A light source radiates 60.0 W of single-wavelength sinusoidal light uniformly in all directions. What is the average intensity of the light from this bulb at a distance of 0.400 m from the bulb

Answers

Answer 1

Answer: [tex]29.85\ W/m^2[/tex]

Explanation:

Given

Power [tex]P=60\ W[/tex]

Distance from the light source [tex]r=0.4\ m[/tex]

Intensity is given by

[tex]I=\dfrac{P}{4\pi r^2}[/tex]

Inserting values

[tex]\Rightarrow I=\dfrac{60}{4\pi (0.4)^2}\\\\\Rightarrow I=\dfrac{60}{2.010}\\\\\Rightarrow I=29.85\ W/m^2[/tex]

Answer 2

Answer:

29.85 W/ m^2

Explanation:

A Light Source Radiates 60.0 W Of Single-wavelength Sinusoidal Light Uniformly In All Directions. What

Related Questions

A double-slit experiment is performed with light of wavelength 550 nm. The bright interference fringes are spaced 2.3 mm apart on the viewing screen. What will the fringe spacing be if the light is changed to a wavelength of 360 nm?

Answers

Answer:

[tex]d_2=1.5*10^-3m[/tex]

Explanation:

From the question we are told that:

Initial Wavelength [tex]\lambda_1=550nm=550*10^{-9}[/tex]

Space 1 [tex]d_1=2.3*10^{-3}[/tex]

Final wavelength [tex]\lambda_2=360*10^{-9}[/tex]

Generally the equation for Fringe space at [tex]\lambda _2[/tex] is mathematically given by

 [tex]d_2=\frac{d_1}{\lambdaI_1}*\lambda_2[/tex]

 [tex]d_2=\frac{2.3*10^{-3}}{550*10^{-9}}*360*10^{-9}[/tex]

 [tex]d_2=1.5*10^-3m[/tex]

1. On each of your equipotential maps, draw some electric field lines with arrow heads indicating the direction of the field. (Hint: At what angle do field lines intersect equipotential lines?) Draw sufficient field lines that you can "see" the electric field.

Answers

Answer:

The angle between the electric field lines and the equipotential surface is 90 degree.

Explanation:

The equipotential surfaces are the surface on which the electric potential is same. The work done in moving a charge from one point to another on an equipotential surface is always zero.

The electric field lines are always perpendicular to the equipotential surface.

As

[tex]dV = \overrightarrow{E} . d\overrightarrow{r}\\\\[/tex]

For equipotential surface, dV = 0 so

[tex]0 = \overrightarrow{E} . d\overrightarrow{r}\\\\[/tex]

The dot product of two non zero vectors is zero, if they are perpendicular to each other.

Physical quantities expresed only by their magnitude is​

Answers

Answer:

Scalar quantity is the Physical quantity expresed only by their magnitude.

The earth's radius is about 4000 miles. Kampala, the capital of Uganda, and Singapore are both nearly on the equator. The distance between them is 5000 miles as measured along the earth's surface.
a. Through what angle do you turn, relative to the earth, if you fly from Kampala to Singapore? Give your answer in both radians and degrees.
b. The flight from Kampala to Singapore take 9 hours. What is the plane's angular speed relative to the earth?

Answers

Answer:

a) the required angle in both radian and degree is  1.25 rad and 71.6°

b) the plane's angular speed relative to the earth is 3.86 × 10⁻⁵ rad/sec

Explanation:

Given the data in the question;

a)

we know that The expression for the angle subtended by an arc of circle at the center of the circle is,

θ = Length / radius

given that Length is 5000 miles and radius is 4000 miles

we substitute

θ = 5000 miles / 4000 miles

θ = 1.25 rad

Radian to Degree

θ = 1.25 rad × ( 180° / π rad )

θ =  71.6°

Therefore, required angle in both radian and degree is  1.25 rad and 71.6°

b)

The flight from Kampala to Singapore take 9 hours.

the plane's angular speed relative to the earth = ?

