If at a particular instant and at a certain point in space the electric field is in the x-direction and has a magnitude of 3.80 V/m , what is the magnitude of the magnetic field of the wave at this same point in space and instant in time?

Answers

Answer 1

Answer:

The value is  [tex]|\vec B| = 1.267 *0^{-8} \ T[/tex]

Explanation:

From the question we are told that  

   The magnitude of the electric fields is  [tex]E = 3.80 V/m[/tex]

Generally speed of light is mathematically represented as

        [tex]c = \frac{|\vec E|}{ |\vec B|}[/tex]

Here c is the speed of light with value  [tex]c = 3.0*10^{8} \ m/s[/tex]

        [tex]|\vec B |[/tex]  is the  magnitude of the magnetic field so  

=>         [tex]|\vec B| = \frac{|\vec E|}{c}[/tex]

=>         [tex]|\vec B| = \frac{ 3.80 }{3.0*10^{8}}[/tex]

=>         [tex]|\vec B| = 1.267 *0^{-8} \ T[/tex]


Related Questions

A diffraction grating with 68 slits per cm is used to measure the wavelengths emitted by hydrogen gas.
A. At what angles in the fourth-order spectrum would you expect to find the two violet lines of wavelength 434 nm and of wavelength 410 nm?
B. What are the angles if the grating has 12,800 slits per cm?

Answers

Answer:

a

  [tex]\theta _1 =0.687 ^o[/tex]

  [tex]\theta _2 =0.630 ^o[/tex]

b

 Generally given that the domain arcsine function is  between -1 and 1 then the arcsine of 2.22 will not be valid

Generally given that the domain arcsine function is  between -1 and 1 then the arcsine of 2.1 will not be valid

Explanation:

From the question we are told that

     The slit grating is  [tex]N = 68 \ slits / cm = 6800 \ slits / m[/tex]

      The order of spectrum is [tex]n = 4[/tex]

Generally the width of the slit is mathematically represented as  

               [tex]a = \frac{1}{ 6800}[/tex]      

=>            [tex]a = 0.000147 \ m[/tex]  

Generally the condition for constructive interference is

              [tex]asin\theta = n * \lambda[/tex]

Now for the first wavelength the angle is evaluated as

             [tex]\theta _1 = sin ^{-1} [ \frac{n \lambda_1 }{a} ][/tex]

    =>     [tex]\theta _1 = sin ^{-1} [ \frac{4* 434 *10^{-9} }{ 0.000147 } ][/tex]

    =>    [tex]\theta _1 =0.687 ^o[/tex]

Now for the second wavelength the angle is evaluated as

             [tex]\theta _2 = sin ^{-1} [ \frac{n \lambda_2 }{a} ][/tex]

    =>     [tex]\theta _2 = sin ^{-1} [ \frac{4* 410 *10^{-9} }{ 0.000147 } ][/tex]

    =>    [tex]\theta _2 =0.630 ^o[/tex]

Gnerally if grating is   [tex]N = 12800 \ slits per cm = 1280000 \ slits / m[/tex]

Generally the width of the slit is mathematically represented as  

               [tex]a = \frac{1}{ 1280000}[/tex]      

=>            [tex]a = 7.813 *10^{-7} \ m[/tex]  

Generally the condition for constructive interference is

              [tex]asin\theta = n * \lambda[/tex]

Now for the first wavelength the angle is evaluated as

             [tex]\theta _1 = sin ^{-1} [ \frac{n \lambda_1 }{a} ][/tex]

            [tex]\theta _1 = sin ^{-1} [ \frac{4* 434 *10^{-9} }{ 7.813*10^{-7} } ][/tex]

    =>     [tex]\theta _1 = sin ^{-1} [ 2.22][/tex]

Generally given that the domain arcsine function is  between -1 and 1 then the arcsine of 2.22 will not be valid

    =>    [tex]\theta _1 =0.687 ^o[/tex]

Now for the second wavelength the angle is evaluated as

             [tex]\theta _2 = sin ^{-1} [ \frac{n \lambda_2 }{a} ][/tex]

