Answer:
The correct answer is d
Explanation:
In this exercise they ask us which statement is correct, for this we plan the solution of the problem, this is a Doppler effect problem, it is the frequency change due to the relative speed between the emitter and the receiver of sound.
The expression for the Doppler effect of a moving source is
f ’= (v / (v- + v_s) f
From this expression we see that if the speed the sound source is different from zero feels a change in the frequency.
The correct answer is d
A book sits on a bookshelf without moving until a student picks it up. Which law best explains why the book remains at rest until the student picks it up? (AKS 3a1 DOK 1) Newton's Second Law log Newton's Third Law Newton's Law of Universal Gravitation Newton's First Law
Answer:
Newtons first law
Explanation:
object in rest stays at rest
object in motion stays in motion
For each statement about the nuclear force select True or False. In nuclei, it is stronger than the Coulomb force. It acts between a proton and an electron. Its strength goes as 1/distance2, like gravity. It acts between a proton and a neutron. It acts between an electron and a neutron. It acts between two neutrons and also between two protons.
Answer:
a) True. The Strong interaction is the strong Coulomb force
e) True. It acts between protons, neutrons and between a proton and a neutron
Explanation:
In this exercise you must answer if there are some statements about the nuclei, let's write some of their characteristics
* at the core there is an attractive force called Strong Interaction
* This interaction acts on protons and neutrons alike
* There is a Coulombian-type repulsive force that acts between protons.
Let's review the claims
a) True. The Strong interaction is the strong Coulomb force
b) False. It acts on the elements of the nucleus and here there are no electrons
c) False. It is much greater than gravity and generally described by an exponential
d) False. It acts between components of the nucleus and here there are no electrons
e) True. It acts between protons, neutrons and between a proton and a neutron
A 50 kg laboratory worker is exposed to 30 mJ of neutron radiation with an RBE of 10. Part A What is the dose in mSv
Answer:
The dose is 6 mSV
Explanation:
The absorbed dose (in gray - Gy) is the amount of energy that ionizing radiation deposits per unit mass of tissue. That is,
Absorbed dose = Energy deposited / Mass
while Dose equivalent (DE) (in Seivert -Sv) is given by
DE = Absorbed dose × RBE (Relative biological effectiveness)
First, we will determine the Absorbed dose
From the question, Energy deposited = 30mJ and Mass = 50kg
From,
Absorbed dose = Energy deposited / Mass
Absorbed dose = 30mJ/50kg
Absorbed dose = 0.6 mGy
Now, for the Dose equivalent (DE)
DE = Absorbed dose × RBE
From the question, RBE = 10
Hence,
DE = 0.6mGy × 10
DE = 6 mSv
The dose will be 6 mSv. The absorbed dose (in gray - Gy) is the amount of energy that ionizing radiation deposits per unit mass of tissue.
What is the dose?A unit is a measurable quantity of a drug, vitamin, or pathogen provided in a single dose. A medication's dosage form is a blend of active and inactive ingredients used to give it.
The given devotion is;
[tex]D_a=[/tex]Absorbed dose
[tex]\rm E_D[/tex] = energy deposited
[tex]\rm D_E[/tex] is the dose equivalent
RBE= Relative biological effectiveness
The absorbed dose is found as;
[tex]\rm D_a = \frac{E_d}{m}\\\\ D_a = \frac{30 \ mJ}{50}\\\\ \rm D_a = 0.6 \ mGy[/tex]
The dose equivalent is found as;
[tex]\rm D_E=D_A \times RBE \\\\ D_E=0.6 \times 10 \\\\ \rm D_E=6 mSV[/tex]
Hence the dose will be 6 mSv.
To learn more about the dose refer to the link;
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Waves on a String: The speed of sound in steel is 5000 m/s. What is the wavelength of a sound wave of frequency 660 Hz in steel
Answer:
The wavelength of the sound wave is 7.58 m
Explanation:
Given;
speed of wave in steel, v = 5000 m/s
frequency of the wave in steel, f = 660 Hz
The wavelength of a sound wave is given by;
λ = v / f
where;
v is the speed of sound wave in steel
f is the frequency of sound wave in steel
λ is the wavelength of the sound wave in steel
λ = 5000 / 660
λ = 7.58 m
Therefore, the wavelength of the sound wave is 7.58 m
Describe an experiment to show that the pressure in a liquid increases with depth.