we know that, the relation between angular velocity and angular displacement is;

ω = θ / t

given that θ is 1.25 rads and time t is 9 hours or ( 9 × 3600 sec ) = 32400 sec

we substitute

ω = 1.25 rad / 32400 sec

ω = 3.86 × 10⁻⁵ rad/sec

Therefore, the plane's angular speed relative to the earth is 3.86 × 10⁻⁵ rad/sec

Suppose a power plant uses a Carnot engine to generate electricity, using the atmosphere at 300 K as the low-temperature reservoir. Suppose the power plant produces an amount of electric energy with the hot reservoir at 500 K during Day One and then produces the same amount of electric energy with the hot reservoir at 600 K during Day Two. The thermal pollution was:

Answers

Answer: hello your question lacks some vital information below is the complete question

Suppose a power plant uses a Carnot engine to generate electricity, using the atmosphere at 300 K as the low-temperature reservoir. Suppose the power plant produces 1 × 106 J of electricity with the hot reservoir at 500 K during Day One and then produces 1 × 106 J of electricity with the hot reservoir at 600 K during Day Two. The thermal pollution was

answer:

Total thermal pollution = 2.5 * 10^6 J

Explanation:

Low temperature reservoir = 300 K

hot reservoir temperature = 500 K

Electrical energy produced by plant ( W ) = 1 * 10^6 J

lets assume ; Q1 = energy absorbed , Q2 = energy emitted

W = Q1 - Q2  or  Q2 = Q1 - W  ( we will apply this as the formula for determining thermal pollution )

For day 1

T1 = 500k , T2 = 300k

applying Carnot engine formula

W / Q1 = 1 - T2/T1

∴ Q1 = 10^6 / ( 1 - (300/500)) = 2.5 * 10^6 J

thermal pollution ; Q2 = Q1 - W = ( 2.5 * 10^6 - 1 * 10^6 ) = 1.5 * 10^6 J

for Day 2

T1 = 600k,  T2 = 300k

Q1 = 10^6 / ( 1 - (300/600)) = 2 * 10^6 J

Thermal pollution; Q2 = Q1 - W  = 1 * 10^6 J

Therefore the Total thermal pollution =  1 * 10^6  + 1.5 * 10^6  = 2.5 * 10^6 J

A uniformly charged thin rod of length L and positive charge Q lies along the x-axis with its left end at the origin as shown in Figure 1.

a. Set up a correct integral expression for the potential at point A,which lies a distance H above the right end of the rod. Point A has coordinates (L, H). You need to give appropriate limits of integration and expressions for r and dq

b. Set up a correct integral expression for the potential at point B on the x-axis, a distance D from the left end of the rod with the appropriate limits of integration.You need to give appropriate limits of integration and expressions for r and dq.

Answers

Answer:

b)

Explanation:

What is the effect on range and maximum height of a projectile as the launch height, launch speed, and launch angle are increased?

Answers

Answer:

The highest point in the trajectory occurs at the midpoint of the path. This highest point increases as the angle increases. At a 75° launch angle, the maximum height is approximately 76 meters. However, a further increase in launch angle beyond this 75° angle will increase the peak height even more.

1. What is the total distance the car moves until it stops?
a. 250 m
b. 450 m
c. 300 m
d. 600 m.​

Answers

B
Just took the quiz bro it was easy

The cannon on a battleship can fire a shell a maximum distance of 33.0 km.
(a) Calculate the initial velocity of the shell.

Answers

Answer:

v = 804.23 m/s

Explanation:

Given that,

The maximum distance covered by a cannon, d = 33 km = 33000 m

We need to find the initial velocity of the shell. Let it is v. It can be calculated using the conservation of energy such that,

[tex]v=\sqrt{2gh} \\\\v=\sqrt{2\times 9.8\times 33000} \\\\v=804.23\ m/s[/tex]

So, the initial velocity of the shell is 804.23 m/s.

A 1500 kg car traveling at 20 m/s suddenly runs out of gas while approaching the valley shown in the figure. The alert driver immediately puts the car in neutral so that it will roll.

