A Sound Wave Has A Frequency Of 228 And A Wavelength Of 1.62. What Is The Speed Of The Sound? (2024)

The time for 10 trips along the length of the string is 110 sec.

The rate of speed of function of an object in any path. speed is measured because of the ratio of distance to the time wherein the space was protected. speed is a scalar quantity as it has the best path and no importance.

Length of string = 7.510

time of 1 round = 11sec

Time for 10 trips = 11sec × 10

= 110 sec

Speed is the time price at which an object is transferring along a direction, at the same time as the pace is the rate and course of an object's motion. positioned any other way, velocity is a scalar price, at the same time as the pace is a vector. Velocity is defined because of the charge of trade of distance with time. It has the measurement of distance by time.

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From the question, we can deduce the following:

Length of rectangle = 4 meters less than 3 times the width.

Perimeter of rectangle = 128 meters

Let'd find the length and width of the rectangle.

Let L represent the length and let W represent the width.

Thus, we have:

L = 3W - 4.......................equation 1

Apply the formula for perimeter of a rectangle:

P = 2W + 2L........................equation 2

Substitute 128 for P, and (3W - 4) for L in equation 2:

128 = 2W + 2(3W - 4)

Let's solve for W in the equation above.

Apply distributive property:

128 = 2W + 2(3W) + 2(-4)

128 = 2W + 6W - 8

128 = 8W - 8

Add 8 to both sides of the equation:

128 + 8 = 8W - 8 + 8

136 = 8W

Divide both sides by 8:

[tex]\begin{gathered} \frac{136}{8}=\frac{8W}{8} \\ \\ 17=W \\ \\ W=17 \end{gathered}[/tex]

Therefore, the width of the rectangle is 17 meters.

To find the length of the rectangle, substitute 17 for W in equation 1:

L = 3W - 4

L = 3(17) - 4

L = 51 - 4

L = 47

Therefore, the length of the rectangle is 47 meters.

ANSWER:

• Width = 17 meters

,

• Length = 47 meters

Let's make a diagram to visualize the situation.

As you can observe in the diagram above, the distance is 100 meters. To find the average velocity for the sprint, we have to divide the distance by the time.

[tex]\begin{gathered} \bar{v_{\text{sprint}}}=\frac{100m}{10\sec} \\ \bar{v_{}}=10(\frac{m}{s}) \end{gathered}[/tex]

(a) The average velocity for the sprint is 10 meters per second.

We know that the athlete takes 25 seconds to walk back the 100 meters. Let's divide again to find the average velocity for the walk.

[tex]\begin{gathered} v_{\text{walk}}=\frac{-100m}{25\sec } \\ v_{\text{walk}}=-4(\frac{m}{s}) \end{gathered}[/tex]

(b) The average velocity for the walk is -4 meters per second. Observe that this velocity is negative, that is because the athlete is going in the opposite direction (left side).

To find the average velocity we have to divide the total displacement by the total time spent. However, the total displacement, in this case, is zero because the athlete is going back to the starting point.

[tex]v=\frac{0m}{10\sec +25\sec }=0(\frac{m}{s})[/tex]

Therefore, the average velocity is zero for the entire round trip.

To find the average speed, we have to divide the total distance travel by the time spent.

[tex]\begin{gathered} \bar{v}=\frac{200m}{10\sec+25\sec}=\frac{200m}{35\sec } \\ \bar{v}\approx5.71(\frac{m}{s}) \end{gathered}[/tex]

Therefore, the average speed is 5.71 meters per second.

ANSWER

[tex]\begin{equation*} 100N \end{equation*}[/tex]

EXPLANATION

We want to find the weight of the bag of rocks here on Earth.

To do that, we have to find the product of the bag's mass and acceleration due to gravity:

[tex]W=m*g[/tex]

where g = acceleration due to gravity

Therefore, the weight of the bag of rocks is:

[tex]\begin{gathered} W=10.0*10 \\ W=100N \end{gathered}[/tex]

That is the answer.

