The fundamental frequency of a piano wire can be calculated using the formula f = sqrt(T/(4L^2*m)). Plugging in the given values, the fundamental frequency of the piano wire is 116.6 Hz.
The fundamental frequency of a piano wire can be calculated using the formula:
f = √(T/(4L2m))
Where f is the fundamental frequency, T is the tension, L is the length, and m is the mass per unit length.
Plug in the given values:
T = 2030 N
L = 1.3 m
m = 0.003 kg/m
f = √(2030/(4*1.32*0.003))
Solving this equation gives the fundamental frequency f = 116.6 Hz.
When point charges q1 = +7.9 μC and q2 = +6.0 μC are brought near each other, each experiences a repulsive force of magnitude 0.75 N. Determine the distance between the charges.
Answer:0.754m
Explanation:
F=kq1q2/r^2
R^2= kq1q2/f
R^2= 9*10^9*7.9*10^-6*6.0*10^-6/0.75
R^2= 9*7.9*6*10^-3/0.75
R^2=0.4266/0.75
R^2=0.5688
R=√0.5688
R=0.754m
If an athlete expends 3480. kJ/h, how long does she have to play to work off 1.00 lb of body fat? Note that the nutritional calorie (Calorie) is equivalent to 1 kcal, and one pound of body fat is equivalent to about 4.10 × 103 Calories.
Answer:
The time required by the Athlete to work off 1.00 lb of body fat = 0.296 minute
Explanation:
1 lb of body fat = 4.1 k cal
1 k cal = 4.184 Kilo joule
1 lb of body fat = 4.1 × 4.184 = 17.1544 Kilo joule
Athlete expends 3480 Kilo joule in one hour
⇒ Time required to expand 3480 Kilo joule = 60 minute
⇒ Time required to expand 1 Kilo joule = [tex]\frac{60}{3480}[/tex] [tex]\frac{min}{KJ}[/tex]
⇒ Time required to expand 17.1544 Kilo joule = [tex]\frac{60}{3480}[/tex] × 17.1544 = 0.296 min
Therefore the time required by the Athlete to work off 1.00 lb of body fat = 0.296 minute
A supply plane needs to drop a package of food to scientists working on a glacier in Greenland. The plane flies 80.0 m above the glacier at a speed of 120 m/s.How far short of the target should it drop the package?
Explanation:
Below is an attachment containing the solution
Consider a vibrating system described by the initial value problem. (A computer algebra system is recommended.) u'' + 1 4 u' + 2u = 2 cos ωt, u(0) = 0, u'(0) = 2 (a) Determine the steady state part of the solution of this problem.
Answer:
Therefore the required solution is
[tex]U(t)=\frac{2(2-\omega^2)^2}{(2-\omega^2)^2+\frac{1}{16}\omega} cos\omega t +\frac{\frac{1}{2}\omega}{(2-\omega^2)^2+\frac{1}{16}\omega} sin \omega t[/tex]
Explanation:
Given vibrating system is
[tex]u''+\frac{1}{4}u'+2u= 2cos \omega t[/tex]
Consider U(t) = A cosωt + B sinωt
Differentiating with respect to t
U'(t)= - A ω sinωt +B ω cos ωt
Again differentiating with respect to t
U''(t) = - A ω² cosωt -B ω² sin ωt
Putting this in given equation
[tex]-A\omega^2cos\omega t-B\omega^2sin \omega t+ \frac{1}{4}(-A\omega sin \omega t+B\omega cos \omega t)+2Acos\omega t+2Bsin\omega t = 2cos\omega t[/tex]
[tex]\Rightarrow (-A\omega^2+\frac{1}{4}B\omega +2A)cos \omega t+(-B\omega^2-\frac{1}{4}A\omega+2B)sin \omega t= 2cos \omega t[/tex]
Equating the coefficient of sinωt and cos ωt
[tex]\Rightarrow (-A\omega^2+\frac{1}{4}B\omega +2A)= 2[/tex]
[tex]\Rightarrow (2-\omega^2)A+\frac{1}{4}B\omega -2=0[/tex].........(1)
and
[tex]\Rightarrow -B\omega^2-\frac{1}{4}A\omega+2B= 0[/tex]
[tex]\Rightarrow -\frac{1}{4}A\omega+(2-\omega^2)B= 0[/tex]........(2)
Solving equation (1) and (2) by cross multiplication method
[tex]\frac{A}{\frac{1}{4}\omega.0 -(-2)(2-\omega^2)}=\frac{B}{-\frac{1}{4}\omega.(-2)-0.(2-\omega^2)}=\frac{1}{(2-\omega^2)^2-(-\frac{1}{4}\omega)(\frac{1}{4}\omega)}[/tex]
[tex]\Rightarrow \frac{A}{2(2-\omega^2)}=\frac{B}{\frac{1}{2}\omega}=\frac{1}{(2-\omega^2)^2+\frac{1}{16}\omega}[/tex]
[tex]\therefore A=\frac{2(2-\omega^2)^2}{(2-\omega^2)^2+\frac{1}{16}\omega}[/tex] and [tex]B=\frac{\frac{1}{2}\omega}{(2-\omega^2)^2+\frac{1}{16}\omega}[/tex]
Therefore the required solution is
[tex]U(t)=\frac{2(2-\omega^2)^2}{(2-\omega^2)^2+\frac{1}{16}\omega} cos\omega t +\frac{\frac{1}{2}\omega}{(2-\omega^2)^2+\frac{1}{16}\omega} sin \omega t[/tex]
The spring of a spring gun has force constant k =400 N/m and negligible mass. The spring is compressed6.00 cm and a ball with mass 0.0300 kg isplaced in the horizontal barrel against the compressed spring. Thespring is then released, and the ball is propelled out the barrelof the gun. The barrel is 6.00 cm long, so the ballleaves the barrel at the same point that it loses contact with thespring. The gun is held so the barrel is horizontal.
