For a certain transverse standing wave on a long string, an antinode is at x = 0 and an adjacent node is at x = 0.20 m. The displacement y(t) of the string particle at x = 0 is shown in the figure, where the scale of the y axis is set by ys = 4.3 cm. When t = 0.90 s, what is the displacement of the string particle at (a) x = 0.30 m and (b) x = 0.40 m ? What is the transverse velocity of the string particle at x = 0.30 m at (c) t = 0.90 s and (d) t = 1.3 s?

For A Certain Transverse Standing Wave On A Long String, An Antinode Is At X = 0 And An Adjacent Node

Answers

Answer 1

The expressions for the traveling and standing wave to find the results for the questions about the displacement and speed of the particle are:

       a) For time zero, the displacement at position x = 0.30 m is y = 3.04 cm

     b) For time zero, the displacement at position x = 0.40 m is: y = 0

      c) For the point x = 0.30 and time t = 0.9s, the velocity of the particle is:

          v = 9.11 cm / s

      d) For the point x = 0.30 and time t = 1.3s, the velocity of the particle is:

          v = 9.65 cm / s

The traveling wave is a disturbance in the medium that moves at constant speed, in the case of a transverse wave the expression for the perpendicular oscillation is:

         y = A sin (kx - wt)

Where y is the oscillation perpendicular to the direction of the displacement, A the amplitude, k in wave number and w the angular velocity.

Standing waves are formed when a traveling wave collides with an obstacle and is reflected, in this case the sum of the two waves gives a wave that does not shift in time and fulfills the relationship

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

Where λ is the wavelength, L the distance between the reflection points and n the number of nodes.

Indicates that for the standing wave the distance between an antinode and the node is x = 0.20 m, therefore

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

              λ = 4L

              λ = 4 0.20

              λ = 0.80 m

The wave number.

              k = [tex]\frac{2\pi }{\lambda }[/tex]  

              k = [tex]\frac{2 \pi }{0.80 }[/tex]  

              k = 2.5π i m⁻¹

In the associated traveling wave, from the graph we can see that the period of the wave is:

             T = 2.8 s

the angular velocity is related to the period.

             [tex]w=\frac{2\pi}{T} \\w = \frac{2\pi }{2.8}[/tex]  

             w = 0.714π  rad/s

indicate the maximum displacement that is the amplitude of the wave.

              A = [tex]y_s[/tex]  

             A = 4.3 cm

Let's write the equation of the traveling wave.

              y = 4.3 sin [π (2.5 x - 0.714 t)]

with this expression we can answer the questions.

a) the displacement of the particle for x = 0.30 m

            y = 4.3 sin (π (2.5 0.30 - 0.714 t))

            y = 4.3 sin π( 0.75 - 0.714 t(

Remember that the angles must be in radians.  For time t = 0 the displacement is

              y = 4.3  0.707

              y = 3.04 cm

 

b) The displacement for x = 0.4m

              y = 4.3 sin (π 2.5 0.4)

              y = 0 cm

c) the transverse velocity of the wave at x = 0.30 m for the time of t = 0.90s

the speed of the wave is

              [tex]v= \frac{dy}{dt} \\v= A w cos ( kx - wt)[/tex]  

              v = 4.3 0.714π cos π(2.5 0.3 - 0.714 t)

              v = 9.65 cos π(0.75 - 0.714 t)

For time t = 0.90 s the velocity is:

            v = 9.65 cos π(0.75 - 0.714 0.9)

            v = 9.65 0.9436

            v = 9.11 cm / s

d) The velocity for time t = 1.3 s

           v = 9.65 cos π(0.75 - 0.714 1.3)

           v = 9.65 0.9999

           v = 9.65 cm / s

In conclusion, using the expressions for the traveling and standing wave, we can find the results for the questions about the displacement and speed of the particle are:

      a) For time zero, the displacement at position x = 0.30 m is y = 3.04 cm

     b) For time zero, the displacement at position x = 0.40 m is: y = 0

      c) For the point x = 0.30 and time t = 0.9s, the velocity of the particle is:

          v = 9.11 cm / s

      d) For the point x = 0.30 and time t = 1.3s, the velocity of the particle is:

          v = 9.65 cm / s

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Related Questions

Croquet ball A moving at 8.3 m/s makes a head on collision with ball B of equal mass and initially at rest. Immediately after the collision ball B moves forward at 6.4 m/s .

What fraction of the initial kinetic energy is lost in the collision?

Answers

Answer:

0.25

Explanation:

That is the right answer.

The fraction of the initial kinetic energy is lost in the collision is 35.3%.