What will be the car’s speed as it coasts into the gas station on the other side of the valley?

Answers

Answer:

v_f = 17.4 m / s

Explanation:

For this exercise we can use conservation of energy

starting point. On the hill when running out of gas

          Em₀ = K + U = ½ m v₀² + m g y₁

final point. Arriving at the gas station

         Em_f = K + U = ½ m v_f ² + m g y₂

energy is conserved

         Em₀ = Em_f

         ½ m v₀ ² + m g y₁ = ½ m v_f ² + m g y₂

        v_f ² = v₀² + 2g (y₁ -y₂)

         

we calculate

        v_f ² = 20² + 2 9.8  (10 -15)

        v_f = √302

         v_f = 17.4 m / s

Calculate the change in length of a 90.5 mm aluminum bar that has increased in temperature by from -14.4 oC to 154.6 oC
Take the coefficient of expansion to be 25 x 10-6 (oC)-1 . Write the answer in meters with three significant figures

Answers

Answer:

 ΔL = 3.82 10⁻⁴ m

Explanation:

This is a thermal expansion exercise

          ΔL = α L₀ ΔT

          ΔT = T_f - T₀

where ΔL is the change in length and ΔT is the change in temperature

Let's reduce the length to SI units

          L₀ = 90.5 mm (1m / 1000 mm) = 0.0905 m

let's calculate

          ΔL = 25.10⁻⁶ 0.0905 (154.6 - (14.4))

          ΔL = 3.8236 10⁻⁴ m

     

using the criterion of three significant figures

          ΔL = 3.82 10⁻⁴ m

When air expands adiabatically (without gaining or losing heat), its pressure P and volume V are related by the equation PV1.4=C where C is a constant. Suppose that at a certain instant the volume is 420 cubic centimeters and the pressure is 99 kPa and is decreasing at a rate of 7 kPa/minute. At what rate in cubic centimeters per minute is the volume increasing at this instant?

Answers

Answer:

[tex]\frac{dV}{dt}=21.21cm^3/min[/tex]

Explanation:

We are given that

[tex]PV^{1.4}=C[/tex]

Where C=Constant

[tex]\frac{dP}{dt}=-7KPa/minute[/tex]

V=420 cubic cm and P=99KPa

We have to find the rate at which the  volume increasing at this instant.

Differentiate w.r.t t

[tex]V^{1.4}\frac{dP}{dt}+1.4V^{0.4}P\frac{dV}{dt}=0[/tex]

Substitute the values

[tex](420)^{1.4}\times (-7)+1.4(420)^{0.4}(99)\frac{dV}{dt}=0[/tex]

[tex]1.4(420)^{0.4}(99)\frac{dV}{dt}=(420)^{1.4}\times (7)[/tex]

[tex]\frac{dV}{dt}=\frac{(420)^{1.4}\times (7)}{1.4(420)^{0.4}(99)}[/tex]

[tex]\frac{dV}{dt}=21.21cm^3/min[/tex]

Answer:

[tex]\dot V=2786.52~cm^3/min[/tex]

Explanation:

Given:

initial pressure during adiabatic expansion of air, [tex]P_1=99~kPa[/tex]

initial volume during the process, [tex]V_1=420~cm^3[/tex]

The adiabatic process is governed by the relation [tex]PV^{1.4}=C[/tex] ; where C is a constant.

Rate of decrease in pressure, [tex]\dot P=7~kPa/min[/tex]

Then the rate of change in volume, [tex]\dot V[/tex] can be determined as:

[tex]P_1.V_1^{1.4}=\dot P.\dot V^{1.4}[/tex]

[tex]99\times 420^{1.4}=7\times V^{1.4}[/tex]

[tex]\dot V=2786.52~cm^3/min[/tex]

[tex]\because P\propto\frac{1}{V}[/tex]

[tex]\therefore[/tex] The rate of change in volume will be increasing.

Can you think of reasons why the charge on each ball decreases over time and where the charges might go

Answers

Answer:

By the principle of corona discharge.