Calculate the speed with which the ballleaves the barrel if you can ignore friction.
Calculate the speed of the ball as it leavesthe barrel if a constant resisting force of 6.00 Nacts on the ball as it moves along the barrel.
For the situation in part (b), at whatposition along the barrel does the ball have the greatest speed?(In this case, the maximum speed does not occur at the end of thebarrel.)
What is that greatest speed?
a) 6.9 m/s
b) 4.9 m/s
c) After 4.50 cm
d) 5.2 m/s
Explanation:
a)
In this case, there is no resistive force. Therefore, according to the law of conservation of energy, all the initial elastic potential energy stored in the spring when it is compressed is converted into kinetic energy of the ball as it leaves the barrel; so we can write:
[tex]\frac{1}{2}kx^2=\frac{1}{2}mv^2[/tex]
where:
k = 400 N/m is the spring constant
x = 6.00 cm = 0.06 m is the compression of the spring
m = 0.03 kg is the mass of the ball
v is the velocity of the ball as it leaves the barrel
Solving the equation, we can find the speed of the ball:
[tex]v=\sqrt{\frac{kx^2}{m}}=\sqrt{\frac{(400)(0.06)^2}{0.03}}=6.9 m/s[/tex]
b)
In this case, there is a constant resistive force of
[tex]F_r = 6.00 N[/tex]
acting on the ball as it moves along the barrel.
The work done by this force on the ball is:
[tex]W=-F_rd[/tex]
where
d = 6.00 cm = 0.06 m is the length of the barrel
And where the negative sign is due to the fact that the force is opposite to the motion of the ball
Substituting,
[tex]W=-(6.00)(0.06)=-0.36 J[/tex]
Therefore, part of the initial elastic potential energy stored in the spring has been converted into thermal energy due to the resistive force. So, the final kinetic energy of the ball will be less than before:
[tex]\frac{1}{2}kx^2+W=\frac{1}{2}mv^2[/tex]
And solving for v, we find the new speed:
[tex]v=\sqrt{\frac{kx^2}{m}+\frac{2W}{m}}=4.9 m/s[/tex]
c)
The (forward) force exerted by the spring on the ball is
[tex]F=k(x_0-x)[/tex]
where
k = 400 N/m
x is the distance covered by the ball
[tex]x_0=0.06 m[/tex] is the maximum displacement of the spring
While the (backward) resistive force is instead
[tex]F_r=6.0 N[/tex]
So the net force on the ball is
[tex]F=k(x-x_0)-F_r[/tex]
The term [tex]k(x-x_0)[/tex] prevails at the beginning, so the ball continues accelerating forward, until this term becomes as small as [tex]F_r[/tex]: after that point, the negative resistive force will prevail, so the ball will start delecerating. Therefore, the greatest speed is reached when the net force is zero; so:
[tex]k(x_0-x)-F_r=0\\x=x_0-\frac{F_r}{k}=0.06-\frac{6.00}{400}=0.045 m[/tex]
d)
After the ball has covered a distance of
x = 0.045 m
The work done by the resistive force so far is:
[tex]W=-F_r x =-(6.00)(0.045)=-0.27 J[/tex]
The total elastic potential energy of the spring at the beginning was
[tex]E_e=\frac{1}{2}kx_0^2 = \frac{1}{2}(400)(0.06)^2=0.72 J[/tex]
While the elastic potential energy left now is
[tex]E_e'=\frac{1}{2}k(x_0-x)^2=\frac{1}{2}(400)(0.015)^2=0.045 J[/tex]
So, the kinetic energy now is:
[tex]K=E_e-E_e'+W=0.72-0.045-0.27=0.405 J[/tex]
And by using the equation
[tex]K=\frac{1}{2}mv^2[/tex]
We can find the greatest speed:
[tex]v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(0.405)}{0.03}}=5.2 m/s[/tex]
A parallel plate capacitor is connected to a battery and charged to voltage V. Leah says that the charge on the plates will decrease if the distance between the plates is increased while they are still connected to the battery. Gertie says that the charge will remain the same. Which one, if either, is correct, and why?