    =>     [tex]\theta _2 = sin ^{-1} [ \frac{4* 410 *10^{-9} }{ 7.813*10^{-7} } ][/tex]

     =>  [tex]\theta _2 = sin ^{-1} [2.1 ][/tex]

Generally given that the domain arcsine function is  between -1 and 1 then the arcsine of 2.22 will not be valid

An object with a mass of 32 kg has an initial energy of 500). At the end of the experimentthe velocity of the object is recorded as 5.1 m/s . the object travelled 50 m to get to this point, what was the average force of friction on object during the tripAssume no potential energy Show all work

Answers

Answer:

 F = 1.68 N

Explanation:

Let's solve this exercise in parts.

Let's use the concept of conservation of the mechanical nerve

initial

    Em₀ = 500 J

The energy is totally kinetic

     Em₀ = K = ½ m v₀²

     v₀ = [tex]\sqrt{\frac{2 Em_{o} }{m} }[/tex]

     v₀ = √ (2 500/32)

     v₀ = 5.59 m / s

now with kinematics we can find a space

      v² = v₀² - 2 a x

the negative sign is because the body is stopping

       a =[tex]( \frac{v_{o}^{2} - v^{2} }{2x} )[/tex]  

let's calculate

       a = (5.59² - 5.1²) / 2 50  

       a = 0.0524 m / s²

Finally let's use Newton's second law

     F = ma

     F = 32 0.0524

     F = 1.68 N

A bullet with an initial kinetic energy of 400 J strikes a wooden block where a 8000 N resistive force stops the bullet. What is the distance the bullet travels into the block?

How do you answer this question?

Answers

Answer:

d = 0.05 [m] = 50 [mm]

Explanation:

We must remember the principle of conservation of energy which tells us that energy is transformed from one way to another. For this case, the initial kinetic energy is transformed into useful work that is equal to the product of force by distance.

[tex]E_{k}=F*d\\400 = 8000*d\\d = 0.05 [m] = 50 [mm][/tex]

Galaxy B moves away from galaxy A at 0.577 times the speed of light. Galaxy C moves away from galaxy B in the same direction at 0.731 times the speed of light. How fast does galaxy C recede from galaxy A?

Answers

Answer:

The value is  [tex]p = 0.7556 c[/tex]

Explanation:

From the question we are told that

   The speed at which galaxy B moves away from galaxy A is  [tex]v = 0.577c[/tex]

Here c is the speed of light with value  [tex]c = 3.0 *10^{8} \ m/s[/tex]

     The speed at which galaxy C moves away from galaxy B is  [tex]u = 0.731 c[/tex]

Generally from the equation of  relative speed we have that  

     [tex]u = \frac{p - v}{ 1 - \frac{ p * v}{c^2} }[/tex]

Here p is the velocity at which galaxy C recede from galaxy A so

     [tex]0.731c = \frac{p - 0.577c }{ 1 - \frac{ p * 0.577c}{c^2} }[/tex]

=>   [tex]0.731c [1 - \frac{ p * 0.577}{c}] = p - 0.577c[/tex]

=>   [tex]0.731c - 0.4218 p = p - 0.577c[/tex]

=>   [tex]0.731c + 0.577c = p + 0.4218 p[/tex]

=>   [tex]1.308 c = 1.731 p[/tex]

=>    [tex]p = 0.7556 c[/tex]

If an ocean wave passes a stationary pointevery 4 s and has a velocity of 7 m/s, what isthe wavelength of the wave?Answer in units of m.

Answers

Answer:

28m

Explanation:

Step one:

given data

period T= 4seconds

velocity v= 7m/s

wave lenght λ=?

Step two:

we know that f=1/T

the expression relating period and wave lenght is

v=λ/T

λ=v*T

λ=7*4

λ=28m

The wavelength of the wave is 28m

A 12 N net force is applied to an object as it moves a distance of 3.0 m: Use the
Work-Kinetic Energy Theorem to determine the object's change in kinetic energy.
Enter your answer in Joules.

