Answer:
Velocity of sound is 512 m/s
Explanation:
for first resonance condition we know that
[tex]\frac{\lambda}{4} = L[/tex]
[tex]\frac{\lambda}{4} = 0.40[/tex]
[tex]\lambda = 1.6 m[/tex]
now the frequency is given as
[tex]f = 320 Hz[/tex]
now to find the velocity of sound we have
[tex]v = \lambda f[/tex]
[tex]v = 1.6 (320)[/tex]
[tex]v = 512 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.
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.
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.
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 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.
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.
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]
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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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.
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
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 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
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.
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.The battery is now disconnected from the plates and the separation of the plates is doubled ( = 0.78 cm). What is the energy stored in this new capacitor?
Answer:
The energy stored in this new capacitor is [tex]4.4514\times10^{-9}\ J[/tex]
Explanation:
Suppose, Two parallel plates, each having area A = 2180 cm² are connected to the terminals of a battery of voltage [tex]V_{b}= 6\ V[/tex] as shown. The plates are separated by a distance d = 0.39 cm.
We need to calculate the charge
Using formula of capacitance
[tex]C=\dfrac{Q}{V}[/tex]
[tex]\dfrac{Q}{V}=\dfrac{\epsilon_{0}A}{d}[/tex]
[tex]Q=V\times\dfrac{\epsilon_{0}A}{d}[/tex]
Put the value into the formula
[tex]Q=6\times\dfrac{8.85\times10^{-12}\times2180\times10^{-4}}{0.39\times10^{-2}}[/tex]
[tex]Q=2.968\times10^{-9}\ C[/tex]
The distance between the plates is doubled.
We need to calculate the new capacitance
Using formula of capacitance
[tex]C'=\dfrac{\epsilon_{0}A}{d}[/tex]
Put the value into the formula
[tex]C'=\dfrac{8.85\times10^{-12}\times2180\times10^{-4}}{0.78\times10^{-2}}[/tex]
[tex]C'=2.473\times10^{-10}\ F[/tex]
We need to calculate the energy stored in this new capacitor
Using formula of energy
[tex]U=\dfrac{1}{2}C'V^2[/tex]
Put the value into the formula
[tex]U=\dfrac{1}{2}\times2.473\times10^{-10}\times(6)^2[/tex]
[tex]U=4.4514\times10^{-9}\ J[/tex]
Hence, The energy stored in this new capacitor is [tex]4.4514\times10^{-9}\ J[/tex]
Point sources of air pollution are __________. Question 9 options: all the hydrocarbons produced by trees in the Smoky Mountains specific spots--such as a factory's smokestacks--where large quantities of pollution are discharged the diffuse release of pollution from autos and homes into the atmosphere the release of pollution from many unidentifiable sources
Answer:
Smoky Mountains specific spots--such as a factory's smokestacks--where large quantities of pollution are discharged
Explanation:
Point source pollution is characterized by the following components:
It is a single sourceThe source is identifiableThe source is known to release pollutants into the environmentFrom the options, Smoky Mountains specific spots--such as a factory's smokestacks--where large quantities of pollution are discharged ticked all the necessary box for a point source pollution.
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
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
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.
______ 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.
Which of the following types of reactions would decrease the entropy within a cell?
digestion
hydrolysis
respiration
dehydration reactions
catabolism
Answer:
dehydration reactions
Final answer:
Dehydration reactions decrease entropy within a cell by building larger molecules from smaller units, unlike catabolic reactions like digestion and respiration which increase entropy by breaking down molecules.
Explanation:
The type of reaction that would decrease the entropy within a cell are dehydration reaction. Entropy is a measure of disorder or randomness in a system, and dehydration reactions are anabolic processes that build larger molecules from smaller ones, thus decreasing entropy. In contrast, catabolic reactions, such as digestion, hydrolysis, and respiration, generally increase entropy in a system by breaking down complex molecules into simpler ones and releasing energy in the process.
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.
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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
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]
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.
The octet rule states that in chemical compounds atoms tend to have
Answer:
In chemical compounds, atoms tends to have the electron configuration of a noble gas.
Explanation:
The noble gases are unreactive because of their electron configurations. This noble gas neon has the electron configuration of 1s22s22p6 . It has a full outer shell and cannot incorporate any more electrons into the valence shell.
The octet rule states that atoms tend to form compounds in ways that give them eight valence electrons and thus the electron configuration of a noble gas. An exception to an octet of electrons is in the case of the first noble gas, helium, which only has two valence electrons.
The octet rule states that atoms (excluding some exceptions like hydrogen and helium) in a compound strive for eight electrons in their valence shell, making it stable. This is achieved by sharing, accepting or donating electrons. For instance, oxygen in a water molecule gains two electrons from two hydrogen atoms through covalent bonding.
Explanation:The octet rule is a principle in chemistry which states that atoms in chemical compounds tend to achieve a stable electron configuration with eight electrons, a complete 'octet', in their valence shell. Atoms will donate, accept, or share electrons to fulfill this rule. For example, oxygen, which has six electrons in its valence shell, will react with other atoms so as to gain two more electrons, completing its octet. This is achieved through covalent bonding, where electrons are shared between atoms, such as in a water molecule H2O. It is important to note that there are exceptions to this rule, notably hydrogen and helium which are stable with two and helium with two electrons in their valence shell respectively.
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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.
Before using a string in a comparison, you can use either the To Upper method or the To Lower method to convert the string to upper case or lower case, respectively, and then use the converted string in the comparison.1. True2. False
Answer:
True
Explanation:
If there's no preference over the string case (upper case or lower case), one can convert both strings to upper case or to lowercase and then compare the converted strings to test if they're equal or not.
An Illustration is
string a = "Boy"
string b = 'bOy"
if(a.ToUpper() == b.ToUpper() || a.ToLower() == b.ToLower())
{
Print "Equal Strings"
}
else
{
Print "Strings are not equal";
}
The above will first convert both strings and then compare.
Since they are the same (after conversion), the statement "Equal Strings" will be printed, without the quotes
The statement is true. Most programming languages include methods to convert strings to either lower or upper case. This feature is useful to make comparisons case-insensitive.
Explanation:
This statement is true. In many programming languages, there are methods to convert a string to either lower case or upper case. These methods are often utilized to make string comparisons case-insensitive, eliminating any discrepancy caused by case differences.
For example, in Java, the methods are toLowerCase() and toUpperCase(), while in Python, they are lower() and upper(). This is particularly useful when a program is designed to interact with human input, as it allows the program to accept input regardless of the case used.
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We have seen that the heart produces a magnetic field, and that this can be used to diagnose problems with the heart. The magnetic field of the heart is a dipole field that is produced by a loop current in the outer layers of the heart. Suppose the field at the center of the heart is 90 pT (a pT is 10−12T ) and that the heart has a diameter of approximately 12 cm. What current circulates around the heart to produce this field?
Explanation:
The field at the center of circular current can be calculated by
B = [tex]\frac{\mu _0 I}{r}[/tex]
here μ₀ is permeability constant and is equal to 4π x 10⁻⁷
I is the current in circular path and r is the radius of circle
Thus I =[tex]\frac{90x10^-^1^2x12x10^-^2}{4\pi x 10^-^7 }[/tex] = 8.8 x 10⁻⁵ A
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