Explanation:
Below is an attachment containing the solution.
What is the amount of electric field passing through a surface called? A. Electric flux.B. Gauss’s law.C. Electricity.D. Charge surface density.E. None of the above.
Answer:
Electric flux
Explanation:
The electric flux measures the amount of electric field passing through a surface. For any closed surface, the electric field passing through it (electric flux) is given by Guass law. The mathematical relation between electric flux and the enclosed charge is known as Gauss law for the electric field. Electric flux may also be visualised as the amount of electric lines of force passing through an area.
Studying climate on other planets is important to understanding climate on earth because
Answer:
yes it is important.
Explanation:
It is important to study the climate of other planet to understand the climate of the earth because it is useful to know which things are harmful for our earth and which are important for our earth climate comparing with the other planets.
Three different orientations of a magnetic dipole moment in a constant magnetic field are shown below. Which orientation results in the largest magnetic torque on the dipole ?
Answer:
The orientation b has the largest magnetic torque.
Explanation:
As the complete question is not given, the complete question is attached herewith
From the diagram
[tex]|\mu_a|=|\mu_b|=|\mu_c|=|\mu|[/tex]
Also the angles for the 3 orientations are given as
[tex]\theta_a>90\\\theta_b=90\\\theta_c<90\\[/tex]
Now as the torque τ is given as
[tex]\tau=|\mu||B|sin\theta[/tex]
As the value of μ and B is same so value of τ is maximum for sin θ is maximum so
[tex]sin \theta_{max}=1\\\theta_{max}=90[/tex]
So the orientation b has the largest magnetic torque.
The orientation with the largest magnetic torque on the dipole is Orientation B (See attached image).
What is Magnetic Torque?
The torque on the dipole is defined as:
τ = µ×B,
where B is the external magnetic field.
The magnitude of this torque is µB sinθ, where θ is the angle between B and µ
Magnetic Torque is highest when;
→ →
µ ⊥ β
That is when θ = 90°. Hence B is the correct answer. Please see attached image.
Learn more about Magnetic Torque at:
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The amount of kinetic energy an object has depends on its mass and its speed. Rank the following sets of oranges and cantaloupes from least kinetic energy to greatest kinetic energy. If two sets have the same amount of kinetic energy, place one on top of the other. 1. mass: m speed: v2. mass: 4 m speed: v3. total mass: 2 m speed: 1/4v4. mass: 4 m : speed: v5. total mass: 4 m speed: 1/2v
mass₃<mass₁=mass₅<mass₂=mass₄
Explanation:
Given data :-
1. mass: m speed: v
2. mass: 4 m speed: v
3. mass: 2 m speed: ¼ v
4. mass: 4 m speed: v
5. mass: 4 m speed: ½ v
We know that Kinetic energy (KE) = ½ mv²
Where m= mass of the body
v=velocity of the body
Substituting the values of respective mass and velocity from the above given data-
KE of Body 1(mass₁) = ½*m*v² = mv²/2
KE of Body 2(mass₂) = ½*4m*v² = 2mv²
KE of Body 3(mass₃) = ½*2m*(1/4v)² = mv²/16
KE of Body 4(mass₄) = ½*4m*v ² = 2mv ²
KE of Body 5(mass₅) = ½*4m*(1/2v)² = mv²/2
A ? is a conductor installed on the supply side of a service or separately derived system to ensure the required electrical conductivity between metal parts required to be electrically connected.
Answer:
Supply-side bonding jumper.
Explanation:
A supply-side bonding jumper is a conductor installed on the supply side of a service or separately derived system to ensure the required electrical conductivity between metal parts required to be electrically connected.
The supply side bonding jumper was referred to as the equipment bonding jumper prior to 2011 NEC when its name changed.
The wire runs from the source of the separately derived system to the first disconnecting means.
______ Made geocentric model of the solar system using epicycles
Answer:
Ptolemy made geocentric model of the solar system using epicycles
Explanation:
Ptolemy made geocentric model of the solar system using epicycles.
This model accounted for the apparent motions of the planets in a very direct way, by assuming that each planet moved on a small circle, called an epicycle, which moved on a larger circle, called a deferent.
Therefore, Ptolemy is the answer.