ANSWER:

False

STEP-BY-STEP EXPLANATION:

We have that solar energy is created by nuclear fusion that takes place in the sun. Fusion occurs when protons from hydrogen atoms collide violently in the sun's core and fuse together to create a helium atom. This process, known as the PP (proton-proton) chain reaction, gives off a huge amount of energy.

Therefore, the statement is false, since it is nuclear fusion and not atomic function.

We are given that the Titanic hit an iceberg half of its mass. To determine the velocity of the iceberg after the collision we have to do a balance of momentum:

[tex]m_Tv_{1T}+m_Iv_{1I}=m_Tv_{2T}+m_Iv_{2I}[/tex]

Where:

[tex]\begin{gathered} m_T=\text{ mass of the titanic} \\ v_{1T}=\text{ initial velocity of the titanic} \\ m_I=\text{ mass of the Iceberg} \\ v_{1I}=\text{ initial velocity of the iceberg} \\ v_{2T}=\text{ final velocity of the titanic} \\ v_{2I}=\text{ final velocity of the iceberg} \end{gathered}[/tex]

Now, since the iceberg is initially at rest, we have:

[tex]v_{1I}=0[/tex]

Substituting in the balance of momentum we get:

[tex]\begin{gathered} m_Tv_{1T}+m_I(0)_{}=m_Tv_{2T}+m_Iv_{2I} \\ m_Tv_{1T}=m_Tv_{2T}+m_Iv_{2I} \end{gathered}[/tex]

We are given that the mass of the iceberg is half of the mass of the Titanic, therefore, we have:

[tex]m_I=\frac{m_T}{2}[/tex]

Substituting in the balance of momentum:

[tex]m_Tv_{1T}=m_Tv_{2T}+\frac{m_T}{2}v_{2I}[/tex]

Now, we can cancel out the mass of the Titanic:

[tex]v_{1T}=v_{2T}+\frac{1}{2}v_{2I}[/tex]

Now we solve for the final velocity of the iceberg. We subtract the final velocity of the Titanic from both sides:

[tex]v_{1T}-v_{2T}=\frac{1}{2}v_{2I}[/tex]

Now we multiply both sides by 2:

[tex]2(v_{1T}-v_{2T})=v_{2I}[/tex]

Substituting the values we get:

[tex]2(11.3\frac{m}{s}-3.1\frac{m}{s})=v_{2I}[/tex]

Solving the operations we get:

[tex]16.4\frac{m}{s}=v_{2I}[/tex]

Therefore, the final velocity of the iceberg is 16.4 meters per second.

The rock will appear red in color to our eyes.

The number one additive colors are red, inexperienced, and blue, which means that any coloration can be constituted of a linear superposition of these shades. according to this RGB (red, inexperienced, Blue) refers to the machine for representing shades on a computer display.

Color is the spelling used inside the united states. color is used in other English-speaking international locations. The phrase color has its roots (unsurprisingly) in the Latin phrase color. It entered Middle English via the Anglo-Norman color, which became a model of the old French color.

Colorings hold importance for people around the sector. not simplest do colors affect emotion, but they also maintain which means in faith and numerous cultures.

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Given:

The speed of light in a material, v=1.33×10⁸ m/s

To find:

The index of refraction of the material.

Explanation:

The speed of light in the vacuum is c=3×10⁸ m/s

The index of refraction of a material is given by the ratio of the speed of light in the vacuum to that in the material.