Answers

Answer:

4 joules

Explanation:

what is the potential energy of a 30kg rock that falls 15 meters

Answers

Answer:

4500 J

Explanation:

The potential energy of a body can be found by using the formula

PE = mgh

where

m is the mass

h is the height

g is the acceleration due to gravity which is 10 m/s²

From the question we have

PE = 30 × 10 × 15

We have the final answer as

4500 J

Hope this helps you

A 500 kg car is moving at 30 m/s. The driver sees a barrier ahead. If the car takes 100 m to come to rest, what is the magnitude of the force necessary to stop the car?

How do you solve this question?

Answers

Answer:

F = 2250 [N]

Explanation:

In order to solve this problem, we must first use the following equation of kinematics.

[tex]v_{f}^{2} =v_{o}^{2}-2*a*x[/tex]

where:

Vf = final velocity = 0 (come to rest)

Vo =  initial velocity = 30 [m/s]

a = acceleration or desaceleration [m/s²]

x = distance = 100 [m]

[tex](0)=30^{2} -2*a*100\\900 = 200*a\\a = 4.5 [m/s^{2}][/tex]

Now we must use the following equation of kinetics, which is based on Newton's second law that explains that the sum of forces on a body is equal to the product of mass by acceleration.

∑F = m*a

where:

F = force [N]

m = mass = 500 [kg]

a = acceleration = 4.5 [m/s²]

[tex]F = 500*4.5\\F = 2250 [N][/tex]

A monatomic ideal gas with an initial pressure of 500 kPa and an initial volume of 1.80 L expands isothermally to a final volume of 5.20 L. How much work is done on the gas in this process?
A) 1700J
B) 875J
C) 1570J
D) 900J
E) 955J

Answers

Answer:

955 J  

Explanation:

PV = nRT

500 x 10³ x 1.8 x 10⁻³ = nRT

= 900 J

work done by gas in isothermal expansion

= nRT lnV₂ / V₁

= 900 ln 5.2 / 1.8

= 900 x ln 2.89

= 900 x 1.06

= 955 J  

A 1.00-m3 object floats in water with 30.0% of its volume above the waterline. What does the object weigh out of the water?

Answers

Answer:

Object's weight = 6,839.42 N

Explanation:

Given

Above waterline = 30%

Volume of object = 1m^3

Required

Determine the weight of the object

First, we need to calculate its Mass

Mass = Density of Water * Volume of object in water

Density of water = 997kg/m³

If 30% is above waterline, then 70% is in water.

So:

Mass = Density of Water * Volume of object in water

Mass = 997kg/m³ * 70%m³

Mass = 997kg * 70%

Mass = 697.9 kg

The object weight sis then calculated as thus:

Weight = Mass * Acceleration of gravity

Weight = 697.9 kg * 9.8m/s²

Weight = 6 839.42 N

Forces of 70 N at 130 degrees, and 20 N at an angle of 280 degrees, measured counter-clockwise from the positive x-axis, act on an object.
A. What are the components (F1x, F1y) of the first force force (in Newtons)?
B. What are the components (F2x, F2y) of the second force force (in Newtons)?
C. What are the components (Fx, Fy) of the resultant force (in Newtons)?
D. What is the magnitude of the resultant force (in Newtons)?
E. What is the angle of the resultant force with respect to x-axis?

Answers

Answer:

A. ) F₁ₓ = -45.0 N F₁y = 53.6 N

B.)  F₂ₓ = 3.48 N F₂y = -19.7 N

C.)  Fₓ = -41.5 N Fy = 33.9 N

D)  F = 53.6 N

E)  θ = -39. 2º (320.8º)

Explanation:

A)

Applying simple trig, like definitions of cos and sin of an angle, we can get the x- and y- components of F₁, as follows:

       [tex]F_{x1} = F_{1} * cos (130) = 70 N * cos (130) = -45 N (1)\\F_{y1} = F_{1} * sin (130) = 70 N * sin (130) = 53.6 N (2)[/tex]

B)

Repeating for F₂:

       [tex]F_{x2} = F_{2} * cos (280) = 20 N * cos (280) = 3.48 N (3)\\F_{y2} = F_{2} * sin (280) = 20 N * sin (280) = -19.7 N (4)[/tex]

C)  

The x- and y- components of the resultant force, are just the algebraic

        sum of the x- and - y components of F₁ and F₂:

Fₓ = Fₓ₁ + Fₓ₂ = -45 N + 3.48 N =  -41.5 N (5)By the same token, Fy can be written as follows:Fy = Fy₁ + Fy₂ = 53.6 N + (-19.7 N) = 33.9 N (6)

D)

The magnitude of the resultant force can be obtained applying the Pythagorean Theorem to Fx and Fy, as follows:

       [tex]F_{t} =\sqrt{F_{x} ^{2} + F_{y} ^{2} } = \sqrt{(-41.5N)^{2} +(33.9N)^{2}} = 53.6 N (7)[/tex]

E)

Finally the angle regarding the x- axis of the resultant force vector, can be obtained using the definition of the tangent of an angle, as follows:

       [tex]\theta = arc tg \frac{33.9N}{(-41.5N)} = arc tg (-0.817) = -39. 2 \deg[/tex]

The edge of a flying disc with a radius of 0.13 m spins with a tangential speed of 3.3 m/s. The centripetal acceleration of the edge of the disc is ?m/s2.

Answers

Answer:

Centripetal acceleration = 83.77m/s²

Explanation:

Given the following data;

Radius, r = 0.13m

Velocity, v = 3.3m/s

To find centripetal acceleration;

Centripetal acceleration is given by the formula;

[tex] Acceleration, a = \frac {v^{2}}{r}[/tex]

Substituting into the equation, we have;

[tex] Centripetal \; acceleration, a = \frac {3.3^{2}}{0.13}[/tex]

[tex] Centripetal \; acceleration, a = \frac {10.89}{0.13}[/tex]

Centripetal acceleration = 83.77m/s²

Therefore, the centripetal acceleration of the edge of the disc is 83.77 m/s².

Answer:

the answer is 84

Explanation:

Yellow light shines on a sheet of paper containing a blue pigment. Determine the appearance of the paper.

I dont need google answers if i get google answers i will delete it

Answers

Black I think

this is due to the fact that blue lights only reflect blue things so if yellow is shone on it it will reflect black appearance.

how it useful

What is the minimum angular velocity (in rpm) for swinging a bucket of water in a vertical circle without spilling any? The distance from the handle to the bottom of the bucket is 35 cm.

Answers

Complete Question

What is the minimum angular velocity (in rpm) for swinging a bucket of water in a vertical circle without spilling any? The distance from the handle to the bottom of the bucket is 35 cm. The student has 70-cm-long arms

Answer:

The value is  [tex]w__{rpm} } = 29.17 \ rpm[/tex]

Explanation:

From the question we are told

    The distance from the handle to the bottom of the bucket is  [tex]d = 35 \ cm = 0.35 \ m[/tex]

      The length of the students arm is  L = 70 cm  = 0.70  m

   Generally the acceleration due to gravity experienced by the bucket of  water is mathematically represented as

       [tex]g = w^2 * r[/tex]

Here is is the radius of the circle which swinging of the bucket makes and this is mathematically represented as

       [tex]r = L + d[/tex]

So

         [tex]g = w^2 * ( L + d )[/tex]

= >     [tex]w = \sqrt{\frac{g }{ L + d } }[/tex]

= >     [tex]w = \sqrt{\frac{ 9.8}{ 0.7 + 0.35} }[/tex]

= >     [tex]w = 3.055 \ rad/s[/tex]

Generally the angular speed in revolution per minute is mathematically represented as

        [tex]w__{rpm} } = \frac{w * 60 }{2 \pi }[/tex]

=>      [tex]w__{rpm} } = \frac{3.055 * 60 }{2 * 3.142 }[/tex]

=>      [tex]w__{rpm} } = 29.17 \ rpm[/tex]

. What is the barycenter of the Moon and Earth?​

Answers

Answer:

About 1000 miles.

Explanation:

PLEASE HELP!