When a metal rod is heated, its resistance changes both because of a change in resistivity and because of a change in the length of the rod. If a silver rod has a resistance of 1.70 Ω at 21.0°C, what is its resistance when it is heated to 180.0°C? The temperature coefficient for silver is α = 6.1 ✕ 10−3 °C−1, and its coefficient of linear expansion is 18 ✕ 10−6 °C−1. Assume that the rod expands in all three dimensions.
Answer:
[tex]3.34\Omega[/tex]
Explanation:
The resistance of a metal rod is given by
[tex]R=\frac{\rho L}{A}[/tex]
where
[tex]\rho[/tex] is the resistivity
L is the length of the rod
A is the cross-sectional area
The resistivity changes with the temperature as:
[tex]\rho(T)=\rho_0 (1+\alpha (T-T_0))[/tex]
where in this case:
[tex]\rho_0[/tex] is the resistivity of silver at [tex]T_0=21.0^{\circ}C[/tex]
[tex]\alpha=6.1\cdot 10^{-3} ^{\circ}C^{-1}[/tex] is the temperature coefficient for silver
[tex]T=180.0^{\circ}C[/tex] is the current temperature
Substituting,
[tex]\rho(180^{\circ}C)=\rho_0 (1+6.1\cdot 10^{-3}(180-21))=1.970\rho_0[/tex]
The length of the rod changes as
[tex]L(T)=L_0 (1+\alpha_L(T-T_0))[/tex]
where:
[tex]L_0[/tex] is the initial length at [tex]21.0^{\circ}C[/tex]
[tex]\alpha_L = 18\cdot 10^{-6} ^{\circ}C^{-1}[/tex] is the coefficient of linear expansion
Substituting,
[tex]L(180^{\circ}C)=L_0(1+18\cdot 10^{-6}(180-21))=1.00286L_0[/tex]
The cross-sectional area of the rod changes as
[tex]A(T)=A_0(1+2\alpha_L(T-T_0))[/tex]
So, substituting,
[tex]A(180^{\circ}C)=A_0(1+2\cdot 18\cdot 10^{-6}(180-21))=1.00572A_0[/tex]
Therefore, if the initial resistance at 21.0°C is
[tex]R_0 = \frac{\rho_0 L_0}{A_0}=1.70\Omega[/tex]
Then the resistance at 180.0°C is:
[tex]R(180^{\circ}C)=\frac{\rho(180)L(180)}{A(180)}=\frac{(1.970\rho_0)(1.00285L_0)}{1.00572A_0}=1.9644\frac{\rho_0 L_0}{A_0}=1.9644 R_0=\\=(1.9644)(1.70\Omega)=3.34\Omega[/tex]
A pipe has a length of 1.29 m. Determine the frequency of the first harmonic if the pipe is open at each end. The velocity of sound in air is 343 m/s. Answer in units of Hz.
Answer:
265.9Hz
Explanation:
In an open pipe, both ends of the pipes are opened. The fundamental frequency in an open pipe is expressed as fo = V/2L where;
f is the frequency of the wave
V is the velocity of the wave = 343m/s
L is the length of the pipe = 1.29m
Substituting the value to get the fundamental frequency in the open pipe we have;
Fo = 343/2(1.29)
Fo = 343/2.58
Fo = 132.95Hz
Harmonics are integral multiples of the fundamental frequency e.g 2fo, 3fo, 4fo, 5fo...
The first harmonic in the open pipe will be f1 = 2fo
Since f1 =2(132.95)
f1 = 265.9Hz
The frequency of the first harmonic if the pipe is open at each end is 265.9Hz
Answer:
132.95 Hz.
Explanation:
Given:
v = 343 m/s
L = 1.29 m.
Since the pipe is open at both ends,
L = λ/2
λ = v/f = 2L
= 2 × 1.29
= 2.58 m
f = 343/2.58
= 132.95 Hz.
Match the following vocabulary with their definitions. 1 . the distribution of light when passed through a prism or other device that breaks the light into its individual components energy level 2 . the state where one or more electrons is in a higher energy level than ground state due to the addition of energy, often in the form of heat or light emission spectrum 3 . regions located around the nucleus where the electrons are found quantum 4 . energy available or given off in specific quantities excited state
Final answer:
The emission spectrum corresponds to energy emitted by electrons as they fall back from an excited state to a lower energy state. Energy levels are the quantized orbits around the nucleus, and quantum refers to the specific amount of energy in these processes. The excited state is a temporary, higher energy level for an electron.