Thus the index of refraction of the given material is calculated using the equation,

[tex]n=\frac{c}{v}[/tex]

On substituting the known values in the above equation,

[tex]\begin{gathered} n=\frac{3\times10^8}{1.33\times10^8} \\ =2.26 \end{gathered}[/tex]

Final answer:

Thus the index of refraction of the given material is 2.26

We are asked to determine the perihelion distance of Halley's comet. To do that we will use the following formula:

[tex]P=a(1-e)[/tex]

Where:

[tex]\begin{gathered} P=\text{ perihelion distance} \\ a=\text{ distance of semi-major ax}is \\ e=\text{ excentricity} \end{gathered}[/tex]

To determine the distance of the semi-major axis we will use Kepler's third law

[tex]T^2=a^3[/tex]

Where:

[tex]\begin{gathered} T=\text{ period in earth years } \\ a=\text{ distance of semi-major ax}is\text{ in Astronomical Units. } \end{gathered}[/tex]

Now, we solve for "a" by taking the cubic root on both sides:

[tex]\sqrt[3]{T^2}=a[/tex]

Now, we substitute the value of "T":

[tex]\sqrt[3]{(67years)^2}=a[/tex]

Solving the operations:

[tex]16.5AU=a[/tex]

Now we substitute this value in the formula for the perihelion:

[tex]P=(16.5AU)(1-0.97)[/tex]

Solving the operations:

[tex]P=0.5AU[/tex]

Therefore, the perihelion distance is 0.5 Astronomical Units.

Take into account that the total linear momentum must conserve before and after the collision. Then, you have:

[tex]p=p^{\prime}[/tex]

p is the momentum before the collision and p' is the momentum after the collision.

By considering the given information you have:

p = m1*v1 + m2*v2

m1 = 2000 kg

m2 = 2500 kg

v1 = 30 m/s

v2 = 0 m/s

p' = (m1 + m2)v

Replace the previous values into the equation p=p' and solve for v:

[tex]\begin{gathered} m_1v_1+m_2v_2=(m_1+m_2)v \\ v=\frac{m_1v_1+m_2v_2}{m_1+m_2} \\ v=\frac{(2000kg)(30\frac{m}{s})+(2500kg)(0\frac{m}{s})}{2000\operatorname{kg}+2500\operatorname{kg}} \\ v=13.33\frac{m}{s} \end{gathered}[/tex]

Hence, the speed after the collision is 13.33 m/s

a) The diagram representing the scenario is shown below

The spring experiences ossilations. The number of ossilations is the period The formula for calculating the period is expressed as

T = 2 x pi x square root of (m/k)

where

T is the period

m is the mass of the grape

k is the spring constant

From the information given,

T = 0.48

k = 650

pi = 3.14

By substituting these values into the formula, we have

[tex]\begin{gathered} 0.48\text{ = 2}\times3.14\times\sqrt[]{\frac{m}{650}} \\ \sqrt[]{\frac{m}{650}}\text{ = }\frac{0.48}{2\times3.14}=\text{ 0.076} \\ \text{Squaring both sides of the equation, we have} \\ \frac{m}{650}=0.005776 \\ m\text{ = 650 }\times0.005776 \\ m\text{ = }3.75 \end{gathered}[/tex]

The mass of the grape is 3.75kg

b) Weight = mass x acceleration due to gravity

If acceleration due to gravity = 10m/s^2,

Then

Weight = 3.75 x 10 = 37.5N

Answer:

Explanation:

The diagram representing the scenario is shown below

x represents the unknown load

Moment = force x distance

moment about the left arm = 10x

moment about the right arm = 5 x 4 = 20

According to the principle of moments, moment about the left arm = moment about the right arm

Thus,

10x = 20

x = 20/10

x = 2

The unknown load is 2 N

ANSWER

[tex]\begin{equation*} 10.64\text{ }Hz \end{equation*}[/tex]

EXPLANATION

We want to find the frequency of the Voltage.

From the equation of the voltage, the coefficient of t represents the angular frequency. This implies that the angular frequency is:

[tex]\omega=66.883[/tex]

The angular frequency is related to the frequency as follows:

[tex]\begin{gathered} \omega=2\pi f \\ \\ f=\frac{\omega}{2\pi} \end{gathered}[/tex]

Therefore, the frequency of the voltage is:

[tex]\begin{gathered} f=\frac{66.883}{2\pi} \\ \\ f=10.64\text{ }Hz \end{gathered}[/tex]

That is the answer.

