to slightly increase the length of her trumpet, she hears 3 beats per second against the tuning fork. What was the trumpet players' initial frequency?
Frequency of trumpet wire = 445 Hz
Explanation:
The given frequency of tuning fork is 440 Hz . It produces 5 beats with trumpet wire . That means , the frequency of wire can be 440 ± 5
It can be either 445 or 435
Now the length of wire is increased , by which its frequency decreases . Because frequency is inversely proportional to length of wire .
If we decrease the frequency in 435 , the difference between tuning fork frequency and wire frequency will become greater than 5 even . So it cannot produce 3 beats with it .
If we decrease frequency from 445 , it can become 443 Hz . It gives 3 beat with the tuning fork as given .
Thus the initial frequency of wire is 445 Hz
To find the initial frequency of the trumpet player, we can use the concept of beat frequency. The initial frequency is either 442 Hz or 438 Hz.
Explanation:To find the initial frequency of the trumpet player, we can use the concept of beat frequency. The beat frequency is the difference between the frequencies of two sounds, in this case, the trumpet player's note and the tuning fork. Initially, the trumpet player hears 5 beats per second. After adjusting the length of her trumpet, she hears 3 beats per second. The change in beat frequency is 5 - 3 = 2 Hz. Since the tuning fork remains at 440 Hz, the initial frequency of the trumpet player can be calculated by adding or subtracting the beat frequency from the tuning fork frequency. In this case, the initial frequency is either 440 + 2 = 442 Hz or 440 - 2 = 438 Hz.
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When water freezes, its volume increases by 9.05% (that is, ΔV V0 = 9.05 ✕ 10−2). What force per unit area is water capable of exerting on a container when it freezes? (It is acceptable to use the bulk modulus of water in this problem.
Answer:
Δp=2.0×10⁴N/cm²
Explanation:
We can write the bulk modulus formula and we solve pressure difference Δp
So
B=Δp(V₀ / ΔV)
Δp=B( ΔV / V₀)
As ΔV / V₀ is given as 9.05×10⁻²
So
Δp=B(9.05×10⁻²)
Δp=(0.22×10¹⁰)(9.05×10⁻²)
Δp=2.0×10⁸ N/m²
Δp=2.0×10⁸ N/m²×(1m²/10⁴cm²)
Δp=2.0×10⁴N/cm²
Answer:
2.0×10⁸ N/m².
Explanation:
Bulk modulus, B is defined as the ratio of the infinitesimal pressure increase to the resulting relative decrease of the volume
B = -V₀ * (Δp/ΔV)
Given:
ΔV/V₀ = 9.05×10⁻²
Bw = 0.22×10¹⁰
Δp = -B * (ΔV/V₀)
= (0.22×10¹⁰) * (9.05×10⁻²)
= 2.0×10⁸ N/m².
Suppose a spring with spring constant 9 N/m is horizontal and has one end attached to a wall and the other end attached to a mass. You want to use the spring to weigh items. You put the spring into motion and find the frequency to be 0.9 Hz (cycles per second). What is the mass? Assume there is no friction.
To find the mass attached to a spring when the frequency of oscillation is known, use the formula for the frequency of SHM (f = (1/2π) ∙ √(k/m)) and rearrange it to solve for mass.
Explanation:To determine the mass attached to a horizontal spring where the frequency of oscillation is known, we can use the formula for the frequency of a mass-spring system undergoing simple harmonic motion (SHM), which is f = (1/2π) ∙ √(k/m), where f is the frequency, k is the spring constant, and m is the mass. Given a spring constant 9 N/m and a frequency 0.9 Hz, we can rearrange the formula to solve for the mass (m = k / (2πf)^2). Plugging in the values, we get the mass m = 9 N/m / (2π ∙ 0.9 Hz)^2, which we can calculate to find the mass of the object attached to the spring.
Using the frequency equation for a mass-spring system, the mass can be determined to be approximately 0.28 kg.
To find the mass attached to the spring, we can use the equation for the frequency of a mass-spring system:
f = (1 / (2π)) * √(k / m)Where:
f is the frequency (0.9 Hz)k is the spring constant (9 N/m)m is the mass in kgThis equation can be rearranged to solve for mass:
m = k / (4π²f²)Plugging in the provided values:
m = 9 N/m / [4π² * (0.9 Hz)²]First, simplify the expression inside the brackets:
(0.9 Hz)² = 0.814π² * 0.81 ≈ 32.17Then:
m = 9 / 32.17 m ≈ 0.28 kgThus, the mass attached to the spring is approximately 0.28 kg.
Squid use jet propulsion for rapid escapes. A squid pulls water into its body and then rapidly ejects the water backward to propel itself forward. A 1.5 kg squid (not including water mass) can accelerate at 20 m/s2 by ejecting 0.15 kg of water.?