I don't even know what Science is I'm so dumb lol XD

Answers

Answer:

C

Explanation:

Sled A has more potential energy because it's mass is 100 kg, and it is higher up than Sled B. The more high up the sled is and the lighter it is, the faster it gets, it creates more and more potential energy.

Hope this helps!

How long does it take a plane, traveling at a constant speed of 123 m/s, to fly once around a circle whose radius is 4330 m?

Answers

Answer:

3.7 min

Explanation:

Step one:

given data

speed = 123m/s

radius of circle= 4330m

Step two:

We need to find the circumference of the circle, it represents the distance traveled

C=2πr

C= 2*3.142*4330

C= 27209.72m

Step three:

We know that velocity= distance/time

time= distance/velocity

time= 27209.72/123

time=221.2 seconds

in minute = 221.2/60

time= 3.7 min

pplzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz helppppp

Answers

Turn towards East because it will be pushed

Answer:

D. The airplane will turn towards the east

Explanation:

If the airplane is thrown straight towards the north, the window which is moving from left (west) to right (east) the wind will knock the plane towards the right (east) since thats the way it is blowing.

if A=3i +2j+3k ,find the magnitude of A+B and A-B​

Answers

Since vector B was not specified, I'll assume one at random. You can later answer your own question.

Answer:

[tex]\mid\mid \vec A+\vec B \mid \mid=\sqrt{105}[/tex]

[tex]\mid\mid \vec A-\vec B \mid \mid=\sqrt{149}[/tex]

Explanation:

Given:

[tex]\vec A=3\hat i +2\hat j+3\hat k[/tex]

And (assumed):

[tex]\vec B=-5\hat i +8\hat j-4\hat k[/tex]

Find the magnitude of

[tex]\vec A+\vec B[/tex]

[tex]\vec A-\vec B[/tex]

Given a vector

[tex]\vec P=x\hat i +y\hat j+z\hat k[/tex]

The magnitude of the vector is:

[tex]\mid\mid \vec P\mid \mid=\sqrt{x^2+y^2+z^2}[/tex]

First part:

[tex]\vec A+\vec B =3\hat i +2\hat j+3\hat k-5\hat i +8\hat j-4\hat k[/tex]

[tex]\vec A+\vec B =-2\hat i +10\hat j-\hat k[/tex]

The magnitude of the sum is:

[tex]\mid\mid \vec A+\vec B \mid \mid=\sqrt{(-2)^2+10^2+(-1)^2}=\sqrt{4+100+1}[/tex]

[tex]\mathbf{\mid\mid \vec A+\vec B \mid \mid=\sqrt{105}}[/tex]

Second part:

[tex]\vec A-\vec B =3\hat i +2\hat j+3\hat k-(-5\hat i +8\hat j-4\hat k)[/tex]

[tex]\vec A-\vec B =3\hat i +2\hat j+3\hat k+5\hat i -8\hat j+4\hat k[/tex]

[tex]\vec A-\vec B =8\hat i -6\hat j+7\hat k[/tex]

The magnitude of the difference is:

[tex]\mid\mid \vec A-\vec B \mid \mid=\sqrt{8^2+(-6)^2+7^2}=\sqrt{64+36+49}[/tex]

[tex]\mathbf{\mid\mid \vec A-\vec B \mid \mid=\sqrt{149}}[/tex]

Can someone help with my physics homework? please

A go cart engine applies a force of 888N and moves the cart forward 22m.

a) How much work is done?


b) What is doing the work?


c) If the driver wants to go further will the amount of work increase or decrease? Do you need a bigger engine to go

further?


d) We put on a bigger engine (1111N) but the cart still moves forward 22m. How much work is done now?


e) Why would you put on a bigger engine if you are still moving 22m?


f) Work requires a change in energy, which engine uses more gas to go 22m?


g) Even an empty semi truck uses much more gas than a car. Why?Find the soultions

Answers

Answer:

a) 19536 joules of work are done.

b) The work is done by the engine on the structure of the cart.

c) There are three options:  (i) Keeping the engine and changing the travelled distance, (ii) Changing the engine and keeping the travelled distance, (iii) Changing the engine and the travelled distance.

d) 24442 joules of work are done.

e) We may change for a bigger engine if it allows a greater acceleration and higher peak speed.

f) The bigger engine uses more gas to go 22 meters.

g) An empty semi truck uses more gas than a car since the first has much more mass than a car and is designed for moves big loads and for being fast.