Explanation:

The charge on each ball will decreases over time due to the electrical discharge in air.

According to the principle of corona discharge, when the curvature is small, the discharge of the charge takes placed form the pointed ends.

The working substance of a certain Carnot engine is 1.90 of an ideal
monatomic gas. During the isothermal expansion portion of this engine's
cycle, the volume of the gas doubles, while during the adiabatic expansion
the volume increases by a factor of 5.7. The work output of the engine is
930 in each cycle.
Compute the temperatures of the two reservoirs between which this engine
operates.

Answers

Answer:

Explanation:

The energy for an isothermal expansion can be computed as:

[tex]\mathsf{Q_H =nRTIn (\dfrac{V_b}{V_a})}[/tex] --- (1)

However, we are being told that the volume of the gas is twice itself when undergoing adiabatic expansion. This implies that:

[tex]V_b = 2V_a[/tex]

Equation (1) can be written as:

[tex]\mathtt{Q_H = nRT_H In (2)}[/tex]

Also, in a Carnot engine, the efficiency can be computed as:

[tex]\mathtt{e = 1 - \dfrac{T_L}{T_H}}[/tex]

[tex]e = \dfrac{T_H-T_L}{T_H}[/tex]

In addition to that, for any heat engine, the efficiency e =[tex]\dfrac{W}{Q_H}[/tex]

relating the above two equations together, we have:

[tex]\dfrac{T_H-T_L}{T_H} = \dfrac{W}{Q_H}[/tex]

Making the work done (W) the subject:

[tex]W = Q_H \Big(\dfrac{T_H-T_L}{T_H} \Big)[/tex]

From equation (1):

[tex]\mathsf{W = nRT_HIn(2) \Big(\dfrac{T_H-T_L}{T_H} \Big)}[/tex]

[tex]\mathsf{W = nRIn(2) \Big(T_H-T_L} \Big)}[/tex]

If we consider the adiabatic expansion as well:

[tex]PV^y[/tex] = constant

i.e.

[tex]P_bV_b^y = P_cV_c^y[/tex]

From ideal gas PV = nRT

we can have:

[tex]\dfrac{nRT_H}{V_b}(V_b^y)= \dfrac{nRT_L}{V_c}(V_c^y)[/tex]

[tex]T_H = T_L \Big(\dfrac{V_c}{V_b}\Big)^{y-1}[/tex]

From the question, let us recall  aw we are being informed that:

If the volumes changes by a factor = 5.7

Then, it implies that:

[tex]\Big(\dfrac{V_c}{V_b}\Big) = 5.7[/tex]

[tex]T_H = T_L (5.7)^{y-1}[/tex]

In an ideal monoatomic gas [tex]\gamma = 1.6[/tex]

As such:

[tex]T_H = T_L (5.7)^{1.6-1}[/tex]

[tex]T_H = T_L (5.7)^{0.67}[/tex]

Replacing the value of [tex]T_H = T_L (5.7)^{0.67}[/tex] into equation [tex]\mathsf{W = nRIn(2) \Big(T_H-T_L} \Big)}[/tex]

[tex]\mathsf{W = nRT_L In(2) (5.7 ^{0.67 }-1}})[/tex]

From in the question:

W = 930 J and the moles = 1.90

using 8.314 as constant

Then:

[tex]\mathsf{930 = (1.90)(8.314)T_L In(2) (5.7 ^{0.67 }-1}})[/tex]

[tex]\mathsf{930 = 15.7966\times 1.5315 (T_L )})[/tex]

[tex]\mathsf{T_L= \dfrac{930 }{15.7966\times 1.5315}}[/tex]

[tex]\mathbf{T_L \simeq = 39 \ K}[/tex]

From [tex]T_H = T_L (5.7)^{0.67}[/tex]

[tex]\mathsf{T_H = 39 (5.7)^{0.67}}[/tex]

[tex]\mathbf{T_H \simeq 125K}[/tex]