Given

v_avg = 0 m/s

solution

The athlete is stationary over a given interval of time. Average speed is the ratio of the total distance travelled by a body to the total time interval taken to cover that particular distance.

The answer would be The athlete returns to the position where he started.

Given:

Mass of skier = 65.0 kg

Length of Slope, h = 42.0 m

Angle of slope = 37.2°

Let's find how fast she is going at the bottom.

To find the speed at the bottom, we have:

[tex]\begin{gathered} \Delta x=\frac{h}{\sin \theta} \\ \\ \Delta x=\frac{42.0}{\sin 37.2} \\ \\ \Delta x=\frac{42.0}{0.605} \\ \\ \Delta x=64.5m \end{gathered}[/tex]

Now, to find the speed apply the fomula:

[tex](v_x)^2_f=(v_x)_i^2+2a_x\Delta x[/tex][tex]\begin{gathered} \text{Where:} \\ a_x=\frac{\Sigma fx_{}}{m}=\frac{(w\sin \theta)-\mu_k(w\cos \theta)}{m}=\frac{(65\ast9.8\sin37.2)-(0.107(65\ast9.8\cos37.2))}{65} \\ \\ a_x=\frac{330.8389535}{65} \\ \\ a_x=5.089 \end{gathered}[/tex]

Thus, we have:

[tex]\begin{gathered} (v_x)^2_f=0+2(5.089)(42)=427.54 \\ \\ (v_x)_f=\sqrt[]{427.54}=20.7\text{ m/s} \end{gathered}[/tex]

Therefore, the skiers speed from the bottom is 20.7 m/s

ANSWER:

20.7 m/s

The diameter of the pipe is d = 0.768 m

The diameter of smooth constriction is d' = 0.4608 m

The density of oil is

[tex]\rho\text{ = 821 }\frac{kg}{m^3}[/tex]

The pressure in the pipe is, P = 7120 N/m^2

The pressure in the constricted section is P' = 5340 N/m^2

We have to find the rate of the flowing.

The area of the pipe will be

[tex]\begin{gathered} a=\pi\text{ (}\frac{d}{2})^2 \\ =\text{ 3.14}\times(\frac{0.768}{2})^2 \\ =0.463m^2 \end{gathered}[/tex]

The area of the constricted section will be

[tex]\begin{gathered} a^{\prime}=\text{ }\pi\times(\frac{d^{\prime}}{2})^2 \\ =3.14\times(\frac{0.4608}{2})^2 \\ =0.1667m^2 \end{gathered}[/tex]

The formula to find the rate of flow is

[tex]V=\text{ aa'}\sqrt[]{\frac{2(P-P^{\prime})}{\rho(a^2-a^{\prime2})}}[/tex]

Substituting the values, the rate of flow will be

[tex]\begin{gathered} V=0.463\times0.1667\times\sqrt[]{\frac{2\times(7120-5340)}{821\lbrack(0.463)^2-(0.1667)^2\rbrack}} \\ =\frac{0.3721m^3}{s} \end{gathered}[/tex]

The electric field intensity is directly proportional to the separation between the plates.

Thus, the electric field intensity increases with the increase in separation between the plates and decreases with the decrease in separation between the plates.