Explanation:

a) If force applied in the cart is uniform, that is, constant in magnitude and direction and is parallel to distance travelled by the car, the work done on the cart is defined by the following equation:

[tex]W = F\cdot \Delta s[/tex] (1)

Where:

[tex]F[/tex] - Force applied by the cart, measured in newtons.

[tex]\Delta s[/tex] - Distance travelled by the car, measured in meters.

[tex]W[/tex] - Work done on the cart, measured in joules.

If we know that [tex]F = 888\,N[/tex] and [tex]\Delta s = 22\,m[/tex], then the work done on the cart is:

[tex]W =(888\,N)\cdot (22\,m)[/tex]

[tex]W = 19536\,J[/tex]

19536 joules of work are done.

b) The work is done by the engine on the structure of the cart.

c) There are three options:  (i) Keeping the engine and changing the travelled distance, (ii) Changing the engine and keeping the travelled distance, (iii) Changing the engine and the travelled distance.

d) If we know that [tex]F = 1111\,N[/tex] and [tex]\Delta s = 22\,m[/tex] , then the work on the cart is:

[tex]W = (1111\,N)\cdot (22\,m)[/tex]

[tex]W = 24442\,N[/tex]

24442 joules of work are done.

e) We may change for a bigger engine if it allows a greater acceleration and higher peak speed.

f) The gas consumption is directly proportional to the square of velocity and mass of the cart and, hence, to the work done on the cart. In consequence, we conclude that the bigger engine uses more gas to go 22 meters.

g) An empty semi truck uses more gas than a car since the first has much more mass than a car and is designed for moves big loads and for being fast.

a. The amount of work done by the go cart engine is 19,536 Nm.

b. Work is being done by the go cart engine i.e the engine installed on the go cart.

c. The amount of work will increase if the driver wants to go further because work done is directly proportional to the distance covered. No, you don't need a bigger engine to go further.

d. The amount of work done when a bigger engine is used is 24,442 Nm.

e. The reason why a bigger engine is used is, so that the engine can easily do more work in the same amount of distance.

f. Since we know that work requires a change in energy, the bigger engine would use more gas to go 22 meters.

g. An empty semi truck uses much more gas than a car because it has more weight (mass) and as such requires more energy, in accordance with the law of inertia.

Given the following data:

Force A = 888 NewtonDistance = 22 meterForce B = 1111 Newton

a. To determine the amount of work done by the go cart engine:

Mathematically, the work done by an object is given by the formula;

[tex]Work\;done = Force \times distance[/tex]

Substituting the given parameters into the formula, we have;

[tex]Work\;done = 888 \times 22[/tex]

Work done = 19,536 Nm.

b. Work is being done by the go cart engine i.e the engine installed on the go cart.

c. The amount of work will increase if the driver wants to go further because work done is directly proportional to the distance covered. No, you don't need a bigger engine to go further.

d. To determine the amount of work done when a bigger engine is used:

[tex]Work\;done = Force \times distance[/tex]

[tex]Work\;done = 1111 \times 22[/tex]

Work done = 24,442 Nm.

e. The reason why a bigger engine is used is, so that the engine can easily do more work in the same amount of distance.

f. Since we know that work requires a change in energy, the bigger engine would use more gas to go 22 meters.

g. An empty semi truck uses much more gas than a car because it has more weight (mass) and as such requires more energy, in accordance with the law of inertia.

Read more: https://brainly.com/question/22599382

What is the error in this representation of the steps involved in gene therapy?

Answers

ahh i hope this helps lol

Answer:

a

Explanation:

When a neutral atom gains or loses
electrons, it becomes charged
and is called a(n)

Answers

Answer:

It is called an ion.