A mixture of gaseous reactants is put into a cylinder, where a chemical reaction turns them into gaseous products. The cylinder has a piston that moves in or out, as necessary, to keep constant pressure on the mixture of 1 atm. The cylinder is also submerged in a large insulated water bath. The temperature of the water bath is monitored, and it is determined from this data that 133.0 kJ of heat flows into the system during the reaction. The position of the piston is also monitored, and it is determined from the data that the piston does 241.0 kJ of work on the system during the reaction.

a. Does the temperature of the water bath go up or down?
b. Does the piston move in or out?
c. Does heat flow into or out of the gaseous mixture?
d. How much heat flows?

Answers

I feel like it would be B makes the most sense not sure tho

A car is moving at a speed of 60 mi/hr (88 ft/sec) on a straight road when the driver steps on the brake pedal and begins decelerating at a constant rate of 10ft/s2 for 3 seconds. How far did the car go during this 3 second interval?

Answers

Answer:

219 ft

Explanation:

Here we can define the value t = 0s as the moment when the car starts decelerating.

At this point, the acceleration of the car is given by the equation:

A(t) = -10 ft/s^2

Where the negative sign is because the car is decelerating.

To get the velocity equation of the car, we integrate over time, to get:

V(t) = (-10 ft/s^2)*t + V0

Where V0 is the initial velocity of the car, we know that this is 88 ft/s

Then the velocity equation is:

V(t) = (-10 ft/s^2)*t + 88ft/s

To get the position equation we need to integrate again, this time we get:

P(t) = (1/2)*(-10 ft/s^2)*t^2 + (88ft/s)*t + P0

Where P0 is the initial position of the car, we do not know this, but it does not matter for now.

We want to find the total distance that the car traveled in a 3 seconds interval.

This will be equal to the difference in the position at t = 3s and the position at t = 0s

distance = P(3s) - P(0s)

 = ( (1/2)*(-10 ft/s^2)*(3s)^2+ (88ft/s)*3s + P0) - ( (1/2)*(-10 ft/s^2)*(0s)^2 + (88ft/s)*0s + P0)

=  ( (1/2)*(-10 ft/s^2)*9s^2+ (88ft/s)*3s + P0) - ( P0)

=  (1/2)*(-10 ft/s^2)*9s^2+ (88ft/s)*3s = 219ft

The car advanced a distance of 219 ft in the 3 seconds interval.

What does Boyle's Law state about the relationship between the pressure and volume of an ideal gas at constant temperature?
a) The product of pressure and volume increases as pressure decreases.
b) The sum of pressure and volume is constant.
c) The sum of product and volume decreases as volume increases.
d) The product of pressure and volume is constant.

Answers

Answer:

Option (d).

Explanation:

According to the Boyle's law, for a given mass of a gas, the pressure of the gas is inversely proportional to the volume of the gas keeping the temperature of the gas is constant.

So,

Let the pressure is P, volume is V and T is the absolute temperature of the gas.

Pressure proportional to the reciprocal of the volume.

[tex]P \alpha \frac{1}{V}\\\\P V = constant[/tex]

The correct option is (d).

At what rate must a cylindrical spaceship rotate if occupants are to experience simulated gravity of 0.58 g

Answers

Answer:

w = 1,066 rad / s

Explanation:

For this exercise we use Newton's second law

         F = m a

the centripetal acceleration is

         a = w² r

indicate that the force is the mass of the body times the acceleration

        F = m 0.58g = m 0.58 9.8

        F = 5.684 m

we substitute

       5.684 m = m w² r

       w = [tex]\sqrt{5.684/r}[/tex]

To finish the calculation we must suppose a cylinder radius, suppose it has r = 5 m

       w = [tex]\sqrt{ 5.684/5}[/tex]

       w = 1,066 rad / s

which has higher eneergy electron r proton

Answers

Answer:

proton have higher energy than electron

Explanation:

tag me brainliest

Answer:

proton

Explanation:

proton is higher energy than the electron

How does the theory of relativity explain the gravity exerted by massive objects?
A. More massive objects create stronger forces of gravity.
B. More massive objects create shallower curves of space-time.
C. More massive objects pull objects from farther away.
D. More massive objects create larger curves of space-time.