Answers:

a. 8.7 m/s²

b. 10 N

c. 5 N

Explanation:

The free-body diagram for the block is

Part (a)

By the second law of Newton, the net horizontal force is equal to mass times horizontal acceleration. In this case, the net horizontal force is the horizontal component of the tension, sp

Fnet = ma

Tx = ma

Tcos30 = ma

Solving for a, we get:

a = Tcos30/m

So, replacing T = 10N, cos 30 = 0.87 and m = 1.0 kg, we get:

a = 10(0.87)/1 = 8.7 m/s²

Therefore, the horizontal acceleration of the block, to one decimal place, is 8.7 m/s²

Part (b)

The weight of the block is equal to mass times gravity, so

W = mg

W = (1.0 kg)(10 m/s²)

W = 10 N

Then, the weight of the block is 10 N

Part (c)

The block is not accelerating in the vertical direction, so the net vertical force is 0 N, then

[tex]\begin{gathered} Fn+T_y-W=0 \\ Fn+T\sin 30-W=0 \end{gathered}[/tex]

Solving for the normal reaction Fn, we get:

[tex]Fn=W-T\sin 30[/tex]

Replacing the weight W = 10N, T = 10N and sin30 = 0.5, we get:

[tex]\begin{gathered} Fn=10N-10N(0.5) \\ Fn=10N-5N \\ Fn=5N \end{gathered}[/tex]

Therefore, the normal reaction is 5N

Answer:

See below

Explanation:

So the tennis ball has traveled 1.1 m after it left your hand....what is its speed then?

df = do + vo t + 1/2 a t^2

df = final position= 1.1 do = original position = 0 vo = 8.2 a = - 9.81

1.1 = 8.2t - 4.95 t^2 Solve for t = .147 s

Vf = vo - at = 8.2 - 9.81 (.147) = 6.76 m/s

Given:

There are two equal and opposite charged paticles.

The force of attraction is F = 32.38 N

The distance between the charged particles is r = 44.3 cm =0.443 m

Required: Magnitude of each charge in micro coulombs.

Explanation:

The magnitude of charge can be calculated by the formula

[tex]\begin{gathered} F=\frac{kq1q2}{r^2} \\ F=\frac{kq^2}{r^2} \\ q\text{ =}\sqrt{\frac{Fr^2}{k}} \end{gathered}[/tex]

Here, k is the Coulomb's constant whose value is

[tex]k\text{ = 9}\times10^9\text{ N m}^2\text{ /C}^2[/tex]

On substituting the values, the magnitude of charge can be calculated as

[tex]\begin{gathered} q=\text{ }\sqrt{\frac{32.38\times(0.443)^2}{9\times10^9}} \\ =\text{ 2.67}\times10^{-5}\text{ C} \\ =26.7\times10^{-6}\text{ C} \\ =26.7\text{ micro Coulomb} \end{gathered}[/tex]

Final Answer: The magnitude of charges is 26.7 microcoulomb.

Given,

The atomic number is Z=51

The number of the neutrons, N=70

The atom which has the atomic number 51 is the antimony.

The atomic mass number is the sum of the protons and the neutrons in the atom. Thus the atomic mass of the given element is

[tex]\begin{gathered} A=Z+N \\ =51+70 \\ =121 \end{gathered}[/tex]

Thus the symbol of the given element is

[tex]^{51}_{121}Sb[/tex]

Given data:

* The fundamental length of pipe is,

[tex]L_1=0.25\text{ m}[/tex]

* The next resonant length of pipe is,

[tex]L_2=0.75\text{ m}[/tex]

* The speed of the sound is v = 338 m/s.

Solution:

The fundamental length of the pipe in terms of the wavelength is,

[tex]L_1=\frac{1}{4}\lambda[/tex]

The second resonant length of the pipe in terms of wavelength is,

[tex]L_2=\frac{3}{4}\lambda[/tex]

Substracting both the values,

[tex]\begin{gathered} L_2-L_1=\frac{3}{4}\lambda-\frac{1}{4}\lambda \\ L_2-L_1=\frac{2}{4}\lambda \\ L_2-L_1=\frac{1}{2}\lambda \end{gathered}[/tex]

Substituting the known values,

[tex]\begin{gathered} 0.75-0.25=\frac{1}{2}\times\lambda \\ 0.5=\frac{1}{2}\times\lambda \\ \lambda=1\text{ m} \end{gathered}[/tex]

Thus, the value of the wavelength of sound emitted by the tuning fork is 1 meter.