The given parameters:

Initial velocity of ball A = 8.3 m/sInitial velocity of ball B = 0Final velocity of ball B = 6.4 m/s

The initial kinetic energy of the system collision is calculated as follows;

[tex]K.E_i = \frac{1}{2} mv_1_i^2 + \frac{1}{2} mv_2_i^2\\\\K.E_i = \frac{1}{2} (m)(8.3)^2 + \frac{1}{2} (m) (0)^2\\\\K.E_i = 34.445 m[/tex]

The final velocity of ball A after collision is calculated as follows;

[tex]u_1 + v_1 = u_2 + v_2\\\\8.3 + v_1 = 0 + 6.4\\\\v_1 = 6.4 - 8.3\\\\v_1 = -1.9 \ m/s[/tex]

The final kinetic energy of the system after collision is calculated as follows;

[tex]K.E_f = \frac{1}{2} m(-1.9)^2 + \frac{1}{2} m(6.4)^2\\\\K.E_f = 22.285 \ m \[/tex]

The fraction of the initial kinetic energy is lost in the collision is calculated as follows;

[tex]= \frac{K_i - K_f}{K_i} \\\\= \frac{34.445 - 22.285}{34.445} \\\\= 0.353\\\\= 35.3\%[/tex]

Learn more about kinetic energy of elastic collision here: https://brainly.com/question/7694106

Starting from rest, a racecar moves 112 m in the first 7 s of uniform acceleration. What is the car's acceleration?​

Answers

Answer:

215mph

Explanation:

multiply 112milws by 2 and multiple that by seven, then divide it by two

If it takes 50.0 seconds to lift 10.0 Newtons of books to a height of 7.0
meters, calculate the power required to lift it?

Answers

Work of the force = 10 N

Time required for the work = 50 sec

Height = 7 m

We are given with the value of work and time in the question.

Substitute the values in the formula of power and then you'll get the power required.

We know that,

w = Work

p = Power

t = Time

By the formula,

Given that,
Work (w) = 7 m = 70 Joules
Time (t) = 50 sec
Substituting their values,

p = 70/50

p = 1.4 watts

Therefore, the power required is 1.4 watts.

Hope it helps!

At what speed must the electron revolve round the nucleus of
the hydrogen in its ground state in order that it may not be pulled into the
nucleus by electrostatic attraction

Answers

Explanation:

I think this is it, give it a try

A stone is allowed to fall from the top of the tower 100m high at the same moment another stone is projected vertically upward with a velocity of 25m/s. Where and when will the two cross each other​

Answers

Answer:

both the stone will meet at a distance of 80 m from the top of tower.

Explanation:

let "t" = time after which both stones meet

"S" = distance travelled by the stone dropped from the top of tower

(100-S) = distance travelled by the projected stone.

◆ i) For stone dropped from the top of tower

-S = 0 + 1/2 (-10) t²

or, S = 5t²

◆ ii) For stone projected upward

(100 - S) = 25t + 1/2 (-10) t²

= 25t - 5t²

Adding i) and ii) , We get

100 = 25t

or t = 4 s

Therefore, Two stones will meet after 4 s.

◆ ¡¡¡) Put value of t = 4 s in Equation i) , we get

S = 5 × 16

= 80 m.

Thus , both the stone will meet at a distance of 80 m from the top of tower.

(Hope this helps can I pls have brainlist (crown)☺️)

In the popular game "Angry Birds" (refer Figure), for the red bird to hit the green bird with their range 55.0 m, what should be the launch angle (with respect to the horizontal) of the slingshot if the time of flight is 2.50 s? ​

Answers

Answer:

Explanation:

What are we to use as gravity in this imaginary world? I will ASSUME 9.81 m/s² and assume no air resistance.

horizontal velocity component

vx = 55.0/2.50 = 22 m/s

vertical initial velocity component

(12.0 - 10.0) = 0 + vy₀(2.50) + ½(-9.81)2.50²

vy₀ = 13.0625 m/s

θ = arctan(vy₀/vx) = arctan(13.0625/22) = 30.6997225... = 30.7°

Entra vapor a una tobera adiabática con un flujo másico de 250 kg/h. Al entrar el vapor
tiene una energía interna específica de 2510 kJ/kg, una presión de 1378 kPa, un
volumen específico de 0.147 m3
/kg y una velocidad de 5 m/s. Las condiciones de salida
son P= 138.7 kPa, volumen específico de 1.099 m3
/kg y energía interna específica de
2263 kJ/kg. Determine la velocidad de salida.

Answers

Este problema describe el funcionamiento de una tobera adiabática (sin flujo de calor), la cual tiene una corriente de entrada que difiere de la de salida espacial y energéticamente, pero que conservan el mismo flujo másico de 250 kg/h. De este modo, usamos un balance de energía con el fin the determinar la velocidad a la que sale el fluido, según es requerido en el problema:

[tex]mu_1+mP_1v_1+mgh_1+\frac{1}{2} mv_1^2=mu_2+mP_2v_2+mgh_2+\frac{1}{2} mv_2^2[/tex]

En la que se tiene que la incógnita es la velocidad de salida, [tex]v_2[/tex], y es posible simplificar el flujo másico, [tex]m[/tex], al estar como factor común en ambos lados y despreciar la energía potencial (mgh), ya que no hay diferencia de altura significativa entre la entrada y salida de la tobera.