Answers

(D)

Explanation:

The more massive an object is, the greater is the curvature that they produce on the space-time around it.

The theory of relativity explain the gravity exerted by massive objects is

more massive objects create larger curves of space-time (option-d).

Do bigger objects exert more gravity?

The term "gravitational force" refers to the attraction between masses. The gravitational force increases in size as the masses get bigger (also called the gravity force). As the distance between masses grows, the gravitational force progressively lessens.

Greater gravitational forces will be used to attract heavier things since the gravitational force is directly proportional to the mass of both interacting objects. Therefore, when two things' respective masses increase, so does their gravitational pull to one another.

To know more about gravity visit:

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Shooting a rock with a slingshot converts
energy to__________,
energy____________,

Answers

Answer:

converts elastic energy to mechanical energy

Explanation:

No esporte coletivo, um dos principais fatores desenvolvidos é o desenvolvimento social. Qual desses não faz parte das virtudes ensinadas no esporte?

Companheirismo
Humildade
Ser justo (Fair Play)
Vencer independente do que precise ser feito

Answers

Answer:

fair palybtgshsisuehdh

Two plastic bowling balls, 1 and 2, are rubbed with cloth until they each carry a uniformly distributed charge of magnitude 0.50 nC . Ball 1 is negatively charged, and ball 2 is positively charged. The balls are held apart by a 900-mm stick stuck through the holes so that it runs from the center of one ball to the center of the other.

Required:
What is the magnitude of the dipole moment of the arrangement?

Answers

Answer:

The right solution is "[tex]4.5\times 10^{-10} \ Cm[/tex]".

Explanation:

Given that,

q = 0.50 nC

d = 900 mm

As we know,

⇒ [tex]P=qd[/tex]

By putting the values, we get

⇒     [tex]=0.50\times 900[/tex]

⇒     [tex]=(0.50\times 10^{-9})\times 0.9[/tex]

⇒     [tex]=4.5\times 10^{-10} \ Cm[/tex]  

Answer:

The dipole moment is 4.5 x 10^-10 Cm.

Explanation:

Charge on each ball, q = 0.5 nC

Length, L = 900 mm = 0.9 m

The dipole moment is defined as the product of either charge and the distance between them.

It is a vector quantity and the direction is from negative charge to the positive charge.

The dipole moment is

[tex]p = q L\\\\p = 0.5 \times 10^{-9}\times 0.9\\\\p = 4.5\times 10^{-10} Cm[/tex]

Complete the following statement: The electromotive force is:______.a. the force that accelerates electrons through a wire when a battery is connected to it.b. the maximum potential difference between the terminals of a battery. c. the force that accelerates protons through a wire when a battery is connected to it.d. the maximum capacitance between the terminals of a battery.e. the potential difference between the terminals of a battery when the battery is not in use.

Answers

Answer:

The electromotive force is the maximum potential difference between the terminals of a battery.

The electromotive force is the maximum potential difference between the terminals of a battery.  The correct option is b.

What is electromotive force?

The electromotive force also called as EMF, is the force which causes current to flow from  the positive to negative terminal of the battery.

The electromotive force is the maximum potential difference between the terminals of a battery.

Thus, the correct option is b.

Learn more about  electromotive force

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A flat coil of wire is placed in a uniform magnetic field that is in the y-direction.
The magnetic flux through the coil is maximum if the coil is:_________.
(a) in the XY plane
(b) in either the XY or the YZ plane
(c) in the XZ plane
(d) in any orientation, because it is constant.

Answers

Answer:

The correct answer is c

Explanation:

Flow is defined by

        Ф =  B . A

bold letters indicate vectors.