Answer:
Explanation:
Water pressure is the result of the weight of all the water above pushing down on the water below. As you go deeper into a body of water, there is more water above, and therefore a greater weight pushing down. This is the reason water pressure increases with depth.
A radio wave transmits 34.5 W/m2 of power per unit area. A flat surface of area A is perpendicular to the direction of propagation of the wave. Assuming the surface is a perfect absorber, calculate the radiation pressure on it.
Answer:
1.15*10^-7 N/m²
Explanation:
Radiation pressure is the pressure exerted on any surface, as a result of the exchange of momentum between the object and its electromagnetic field.
The formula to calculate radiation pressure on a perfect absorber is
P = s/c, where
P = radiation pressure
s = intensity of light
c = speed of light
Now, on substituting the values and plugging it into the equation, we have
P = 34.5 / 3*10^8
P = 1.15*10^-7 N/m²
therefore, radiation pressure is found to be 1.15*10^-7 N/m²
Select all the statements that are true about the dependent variable.
a. It is also the "then" part of your hypothesis.
b. It's sometimes called the "responding variable"
c. It always stays the same.
d. It's often abbreviated as "DV"
Answer:
Hey there!
These are true:
a. It is also the "then" part of your hypothesis.
b. It's sometimes called the "responding variable"
d. It's often abbreviated as "DV"
Let me know if this helps :)
Assume that Earth is in circular orbit around the Sun with kinetic energy K and potential energy U, taken to be zero for infinite separation. Then, the relationship between K and U:
Answer:
K= -U/2
Explanation:
We know that K.E = 1/2mv²
And -GMm/r²= mv²/r
Thus K,E= 1/2(GMm/r)= 1/2(-P.E)
So
P.E = - GMm/r
So K= -U/2
Determine the centroid of the shaded area shown in figure 2. Determine the moment of inertia about y-axis of the shaded area shown in figure 2
Answer:
centroid: (x, y) = (81.25 mm, 137.5 mm)I = 8719.31 mm^2 for unit massExplanation:
Finding the desired measures requires we know a differential of area. That, in turn, requires we have a way to describe a differential of area. Here, we choose to use a vertical slice, which requires we know the area boundaries as a function of x.
The upper boundary is a line with a slope of 125/156.25 = 0.8, and a y-intercept of 125. That is, ...
y1 = 0.8x +125
The lower boundary is given in terms of y, but we can solve for y to find ...
100x = y^2
y2 = 10√x
Then our differential of area is ...
dA = (y1 -y2)dx
__
The centroid is found by computing the first moment about the x- and y-axes, and dividing those values by the area of the figure.
The area will be ...
[tex]\displaystyle A=\int_0^{156.25}{dA}=\int_0^{156.25}{(y_1-y_2)}\,dx[/tex]
The y-coordinate of the centroid is ...
[tex]\displaystyle \overline{y}=\dfrac{S_x}{A}=\dfrac{1}{A}\int_0^{156.25}{\dfrac{y_1+y_2}{2}}\,dA=\dfrac{1}{A}\int_0^{156.25}{\dfrac{y_1+y_2}{2}(y_1-y_2)}\,dx=137.5[/tex]
Similarly, the x-coordinate is ...
[tex]\displaystyle \overline{x}=\dfrac{S_y}{A}=\dfrac{1}{A}\int_0^{156.25}{x}\,dA=\dfrac{1}{A}\int_0^{156.25}{x(y_1-y_2)}\,dx=81.25[/tex]
That is, centroid coordinates are (x, y) = (81.25, 137.5) mm.
__
The moment of inertia is the second moment of the area. If we normalize by the "mass" (area), then the integral looks a lot like the one for [tex]\overline{x}[/tex], but multiplies dA by x^2 instead of x.
The attachment shows that value to be ...
I ≈ 8719.31 mm^2 (normalized by area)
The area is 16276.0416667 mm^2, if you want to "un-normalize" the moment of inertia.
Which contains the most moles: 10 g of hydrogen gas, 100 g of carbon, or 50 g of lead?
Answer:
Carbon
Explanation:
We know that
Number of moles = mass/molecular mass
So
Hydrogen: given as H2 with 10g an atomic number of 1 so MW( 1x 2= 2)
Number of moles = Weight/MW
= 10/2
= 5
Carbon: atomic number 12 C and mass 100
Number of moles = Weight/MW
= 100/12
= 8.33
Lead: atomic num 207 and mW 207 pb
Number of moles = Weight/MW
= 50/207
= 0.241
Therefore Carbon has the most moles.
