human beings can typically detect a difference in sound level of 2.0 dB. What is the ratio of the amplitudes of two sounds whose levels differ by this amount

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.

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.

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

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

Explanation:

Below is an attachment containing the solution.

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.

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.

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]

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.

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.

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

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).

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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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.

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

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

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.

Answer:

a) [tex]v=313.209\ m.s^{-1}[/tex]

b) [tex]v_t=3751.79\ m.s^{-1}[/tex]

Explanation:

Given:

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]

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.

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.

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]

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.

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.

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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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

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

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

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.

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.

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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Explanation:

Below is an attachment containing the solution

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:

Now, to give a more in-depth understanding:

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.