The magnetic field is directed to the y axis, The area of ​​the coil is represented by a vector normal to the plane of the coil, so to have a flux

                i.i = j.j = k.k = 1

and the tori scalar products are zero

a) If the coil must be in the xy plane so that its normal vector is in the Z axis, so there is no flux

b) if the coil is in the plane yz the normal veto is in the x axis, so the flux is zero

C) If the coil is in XZ, the normal vector points in the y direction, usually the scalar product is one and there is a flux in this configuration

The correct answer is c

The cart travels the track again and now experiences a constant tangential acceleration from point A to point C. The speeds of the cart are 10.6 ft/s at point A and 15.6 ft/s at point C. The cart takes 4.00 s to go from point A to point C, and the cart takes 1.80 s to go from point B to point C. What is the cart's speed at point B

Answers

Answer:

a) [tex]a_{avg}=1.25ft/s^2[/tex]

b) [tex]v_b=13.35ft/s[/tex]

Explanation:

From the question we are told that:

Speed at point A [tex]v_A=10.6ft/s[/tex]

Speed at point C [tex]v_C=15.6ft/s[/tex]

Time from Point A to C [tex]T_{ac}=4.00s[/tex]

Time from Point B to C [tex]T_{bc}=1.80s[/tex]

Generally the equation for acceleration From A to B is mathematically given by

 [tex]a_{avg}=\frac{v_c-v_a}{\triangle t}[/tex]

 [tex]a_{avg}=\frac{15.6-10.6}{4.0 }[/tex]

 [tex]a_{avg}=1.25ft/s^2[/tex]

Generally the equation for cart speed at B is mathematically given by

 [tex]a_{avg}=\frac{v_c-v_a}{T_{bc}}[/tex]

 [tex]v_b=v_c-a_{avg}*T_{bc}[/tex]

 [tex]v_b=15.6ft/s-(1.25ft/s^2)(1.80)[/tex]

  [tex]v_b=13.35ft/s[/tex]

the speed of the pulse depends on what?​

Answers

Answer:

The pulse speed depends on the properties of the medium and not on the amplitude or pulse length of the pulse.

Explanation:

Hope this helps.

Topic: Chapter 19: Some wiggle room
A hummingbird flaps its wings up to 70 times per second, producing a 70 Hz
hum as it flies by. If the speed of sound is 340 m/s, how far does the sound
travel between wing flaps?
= 4.86 m
= 58.9 m
= 0.206 m
= 23,800 m

Answers

Answer:

4.86 m

Explanation:

Given that,

The frequency produced by a humming bird, f = 70 Hz

The speed of sound, v = 340 m/s

We need to find how far does the sound  travel between wing flaps. Let the distance is equal to its wavelength. So,

[tex]v=f\lambda\\\\\lambda=\dfrac{v}{f}\\\\\lambda=\dfrac{340}{70}\\\\\lambda=4.86\ m[/tex]

So, the sound travel 4.86 m between wings flaps.

A scientist who studies fossils of ancient life forms .O ornithologist O Paleontologist O Ichthyologist O Marine Biologist .

Hurry !! First one to answer gets points !

Answers

Answer:

paleontologist

Explanation:

Paleontologists are scientists that investigate the fossils of extinct life forms. Thus, the correct option is B.

What is Fossil?

A fossil is defined as the preserved trace, imprint, or proof of a once-living entity from a past geological era. Exoskeletons, bones, shells, impressions of animals or microbes in stone, objects preserved in amber, hair, petrified wood, and genetic traces are a few examples. The collection of all the fossils is called as the fossil record.

An organism from a past geologic era that has been preserved in the Earth's crust is referred to as a fossil. Paleontologists are scientists that investigate the fossils of extinct life forms. The intricate system of fossil records is the main source of information about the evolution of life on Earth.

Therefore, the correct option is B.

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Harmonics a.are components of a complex waveform. b.have frequencies that are integer multiples of the frequency of the complex waveform. c.are pure tones. d.have sinusoidal waveforms. e.all of the above

Answers

Answer:

b.have frequencies that are integer multiples of the frequency of the complex waveform

Explanation:

Please correct me if I am wrong

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