A car traveling at 10 m/s passes over a hill on a road that has a circular cross section of radius 30 m. What is the force exerted by the seat of the car on a 60-kg passenger when the car is passing the top of the hill?
Answer:
F= 389N
Explanation:
Using
Mg-N=mv²/ r
So N= mg- mv²/r
60*9.8- 60*10²/30
=389N
A postal service worker lifts a package into a truck. She exerts 22 J of work lifting the package for 4 seconds. How much power did she user
Calculate the power .
Formula used:-Power = Work done/Time taken
Solution:-We know that,
[tex] \sf{Power = \dfrac{Work \: done}{Time \: taken } }[/tex]
★ Putting the values in the above formula,we get:
[tex] \sf{Power = \dfrac{22}{4} }[/tex]
[tex] \sf{Power = 5.5 \: watts}[/tex]
a telescope is oriting on a spacecraft aroudn the earth. Speed of 3.25 x 10^5 mass 7500 kg. What is the de Broglie wavelength of the telescope. g
Answer:
2.72 x 10^-43 m
Explanation:
mass of the telescope = 7500 kg
speed of the telescope = 3.25 x 10^5 m/s
de Broglie's wavelength of the telescope is given as
λ = h/mv
where
λ is the wavelength of the telescope
h is the plank's constant = 6.63 × 10-34 m^2 kg/s
m is the mass of the telescope = 7500 kg
v is speed of the telescope = 3.25 x 10^5 m/s
substituting value, we have
λ = (6.63 × 10-34)/(7500 x 3.25 x 10^5)
λ = 2.72 x 10^-43 m
Observe: Up until now, all the problems you have solved have involved converting only one unit. However, some conversion problems require you to convert two or more units. Select Speed from the menu. What two units do you need to convert to solve this problem? meter and seconds Think about it: How do you think you can use conversion factors to solve this problem?
Answer:
t is appropriate to clarify that units such as time and angles the transformation is not in base ten, for example:
60 s = 1 min
60 min = 1 h
24 h = 1 day
Therefore, for this transformation, you must be more careful
the length transformation is base 10
Explanation:
In many exercises the units used are transformed by equations into other units called derivatives, in general the transformation of derived units is the product of the transformation of the constituent units.
In the example of velocity, the derivative unit is m / s, which is why it works in the same way that you transform length and time if in the equation it is multiplying it is multiplied and if it is dividing it is divided.
It is appropriate to clarify that units such as time and angles the transformation is not in base ten, for example:
60 s = 1 min
60 min = 1 h
24 h = 1 day
Therefore, for this transformation, you must be more careful
the length transformation is base 10
1000 m = 1 km
hi guys this is my last points and i need some answers :)
1. what is a runoff?
A. A type of precipitation that falls as a mixture of ice and snow B.The water on the surface that soaks deep into the ground C.Tiny droplets of water that condense to form clouds D.Water that moves along the surface of Earth into the oceans
NEXT QUESTION
2.Which statement about oceans is incorrect?
A.Evaporation occurs when water is warmed by the sun.
B.Most evaporation and precipitation occur over the ocean.
C.97 percent of Earth's water is fresh water from the ocean.
D.Water leaves the ocean by the process of evaporation.
Answer:
Number 1: B Number 2: D
Explanation:
Answer:
Hello! Your answer would be D) for one and C) for two.
Explanation:
Run off is when water runs of of things into lakes oceans etc.
So with number two it would be C because not even 3% of our water is fresh.
You are performing a double slit experiment very similar to the one from DL by shining a laser on two nattow slits spaced 7.5 x 103 meters apart. However, by placing a piece of crystal in one of the slits, you are able to make it so that the rays of light that travel through the two slits are Ï out of phase with each other (that is to say, Ao,- ). If you observe that on a screen placed 4 meters from the two slits that the distance between the bright spot clos center of the pattern is 1.5 cm, what is the wavelength of the laser?
Complete Question
You are performing a double slit experiment very similar to the one from DL by shining a laser on two nattow slits spaced [tex]7.5 * 10^{-3}[/tex] meters apart. However, by placing a piece of crystal in one of the slits, you are able to make it so that the rays of light that travel through the two slits are Ï out of phase with each other (that is to say, Ao,- ). If you observe that on a screen placed 4 meters from the two slits that the distance between the bright spot closest to center of the pattern is 1.5 cm, what is the wavelength of the laser?