Answer:
a. FTh = 30 N
b. Fw = 30 N
c. a = 200 m/s2
Explanation:
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The magnitude of squid thrust force, magnitude of force on the water and the acceleration experienced by the water is required.
The magnitude of squid thrust force and magnitude of force on the water is 30 N.
The acceleration is [tex]200\ \text{m/s}^2[/tex]
Newton's Lawsm = Mass
a = Acceleration
m = 1.5 kg
a = [tex]20\ \text{m/s}^2[/tex]
Thrust force
[tex]F=ma\\\Rightarrow F=1.5\times 20\\\Rightarrow F=30\ \text{N}[/tex]
The magnitude of squid thrust force will be equal to the magnitude of force on the water from the third law of motion.
F = 30 N
m = 0.15 kg
Acceleration is given by
[tex]a=\dfrac{F}{m}\\\Rightarrow a=\dfrac{30}{0.15}\\\Rightarrow a=200\ \text{m/s}^2[/tex]
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A 40.0-kg child running at 3.00 m/s suddenly jumps onto a stationary playground merry-go-round at a distance 1.50 m from the axis of rotation of the merry-go-round. The child is traveling tangential to the edge of the merry-go-round just before jumping on. The moment of inertia about its axis of rotation is 600 kg • m2 and very little friction at its rotation axis. What is the angular speed of the merry-go-round just after the child has jumped onto it?
Answer:
0.26087 rad/s
Explanation:
mass of the child (m) = 40 kg
velocity (v) = 3 m/s
distance (r) = 1.5 m
moment of inertia (I) = 600 kg.m^{2}
rotational momentum of the child = Iω
where
moment of inertia of the child (I) = [tex]mr^{2}[/tex] = 40 x 1.5 x 1.5 = 90 kg/m^{2}angular velocity (ω) = velocity / distance = 3 / 1.5 = 2 rad/srotational momentum of the child = Iω = 90 x 2 = 180 kg[tex]m^{2}[/tex]/s
from the conservation of momentum the initial momentum of the child must be the same as the final momentum of the child
initial momentum of the child = final momentum of the child
180 = (90 + 600) ω
180 = 690 ω
ω = 180 / 690 = 0.26087 rad/s
The angular or rotational momentum is the conserved quantity that has both magnitude and direction. It is given by:
[tex]\rm L = mvr[/tex]
Where,
L = rotational/angular momentum, m = mass, v = velocity and r = radius
The angular speed will be 0.26087 rad/s.
The speed can be estimated as:
Given,
Mass of the child (m) = 40 kgVelocity (v) = [tex]3 \rm \;m/s[/tex]Distance (r) = 1.5 mMoment of inertia (I) = [tex]600 \;\rm kgm^{2}[/tex]Rotational momentum is calculated by Iω.
Where,
Moment of inertia of the child (I) = [tex]\rm mr^{2}[/tex][tex]\begin{aligned}&= 40 \times (1.5)^{2} \\&= 90 \rm kg/m^{2}\end{aligned}[/tex]
Angular velocity (ω) = [tex]\rm \dfrac{velocity}{distance}[/tex][tex]\begin{aligned}&= \dfrac{3 }{1.5} \\&= 2 \rm rad/s\end{aligned}[/tex]
Rotational momentum of the child = Iω
[tex]\begin{aligned} \rm I\omega &= 90 \times 2 \\\\&= 180 \;\rm kgm^{2}/s\end{aligned}[/tex]
The final and the initial momentum would be the same according to the conservation law.
Initial momentum = Final momentum
180 = (90 + 600) ω
180 = 690 ω
Solving further:
[tex]\begin{aligned}\omega &= \dfrac{180}{690}\\\\&= 0.26087 \;\rm rad/s\end{aligned}[/tex]
Therefore, 0.26087 rad/s is the angular speed.
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Can an object have kinetic and potential energy at the same time
Answer:
An object can have both kinetic and potential energy at the same time.
Explanation:
For example, an object which is falling, but has not yet reached the ground has kinetic energy because it is moving downwards, and potential energy because it is able to move downwards even further than it already has.
Final answer:
An object can have both kinetic and potential energy at the same time, such as a ball thrown in the air. The sum of these two types of energy in a closed system remains constant unless non-conservative forces are present.
Explanation:
Yes, an object can indeed have kinetic and potential energy at the same time. Consider a ball thrown upward; at any point in its trajectory, except the peak, the ball will have both kinetic energy due to its motion and potential energy due to its height above the ground. To address whether the sum of kinetic energy and potential energy can change without work being done on the object: In a closed system where no external work is applied, the sum of kinetic and potential energies remains constant because of the conservation of mechanical energy. However, if non-conservative forces like friction are involved, they can convert mechanical energy into thermal energy, thus decreasing the total mechanical energy without any external work being performed.