(b). The frequency of the sound is,

[tex]\begin{gathered} v=f\lambda \\ f=\frac{v}{\lambda} \\ f=\frac{338}{1} \\ f=338\text{ Hz} \end{gathered}[/tex]

Thus, the frequency of sound is 338 Hz.

To charge any material, the electrons are added or removed from the material.

The material can be partially charged by unbalanced charge distribution on the material with the charging by induction.

As the electrons are revolving around the nucleus, thus, the amount of energy required to remove the charge is used to overcome the electrostatic attractive force between the nucleus and electrons.

In the case of protons,

The protons are present inside the nucleus under a strong nuclear force.

It is not easy to remove the proton from the nucleus as compared to the electrons from the orbit.

Thus, the charging of the material is carried by the removal or addition of electrons due to their more mobility than protons and their weak holding force.

Answer: A. Friction

Explanation:

When glass is rubbed against silk, there is friction between both materials. Thus, the charge acquired is through friction. The correct option is

A. Friction

Given data:

* The velocity of the car is 13.6 m/s.

* The car stops in 0.321 m.

* The force exerted on the child of mass 21.2 kg is -6110 N.

* The mass of the child is 21.2 kg.

Solution:

The weight of the child is,

[tex]W=mg[/tex]

where m is the mass of the child, and g is the acceleration due to gravity,

Substituting the known values,

[tex]\begin{gathered} W=21.2\times9.8 \\ W=207.76\text{ N} \end{gathered}[/tex]

By dividing the force exerted on the child with the weight of the child,

[tex]\begin{gathered} \frac{F}{W}=\frac{6110}{207.76} \\ \frac{F}{W}=29.4 \\ F=29.4\times W \end{gathered}[/tex]

Thus, the force is 29.4 times the weight of the child.

Certain examples for the conversion of energy from one form to another form are,

From Chemical energy,

(a). To Thermal energy:

1. Coal

2. Natural gas

(b). To radiant energy:

1. Battery power flashlight

2. Lightstick

(c). To light energy,

1. Combustion reaction with coal

2. Combustion reaction with the wood

(d). To nuclear energy,

1. Combustion of oil

2. Combustion of gas

(e). To electrical energy,

1. Battery

(f). To gravitational energy,

1. Food

(g). To mechanical energy,

1. Gasoline

2. Petrol

Given

The speed of light through a medium is

[tex]v=1.24\times10^8\text{ m/s}[/tex]

To find

The type of material

Explanation

The velocity through a medium is given by

[tex]v=\frac{c}{n}[/tex]

where c is the speed of light in .

n is the refractive index of the material

Thus,

[tex]\begin{gathered} 1.24\times10^8=\frac{3\times10^8}{n} \\ \Rightarrow n=2..419 \end{gathered}[/tex]

Thus the refractive index is 2.419

We know diamond has a refractive index of 2.417-2.419

Conclusion

The material is diamond

In order to determine the elongation of the spring, proceed as follow:

Consider that the weight plus the elastic force on the block is equal to the buoyant force of the water:

W + F = E

where:

W: weight of the block = m*g = (5.71kg)(9.8m/s^2) = 55.96N

F: elastic force = kΔL

k: spring constant = 180N/m

E: buoyant force = m*g*ρw/ρb

Then, the net force on the block can be written as follow:

[tex]\begin{gathered} F=mg(\frac{\rho_w}{\rho_b})-W \\ F=(5.71kg)(9.8\frac{m}{s^2})(\frac{1000\frac{kg}{m^3}}{644\frac{kg}{m^3}})-55.96N \\ F=86.89N-55.96N \\ F=30.93N \end{gathered}[/tex]

Next, use the previous value into the formula for the elastic force (Hooke's law) and solve for ΔL:

[tex]\Delta L=\frac{F}{k}=\frac{30.93N}{180\frac{N}{m}}\approx0.172N[/tex]