Explanation:

Atoms of elements can lose or gain electrons making them no longer neutral, they become charged. A charged atom is called an ion. When an atom loses electron(s) it will lose some of its negative charge and so becomes positively charged. A positive ion is formed where an atom has more protons than electrons.

Help ASAP plz and thx u

Answers

Answer:

a). a = F/m

Explanation:

Formula is F=ma

the answer is a ) a=F/m

Pushes and pulls that result from objects that are physically touching
each other

Answers

The answer: A force




Explanation: I looked it up












Answer:

That is false. Take a look at this way. You can push a ball with your own breath, you just need to blow it. And you can pull something from afar with a magnet. It is possible to do both.

Explanation:

Not all physical things can be done only physically. Like I just said, it is possible to use other forces (no, not the dark side one),  such as a magnetic force, displayed by a magnet or anything with a force like so.

One of the harmonics of a column of air in a tube that is open at both ends has a frequency of 448 Hz, and the next higher harmonic has a frequency of 576 Hz. What is the fundamental frequency of the air column in this tube?

Answers

Answer:

The fundamental frequency is  [tex]f_1 =128 \ Hz[/tex]

Explanation:

From the question we are told that

   The frequency of one harmonics is  [tex]f_x= 448 \ Hz[/tex]

    The next higher harmonic is  [tex]f_z = 576 \ Hz[/tex]

Generally the frequency of an air column open at both ends is mathematically represented as

              [tex]f_n = \frac{nv }{ 2 L }[/tex]

Here n  is the order of the harmonics (frequency)

        v is the velocity of the sound

        L  is the length of the column

So for one harmonics we have that

        [tex]f_k = \frac{n v }{2L}[/tex]

Then for the next higher harmonics

       [tex]f_x = \frac{n+1 ) v}{2 L }[/tex]

Generally the difference between these frequencies is mathematically represented as  

       [tex]f_z- f_x = \frac{(n+1 )v}{ 2L} - \frac{(n )v}{ 2L}[/tex]

=>    [tex]576 - 448 = \frac{vn + v - nv }{2L}[/tex]

=>    [tex]\frac{ v }{2L} = 128[/tex]

Generally for fundamental  frequency n =  1

So  

       [tex]f_1 = n * \frac{v}{2L}[/tex]

So

       [tex]f_1 =1 * 128[/tex]

=>    [tex]f_1 =128 \ Hz[/tex]

A statement of the second law of thermodynamics is that:__________.
a) spontaneous reactions are always exothermic.
b) energy is conserved in a chemical reaction that has a decrease in entropy.
c) spontaneous reactions are always endothermic.
d) in a spontaneous process, the entropy of the universe increases.

Answers

Answer:

in a spontaneous process, the entropy of the universe increases.

Explanation:

Entropy is a measure of of the degree of  randomness or disorderliness in a system.

The second law of thermodynamics can be stated as follows; "in any spontaneous process, the entropy of the universe increases."  

The universe here refers to the system's disorder and the disorder of the surroundings.  Therefore, a spontaneous process can occur, in which the entropy of the system decreases, only if the entropy increases in the surroundings.

For instance, when ice freezes, the entropy of liquid water decreases, that is, the entropy of the system decreases. However, heat is given off to the surroundings and the entropy of the surroundings increases. This is an obvious expression of this law.

In the past, Africa used to be further away from Europe than it is now
(shown below). What could explain why Africa is closer to Europe now than
it was before? *

Answers

Answer: Plates shifting

Explanation: After years and years of plates colliding into solid rock, they slowly become closer together. As recent studies have shown, Africa is currently moving closer to Europe one centimeter every year (one inch every 2.5 years).

Answer:

nvudbwasivnjlscv bwbfvsz

Explanation:

friction reduces air resistance?

Answers

Answer:no

Explanation:

No.....................................

Objects accelerate because

Answers

Friction I think soooooooooo

What is the Basic SI unit for distance/length

A. Meters
B. Liters
C. Grams
D. Millimeters

Answers

Meters because distance measures in meters
Meters because distance measure is meters and remeber distancextime measure In meters
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