Answer:
The wavelength is [tex]\lambda = 56250 nm[/tex]
Explanation:
From the question we are told that
The distance of slit separation is [tex]d = 7.5 *10^{-3} \ m[/tex]
The distance of the screen is [tex]D = 4 \ m[/tex]
The distance between the bright spot closest to the center of the interference is [tex]k = 1.5 \ cm = 0.015 \ m[/tex]
Generally the width of the central maximum fringe produced is mathematically represented as
[tex]y = 2 * k = \frac{ D * \lambda}{d}[/tex]
=> [tex]2 * 0.015 = \frac{ \lambda * 4}{ 7.5 *10^{-3}}[/tex]
=> [tex]\lambda = 56250 *10^{-9} \ m[/tex]
=> [tex]\lambda = 56250 nm[/tex]
If you have two substances, one with a density of 2.0 g/cm3 and one with a density of 1.3 g/cm3 and you combined them, which one would float on top
other and why?
Explanation:
Assuming the substances are fluids that do not mix, the lighter substance (ρ = 1.3 g/cm³) will float on top of the heavier substance (ρ = 2.0 g/cm³). This is due to Archimedes' Principle, which explains buoyancy.
A spaceship far from all other objects uses its impulse power system to attain a speed of 104 m/s. The crew then shuts off the power. According to Newton's first law, what will happen to the motion of the spaceship from then on
Answer:
The spaceship will travel in a straight line at constant speed.
Explanation:
According to the Newton's First Law, the spaceship will have an uniform motion in a straight line, that is, the spaceship will travel in a straight line at constant speed.
The spaceship will travel in a straight line at constant speed.
Newton's first law states that if a body is at rest or moving at a constant speed in a straight line, it will remain at rest or keep moving in a straight line at constant speed unless it is acted upon by a force.By applying Newton's First Law, when the spaceship attain a speed of 104 m/s . Then, spaceship will have an uniform motion in a straight line.
Hence, the spaceship will travel in a straight line at constant speed.
Learn more:
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What is the de Broglie wavelength of an electron that strikes the back of the face of a TV screen at 1 15 the speed of light
Answer:
Te wavelength is 3.66 x 10^-11 m
Explanation:
Question is
What is the de Broglie wavelength of an electron that strikes the back of the face of a TV screen at 1/15 the speed of light.
De Broglie's wavelength is gotten as
λ = h/mv
where
λ is the wavelength of the electron
h is the Planck's constant = 6.67 x 10^-34 J-s
m is the mass of an an electron = 9.109 x 10^-31 kg
v is the speed of the electron.
The speed of the electron is given as 1/15 of the speed of light
Speed of electron = 1/15 of 3 x 10^8 = 2.0 x 10^7 m/s
substituting values, we have
λ = (6.67 x 10^-34)/(9.109 x 10^-31)x(2.0 x 10^7) = 3.66 x 10^-11 m
Which of the following statements about electromagnetic radiation is true?
A) It always travels at 3Ã108 m/s.
B) It consists of perpendicularly oscillating electric and magnetic fields, and the direction of propagation is perpendicular to both.
C) It consists of oscillating electric and magnetic fields that are parallel to one another and perpendicular to the direction of propagation.
D) The oscillating electric and magnetic fields are completely out of phase with one another
Answer:
The correct option is B: It consists of perpendicularly oscillating electric and magnetic fields, and the direction of propagation is perpendicular to both.
Explanation:
Electromagnetic radiation travels at the speed of light (c = 3x10⁸ m/s) only in a vacuum. In another propagation medium, the speed of electromagnetic radiation is less than c.
The oscillations of the waves of the electric and magnetic fields are perpendicular to each other, and the direction of propagation is also perpendicular to the two fields.
From all of the above, the correct option is B: It consists of perpendicularly oscillating electric and magnetic fields, and the direction of propagation is perpendicular to both.
I hope it helps you!
The true statement regarding electromagnetic radiation should be option B where it comprises of perpendicularly oscillating electric and magnetic fields.
What is electromagnetic radiation?It should be traveled when the speed of light i.e. (c = 3x10⁸ m/s) should be a vacuum. For the propagation medium, the speed with respect to the electromagnetic radiation should be lower than c. Also, the oscillations of the electric waves and the magnetic field should be perpendicular also the direction should be perpendicular
Hence, the option b is correct.