Explanation:
Lets match the vocabulary with their definitions:
Energy level - The distribution of light when passed through a prism or other device that breaks the light into its individual components.Excited state - The state where one or more electrons is in a higher energy level than ground state due to the addition of energy, often in the form of heat or light.Quantum - Regions located around the nucleus where the electrons are found.Emission spectrum - Energy available or given off in specific quantities.Now, to give a more in-depth understanding:
The emission spectrum of an element is the unique pattern of light emitted when energy is given to an atom and then released as electrons return from an excited state to the ground state.Energy levels are the possible orbits that an electron can occupy around an atom's nucleus. They are quantized, meaning electrons must move between levels in discrete steps, not continuously.Quantum implies that these energy levels and the energy emitted or absorbed are given in specific, quantized amounts.An excited state is when an electron in an atom has absorbed energy and is at a higher energy level than the atom's ground state. This state is temporary and the electron will eventually fall back to the ground state and release a photon.What is the amount of electric field passing through a surface called? A. Electric flux.B. Gauss’s law.C. Electricity.D. Charge surface density.E. None of the above.
Answer:
A. Electric flux
Explanation:
Electric flux is the rate of flow of the electric field through a given area (see ). Electric flux is proportional to the number of electric field lines going through a virtual surface.
Electric flux has SI units of volt metres (V m), or, equivalently, newton metres squared per coulomb (N m2 C−1). Thus, the SI base units of electric flux are kg·m3·s−3·A−1.
A small car of mass m and a large car of mass 4m drive along a highway at constant speed. They approach a curve of radius R. Both cars maintain the same acceleration a as they travel around the curve. How does the speed of the small car vS compare to the speed of the large car vL as they round the curve
Answer:
Explanation:
Given that we have two cars
First car has mass =m
Second car has mass = 4m
They are driving at constant speed
Given that the radius of curve is R
Both cars maintain the same acceleration.
Let velocity of small car be vS
Velocity of big car be vL
From centripetal acceleration
a=V²/R
V²=aR
Then since both car have the same accelerating and bashing through the same curve of radius R
Then, We can say, V² is constant
vL² = vS²
Then taking square root of both side
vL=vS
The speed of the small car vS is greater than the speed of the large car vL as they round the curve due to the difference in their masses and the application of the same acceleration.
Explanation:The speed of the small car vS is greater than the speed of the large car vL as they round the curve. This is because both cars have the same acceleration a, but the small car has less mass than the large car. According to Newton's second law, F = ma, the force required to maintain the same acceleration is greater for the larger car, which means it has a greater speed as it rounds the curve.
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84. Calculate the velocity a spherical rain drop would achieve falling from 5.00 km (a) in the absence of air drag (b) with air drag. Take the size across of the drop to be 4 mm, the density to be 1.00 × 103 kg/m3 , and the surface area to be πr2 .
Answer:
a) [tex]v=313.209\ m.s^{-1}[/tex]
b) [tex]v_t=3751.79\ m.s^{-1}[/tex]
Explanation:
Given:
height of the raindrop, [tex]h=5000\ m[/tex]a)
[tex]v=\sqrt{2g.h}[/tex]
[tex]v=\sqrt{2\times 9.81\times 5000}[/tex]
[tex]v=313.209\ m.s^{-1}[/tex]
b)
given that:
diameter of the drop, [tex]d=4\ mm=0.004\ m[/tex]
density of the air, [tex]\rho=1.18\ kg.m^{-3}[/tex]
the terminal velocity is given as:
[tex]v_t=\sqrt{\frac{2m.g}{\rho.A.c_d} }[/tex]
where:
m = mass
g = acceleration due to gravity
[tex]\rho=[/tex] density of the medium through which the drop is falling (here air)
A = area normal to the velocity of fall
[tex]c_d=[/tex] coefficient of drag = 0.47 for spherical body
[tex]v_t=\sqrt{\frac{2\times 5\times 9.81}{1.18\times \pi\times 0.002^2\times 0.47} }[/tex]
[tex]v_t=3751.79\ m.s^{-1}[/tex]
You are directed to set up an experiment in which you drop, from shoulder height, objects with similar surface areas but different masses, timing how long it takes for each object to hit the floor. Of the following explanations, which best describes your findings?
a. the most dense object hits the ground first
b. the less dense object hits the ground first
c. they will hit the ground at the same time
Answer: c. they will hit the ground at the same time
Explanation:
The volume of both objects is almost the same, so the force of friction will be the same in each one, so we can discard it.
Now, when yo drop an object, the acceleration of the object is always g = 9.8m/s^2 downwards, independent of the mass of the object.
So if you drop two objects with the same volume but different mass, because the acceleration is the same for both of them, they will hit the ground at the same time, this means that the density of the object has no impact in how much time the object needs to reach the floor.
So the correct option is c
A student stands a distance L from a wall and claps her hands. Immediately on hearing the reflection from the wall she claps her hands again. She continues to do this, so that successive claps and the sound of reflected claps coincide. The frequency at which she claps her hands is f. What is the speed of sound in air?