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A soft drink (mostly water) flows in a pipe at a beverage plant with a mass flow rate that would fill 220 0.355 - L cans per minute. At point 2 in the pipe, the gauge pressure is 152kPa and the cross-sectional area is 8.00cm2. At point 1, 1.35m above point 2, the cross-sectional area is 2.00cm2.
a. Find the mass flow rate.
b. Find the volume flow rate.
c. Find the flow speed at point 1.
d. Find the flow speed at point 2.
e. Find the gauge pressure at point
Answer:
A. To find the mass flow rate.
We use= 220 x 0.355/ 60
= 1.3kg/s
B. Volume flowrate is = mass flowrate / density
But density is 1000kg/m³
= 1.3kg/s/ 1000kg/m³
= 0.0013m³/s
C. Flow speead at 1
= 0.0013m³/s / (2 x 10-2m)²
= 6.5m/s
D.flow speed at 2
0.0013m³/s / (8x 10-2m)²
=1.63m/s
E. Gauge pressure at point 1
= 152+ 1/1000 ( 1.63)²- 6.5² + 1000( 9.8) ( 0-1.35)
= 119kpa
A force of 200 N is applied on small piston of a pascal press. What would be the
force applied on the big piston, if the diameter of the small piston is 4.37 cm and the area of the big piston is 98 cm2?
Answer:
The force applied on the big piston is 1306.67 N
Explanation:
Given;
force applied on small piston, F₁ = 200 N
diameter of the small piston, d₁ = 4.37 cm
radius of the small piston, r₁ = d₁/2 = 2.185 cm
Area of the small piston, A₁ = πr₁² = π(2.185 cm)² = 15 cm²
Area of the big piston, A₂ = 98 cm²
The pressure of the piston is given by;
[tex]P = \frac{F}{A} \\\\\frac{F_1}{A_1} = \frac{F_2}{A_2}\\\\ F_2 = \frac{F_1A_2}{A_1}[/tex]
Where;
F₂ is the force on big piston
[tex]F_2 = \frac{200*98}{15} \\\\F_2 = 1306.67 \ N[/tex]
Therefore, the force applied on the big piston is 1306.67 N
If the accuracy in measuring the position of a particle increases, the accuracy in measuring its velocity will
Answer:
Therefore the answer is the precision in the speed DECREASES
Explanation:
In quantum mechanics, we have the uncertainty principle that establishes that when the accuracy of the position increases the accuracy the speed decreases, being related by the expression
Δx Δv ≥ h'/ 2
h' = h/2π
Therefore the answer is the precision in the speed DECREASES
In a compound microscope, the objective has a focal length of 1.0 cm, the eyepiece has a focal length of 2.0 cm, and the tube length is 25 cm. What is the magnitude of the overall magnification of the microscope?
Answer:
The value is [tex]m \approx 310[/tex]
Explanation:
From the question we are told that
The focal length of the objective is [tex]f_o = 1.0 \ cm[/tex]
The focal length of the eyepiece is [tex]f_e = 2.0 \ cm[/tex]
The tube length is [tex]L = 25 \ cm[/tex]
Generally the magnitude of the overall magnification is mathematically represented as
[tex]m = m_o * m_e[/tex]
Where [tex]m_o[/tex] is the objective magnification which is mathematically represented as
[tex]m_o = \frac{L}{f_o }[/tex]
=> [tex]m_o = \frac{25}{1 }[/tex]
=> [tex]m_o = 25[/tex]
[tex]m_e[/tex] is the eyepiece magnification which is mathematically evaluated as
[tex]m_e = \frac{L }{f_e }[/tex]
[tex]m_e = \frac{25 }{ 2}[/tex]
[tex]m_e = 12.5 \ cm[/tex]
So
[tex]m = 25 * 12.5[/tex]
[tex]m \approx 310[/tex]
88. A child pulls a cart to the left along a rough surface. Which one of the free body diagrams correctly depicts all the forces acting on the carrito?
Answer:
Eje x
fr- F = m a
eje y
N-W =0
Explanation:
Para este ejercicio, debemos usar la segunda ley de Newton, donde se necesita fija un sistema de referencia el mas usado es un sistema horizontal y vertical para los ejes x e y
en el adjunto puede ver el diagrama de cuerpo libre correcto,
las fuerza son
Eje x
fr- F = m a
eje y
N-W =0
la ecuacion para la fuerza de roce es
fr = my N
Multiply the number 4.48E-8 by 5.2E-4 using Google. What is the correct answer in scientific notation?