Answer:
Explanation:
The distance of the student from the wall is L
After hearing the reflection she claps her hand again
A complete cycle is the distance to the wall and back to the boy, so the total distance travel then is 2L, then the wave length is 2L
λ=2L
So the frequent is f
Then using wave equation
v=f λ
Since our λ=2L
Then, v=f×2L
v=2fL
The speed of sound in air is 2fL
Final answer:
To find the speed of sound in air based on the frequency of clapping and its reflection off a wall, we can use the rearranged equation V = 2fL, where V is the speed of sound, f is the frequency of clapping, and L is the distance to the wall.
Explanation:
The question involves calculating the speed of sound in air based on the frequency at which a person claps their hands and the time it takes for the sound to reflect off a wall and travel back to them. To solve this, we can use the formula f = ½(V / L), where f is the frequency of clapping (in Hz), V is the speed of sound in air (in m/s), and L is the distance from the person to the wall (in meters). However, the provided formulas from various attempts don't directly address the question but hint at the relationship between the speed of sound, frequency, and distance. Thus, by understanding that the sound needs to travel twice the distance (L) to reach the person again and taking into account the time interval represented by the frequency of clapping, we can rearrange the formula to solve for the speed of sound as V = 2fL. It's understood that the time taken for the sound to travel to the wall and back is inversely related to the frequency of clapping.
A room that has an average ambient sound pressure level of 62 dBA and a maximum sound pressure level lasting more than a minute at 68 dBA must have a public mode signal that is at least ? .
Answer:
Explanation:
A fire alarm notification appliance is an active fire protection component of a fire alarm system. The primary function of the notification appliance is to alert persons at risk.
If want the audible public mode signal to be hear clearly then, we need to have a sound level that is at least 15dB above the average ambient sound level or 5dB above the maximum sound level of at least 1minute
In this case the,
The average ambient sound level is 62dB,
And the maximum sound level is 68dB
Then, the public mode signal should be at least
1. 62dB+ 15dB=77dB
Or
2. 68dB +5dB =73dB.
Then the public mode signal hearing must be at least 77dB.
In a certain time period a coil of wire is rotated from one orientation to another with respect to a uniform 0.38-T magnetic field. The emf induced in the coil is 5.0 V. Other things being equal, what would the induced emf be if the field had a magnitude of 0.55 T
Explanation:
When coil is placed in the magnetic field , the flux attached with it can be found by the relation . Flux Ф = the dot product of magnetic field and area of coil .
Thus Ф = B A cosθ
here B is magnetic field strength and A is the area of coil .
The angle θ is the angle between coil and field direction .
When coil rotates , the angle varies . By which the flux varies . The emf is produced in coil due to variation of flux . The relation for this is
The emf produced ξ = - [tex]\frac{d\phi}{dt}[/tex] = B A sinθ [tex]\frac{d\theta}{dt}[/tex]
Now in the given problem
5 = 0.38 x A x [tex]\frac{d\theta}{dt}[/tex] I
Now if the magnetic field is 0.55 T and all the other terms are same , the emf produced
ξ = 0.55 x A x [tex]\frac{d\theta}{dt}[/tex] Ii
dividing II by I , we have
[tex]\frac{\xi}{5}[/tex] = [tex]\frac{0.55}{0.38}[/tex] = 1.45
or ξ = 7.2 V
A spring is hung vertically with a 425g mass attached to it. The mass is at rest. If the mass causes the spring to stretch 0.67 m, what is the spring constant?
Answer:
6.22 N/m
Explanation:
From Hooke's law we deduce that F=kx where F is the applied force and k is the spring constant while x is the extension or compression of the spring. Making k the subject of the above formula then
[tex]k=\frac {F}{x}[/tex]
We also know that the force F is equal to mg where m is the mass of an object and g is acceleration due to gravity hence substituting F with mg we get that
[tex]k=\frac {mg}{x}[/tex]
Substituting m with 425 g which is equivalent to 0.425 kg and g with 9.81 then 0.67 for x we get that
[tex]k=\frac {mg}{x}=\frac {0.425\times 9.81}{0.67}=6.222761194 N/m\approx 6.22\ N/m[/tex]
Therefore, the spring constant is approximately 6.22 N/m
A throttle position sensor waveform is going to be observed. At what setting should the volts per division be set to see the entire waveform from 0 volts to 5 volts using a DSO with 8 voltage and 10 time divisions?
Answer:
1 V / div
Explanation:
Solution:
- The vertical scale has eight divisions.
- If each division is set to equal 1 volt, the display will show 0 to 8 volts.
- This is okay in a 0 to 5 volt variable sensor such as a throttle position (TP) sensor.
- The volts per division (V/div) should be set so that the entire anticipated waveform can be viewed.
A horizontal spring-mass system has low friction, spring stiffness 205 N/m, and mass 0.6 kg. The system is released with an initial compression of the spring of 13 cm and an initial speed of the mass of 3 m/s.