Answer:
[tex]2.32\times 10^{-11}[/tex]
Explanation:
First number is [tex]4.48\times 10^{-8}[/tex]
Second number is [tex]5.2\times 10^{-4}[/tex]
We need to multiply the two numbers.
[tex]4.48\times 10^{-8}\times 5.2\times 10^{-4}=(4.48\times 5.2)\times 10^{(-8-4)}\\\\=23.296\times 10^{-12}[/tex]
In scientific notation : [tex]2.32\times 10^{-11}[/tex]
Hence, this is the required solution.
blending three primary colors of light
Answer: white light
Explanation:
if you mix red and blue light together, you get magenta and when you mix blue and green light together, you get cyan and when you mix red and green light together, you get yellow but when you mix all three primary colors of light together, you end up with white light or as we see it.
An air track glider is attached to a spring. The glider is pulled to the right and released from rest at t=0 seconds. It then oscillates with a period of 2.40s and a maximum speed of 30.0 cm/s. What is the glider's position at t=0.300s?
Answer:
The glider's position at t=0.300s is 8.1 cm
Explanation:
Given;
maximum speed of the glider, [tex]v_{max}[/tex] = 30.0 cm/s = 0.3 m/s
period of the oscillation, T = 2.4 s
Maximum speed of oscillation is given by;
[tex]v_{max}[/tex] = ωA
Where;
A is the amplitude of the oscillation
ω is the angular speed
[tex]\omega = \frac{2\pi}{T} \\\\\omega =\frac{2\pi}{2.4}\\\\ \omega = 0.833 \pi[/tex]
The amplitude of the oscillation is given by;
[tex]A = \frac{v_{max}}{\omega} \\\\ A = \frac{0.3}{0.833\pi}\\\\ A = 0.1146 \ m[/tex]
The position of the particle is given by;
x(t) = A cosωt
x(0.3) = (0.1146)cos(0.833π x 0.3)
x(0.3) = (0.1146)cos(0.833π x 0.3)
x(0.3) = (0.1146)cos(45)
x(0.3) = 0.081 m
x(0.3) = 8.1 cm
Therefore, the glider's position at t=0.300s is 8.1 cm
If a single circular loop of wire carries a current of 61 A and produces a magnetic field at its center with a magnitude of 1.70 10-4 T, determine the radius of the loop.
Answer:
The radius is [tex]R = 0.22 5 \ m[/tex]
Explanation:
From the question we are told that
The current is [tex]I = 61 \ A[/tex]
The magnetic field is [tex]B = 1.70 *10^{-4} \ T[/tex]
Generally the magnetic field produced by a current carrying conductor is mathematically represented as
[tex]B = \frac{\mu_o * I}{2 * R }[/tex]
=> [tex]R = \frac{\mu_o * I }{ 2 * B }[/tex]
Here [tex]\mu_o[/tex] is the permeability of free space with value [tex]\mu_o = 4\pi * 10^{-7} N/A^2[/tex]
=> [tex]R = \frac{ 4\pi * 10^{-7} * 61 }{ 2 * 1.70 *10^{-4} }[/tex]
=> [tex]R = 0.22 5 \ m[/tex]
A bowling ball traveling with constant speed hits the pins at the end of a bowling lane 16.5 m long. The bowler hears the sound of the ball hitting the pins 2.80 s after the ball is released from his hands. What is the speed of the ball, assuming the speed of sound is 340 m????s?
Answer:
[tex]5.997m/s[/tex]
Explanation:
We were told to calculate the speed of the ball,
Given speed of sound as 340 m
And we know that the sound of the ball hitting the pins is at 2.80 s after the ball is released from his hands.
Speed of ball = distance traveled/(time of hearing - time the sound travels).
Speed= S/t
Where S= distance traveled
t= time of hearing - time the sound travels
time=time for ball to roll+timefor sound to come back.
time of sound=16.5/340
=0.048529secs
solving for speedof ball
Then,Speed of ball = distance traveled/(time of hearing - time the sound travels).
=16.5/(2.80-0.048529) m/s = 5.997m/s
Therefore, the speed of the ball is
5.997m/s