(a) What is the maximum stretch during the motion? m
(b) What is the maximum speed during the motion? m/s
(c) Now suppose that there is energy dissipation of 0.02 J per cycle of the spring-mass system. What is the average power input in watts required to maintain a steady oscillation?
Answer:
(a). The maximum stretch during the motion is 20.7 cm
(b). The maximum speed during the motion is 3.84 m/s.
(c). The energy is 0.060 Watt.
Explanation:
Given that,
Spring stiffness = 205 N/m
Mass = 0.6 kg
Compression of spring = 13 cm
Initial speed = 3 m/s
(a). We need to calculate the maximum stretch during the motion
Using conservation of energy
[tex]E_{initial}=E_{final} [/tex]
[tex]\dfrac{1}{2}kx_{c}^2+\dfrac{1}{2}mv^2=\dfrac{1}{2}kx_{m}^2[/tex]
Put the value into the formula
[tex]\dfrac{1}{2}\times205\times(13\times10^{-2})^2+\dfrac{1}{2}\times0.6\times3^2=\dfrac{1}{2}\times205\times x_{m}^2[/tex]
[tex]x_{m}=\sqrt{\dfrac{4.43\times2}{205}}[/tex]
[tex]x_{m}=20.7\ cm[/tex]
(b). Maximum speed comes when stretch is zero.
We need to calculate the maximum speed during the motion
Using conservation of energy
[tex]E_{initial}=E_{final} [/tex]
[tex]\dfrac{1}{2}kx_{c}^2+\dfrac{1}{2}mv^2=\dfrac{1}{2}mv'^2[/tex]
Put the value into the formula
[tex]\dfrac{1}{2}\times205\times(13\times10^{-2})^2+\dfrac{1}{2}\times0.6\times3^2=\dfrac{1}{2}\times0.6\times v'^2[/tex]
[tex]v'=\sqrt{\dfrac{4.43\times2}{0.6}}[/tex]
[tex]v'=3.84\ m/s[/tex]
(c). Now suppose that there is energy dissipation of 0.02 J per cycle of the spring-mass system
We need to calculate the time period
Using formula of time period
[tex]T=2\pi\sqrt{\dfrac{m}{k}}[/tex]
Put the value into the formula
[tex]T=2\pi\sqrt{\dfrac{0.6}{205}}[/tex]
[tex]T=0.33\ sec[/tex]
We need to calculate the energy
Using formula of energy
[tex]E=\dfrac{P}{t}[/tex]
Put the value into the formula
[tex]E=\dfrac{0.02}{0.33}[/tex]
[tex]E=0.060\ Watt[/tex]
Hence, (a). The maximum stretch during the motion is 20.7 cm
(b). The maximum speed during the motion is 3.84 m/s.
(c). The energy is 0.060 Watt.
A linear accelerator produces a pulsed beam of electrons. The pulse current is 0.50 A, and the pulse duration is 0.10 μs. (a) How many electrons are accelerated per pulse? (b) What is the average current for a machine operating at 500 pulses/s? If the electrons are accelerated to an energy of 50 MeV, what are the (c) average power and (d) peak power of the accelerator?
Answer:
a)N = 3.125 * 10¹¹
b) I(avg) = 2.5 × 10⁻⁵A
c)P(avg) = 1250W
d)P = 2.5 × 10⁷W
Explanation:
Given that,
pulse current is 0.50 A
duration of pulse Δt = 0.1 × 10⁻⁶s
a) The number of particles equal to the amount of charge in a single pulse divided by the charge of a single particles
N = Δq/e
charge is given by Δq = IΔt
so,
N = IΔt / e
[tex]N = \frac{(0.5)(0.1 * 10^-^6)}{(1.6 * 10^-^1^9)} \\= 3.125 * 10^1^1[/tex]
N = 3.125 * 10¹¹
b) Q = nqt
where q is the charge of 1puse
n = number of pulse
the average current is given as I(avg) = Q/t
I(avg) = nq
I(avg) = nIΔt
= (500)(0.5)(0.1 × 10⁻⁶)
= 2.5 × 10⁻⁵A
C) If the electrons are accelerated to an energy of 50 MeV, the acceleration voltage must,
eV = K
V = K/e
the power is given by
P = IV
P(avg) = I(avg)K / e
[tex]P(avg) = \frac{(2.5 * 10^-^5)(50 * 10^6 . 1.6 * 10^-^1^9)}{1.6 * 10^-^1^9}[/tex]
= 1250W
d) Final peak=
P= Ik/e
= [tex]= P(avg) = \frac{(0.5)(50 * 10^6 . 1.6 * 10^-^1^9)}{1.6 * 10^-^1^9}\\2.5 * 10^7W[/tex]
P = 2.5 × 10⁷W
Explanation:
Below is an attachment containing the solution.
In a certain material there is a current of 16 A flowing through a surface to the right, and there is an equal amount of positive and negative charge passing through the surface producing the current. How much negative charge passes through the surface?
Answer:
Explanation:
Give that
I=16A
The current is flowing to the right
Equal amount of positive and negative charges.
i.e, total charge is q= Q+Q=2Q. i.e magnitude
Then, the rate of charge that pass though is dq/dt
Give that,
q=it
2Q=it
Let differentiate with respect to t
2dQ/dt=i
Then, dQ/dt=i/2
Since i=16A
Then, dQ/dt=16/2
dQ/dt= 8 A/s. Toward the left
A ball is thrown upward. At a height of 10 meters above the ground, the ball has a potential energy of 50 joules (with the potential energy equal to zero at ground level) and is moving upward with a kinetic energy of 50 joules. Air friction is negligible. The maximum height reached by the ball is most nearly
Answer: 20m
Explanation:
We will solve this question by applying the law of conservation of energy which states that the sum of potential energy and kinetic energy is always the same.
The PE is 0 at surface and maximum at top while the KE is maximum at surface and 0 at top.
From the question,
PE = mgh = 50 J -(1)
mg* 10 = 50
mg = 50/10
mg = 5
The total energy at that point = PE + KE = 50 + 50 = 100 J
Therefore, at topmost point, the PE will be 100 J
mgH = 100J , H is the needed height
Using the value of mg obtained above, we have
H= 100/5
H = 20 m
The man jumps from the window of a burning hotel and lands in a safety net that stops him fall in 1 mm. Determine the average force that the net exerts on the man if he enters the net at a speed of 28 m/sm/s. Assume that the man's mass is 64 kgkg.
Final answer:
The average force that the net exerts on the man is approximately -5.02 x 10^7 N. The negative sign indicates that the force is in the opposite direction to the motion.
Explanation:
To determine the average force that the net exerts on the man, you can use the equation:
Force = mass x acceleration
First, calculate the acceleration of the man using the formula:
acceleration = change in velocity / time taken
In this case, the final velocity is 0 m/s, the initial velocity is 28 m/s, and the time taken is the distance the net stops the man divided by his initial velocity. Since the net stops the man within a distance of 1 mm (0.001 m), the time taken is:
time taken = distance / initial velocity = 0.001 m / 28 m/s = 3.571 x 10-5 s
Now, you can calculate the acceleration:
acceleration = (0 - 28) m/s / 3.571 x 10-5 s = -7.838 x 105 m/s2
Next, calculate the force:
force = mass x acceleration = 64 kg x (-7.838 x 105 m/s2)
force = -5.02 x 107 N
The average force that the net exerts on the man is approximately -5.02 x 107 N. The negative sign indicates that the force is in the opposite direction to the motion.
To calibrate the calorimeter electrically, a constant voltage of 3.6 V is applied and a current of 2.6 A flows for a period of 350 seconds. If the temperature rises from 20.3°C to 29.1°C, what is the heat capacity of the calorimeter?
Answer:
[tex]372.3 J/^{\circ}C[/tex]
Explanation:
First of all, we need to calculate the total energy supplied to the calorimeter.
We know that:
V = 3.6 V is the voltage applied
I = 2.6 A is the current
So, the power delivered is
[tex]P=VI=(3.6)(2.6)=9.36 W[/tex]
Then, this power is delivered for a time of
t = 350 s
Therefore, the energy supplied is
[tex]E=Pt=(9.36)(350)=3276 J[/tex]
Finally, the change in temperature of an object is related to the energy supplied by
[tex]E=C\Delta T[/tex]
where in this problem:
E = 3276 J is the energy supplied
C is the heat capacity of the object
[tex]\Delta T =29.1^{\circ}-20.3^{\circ}=8.8^{\circ}C[/tex] is the change in temperature
Solving for C, we find:
[tex]C=\frac{E}{\Delta T}=\frac{3276}{8.8}=372.3 J/^{\circ}C[/tex]
Final answer:
The heat capacity of the calorimeter, determined by applying a constant voltage and measuring the temperature change, is calculated to be 372.3 J/°C.
Explanation:
To calculate the heat capacity of the calorimeter, we first need to understand the amount of heat (q) added to the system. This can be determined using the formula q = IVt, where I is the current (2.6 A), V is the voltage (3.6 V), and t is the time (350 seconds). So, the heat added to the system is q = 3.6 V * 2.6 A * 350 s = 3276 J. The temperature change (ΔT) observed in the calorimeter is from 20.3°C to 29.1°C, which is a change of 8.8°C. The heat capacity (C) of the calorimeter can then be calculated as C = q/ΔT = 3276 J / 8.8 °C = 372.3 J/°C. This indicates the amount of heat required to raise the temperature of the calorimeter by one degree Celsius.