De este modo, es posible reemplazar los valores dados para obtener la siguiente expresión:

[tex]2510\frac{kJ}{kg} +1378kPa*0.147\frac{m^3}{kg} +\frac{1}{2} (5\frac{m}{s} )^2=2263\frac{kJ}{kg}+138.7kPa*1.099\frac{m^3}{kg} +\frac{1}{2} v_2^2[/tex]

Y así, hallar la velocidad de salida como sigue:

[tex]2725.066\frac{kJ}{kg}=2415.431\frac{kJ}{kg} +\frac{1}{2} v_2^2\\\\v_2=\sqrt{2(2725.066\frac{kJ}{kg}-2415.431\frac{kJ}{kg} )} \\\\v_2=24.9\frac{m}{s}[/tex]

Para revisar:

https://brainly.com/question/23265263https://brainly.com/question/14279777

2. An auditorium has 58 seats in the first row, 62 seats in the second row, 66 seats in the third row, and so
on.
a)Find the explicit formula of this arithmetic sequence.

B) find the number of seats in the twentieth row.

Answers

The answer is (a) hope it helped!<3






calculate 18% of 2758 correct to 4 significant figure​

Answers

Answer:

......the answer is 496.4

Which of the following methods of pest control used in IPM systems is effective because of the increase in biological diversity it provides? timed crop planting O weed suppression composted soil amendments deep tilling soil​

Answers

Answer:

Decline in soil health is a serious worldwide problem that decreases complexity and stability of agricultural ecosystems, commonly making them more prone to outbreaks of herbivorous insect pests. Potato (Solanum tuberosum L., Solanaceae) and onion (Allium cepa L., Amaryllidaceae) production is currently characterized by high soil disturbance and heavy reliance on synthetic inputs, including insecticides. Evidence suggests that adopting soil conservation techniques often (but not always) increases mortality and decreases reproductive output for the major insect pests of these important vegetable crops. Known mechanisms responsible for such an effect include increases in density and activity of natural enemy populations, enhanced plant defenses, and modified physical characteristics of respective agricultural habitats. However, most research efforts focused on mulches and organic soil amendments, with additional research needed on elucidating effects and their mechanisms for conservation tillage, cover crops, and arbuscular mycorrhizae.

Introduction

Soil erosion is a serious problem worldwide (Amundson et al., 2015). Although it is often overshadowed in public discourse by other concerns, such as climate change and invasive species, soil deterioration receives considerable attention from the scientific community (Montgomery, 2007; Borrelli et al., 2017; Berhe et al., 2018). In addition to the loss of agricultural productivity, soil erosion has been linked to increased emissions of greenhouse gases and reduced water quality (Amundson et al., 2015; Berhe et al., 2018). Global soil erosion is forecasted to increase in the near future because of cropland expansion, especially in the least economically developed areas (Borrelli et al., 2017).

What is the potential energy when a 100 kg object is raised 4.00 m straight up?

Answers

Answer:

100kg x 4 x 10 = 4000J

Explanation:

What is First Aid.



I mark u brainliest answer​

Answers

Answer:

First aid refers to the emergency or immediate care you should provide when a person is injured or ill until full medical treatment is available.

Explanation:

Jimmy walks to school by traveling 2.0 miles east and 3.0 miles north from his starting point. What are the magnitude and direction of Jimmy's displacement with respect to his original position?

Answers

Answer:

3.6 miles at 56.3 degrees north of east.

Explanation:

If you draw a diagram to calculate Jimmy's displacement, you end up with a triangle.

To solve for the magnitude of Jimmy's displacement with respect to his original position, we need to use the Pythagorean Theorem formula.

Pythagorean Theorem Formula = a^2 + b^2 = c^2

Step by Step to solve for the magnitude of Jimmy's displacement:

1) a^2 + b^2 = c^2

2) 2.0^2 + 3^2 = c^2

3) 4 + 9 = c^2

4) 13 = c^2

5) To find what "c" equals to we need to find the square root of 13.

6) √13 = 3.6 | c = 3.6 miles

To solve for the direction of Jimmy's displacement with respect to his original position, we need to use the following formula:

tan^-1 ∅ (3/2)

tan^-1 ∅ (1.5)

∅ = 56.309° north of east

Therefore, your answer is 3.6 miles at 56.3 degrees north of east.

what are people words in english

Answers

Answer: Plenty

Explanation:

some words are Hi, Banana, Dude and many more

What is the volume of 150g of a substance that has a density of 150g of a substance that has a density of 11.3g/cm3

Answers

Answer:

25.0 cm3

Explanation:

The volume is 25.0 cm3 .

Figure 1 shows that the two vectors in different direction respectively. Determine the resultant for the two vectors ​

Answers

Answer:

15-45/2 = -7.5 downwards

Explanation:

cos(60) x 45 = z

z = 45/2

sin (30) x 70 = y

y = 15

15-45/ 2 = -7.5 downwards

What’s Newton’s second law? Explain and mention some examples in daily life

Answers

Newton’s second law of motion is force equals mass times acceleration.

F = m•a

An example of this would be hitting a ball. If you hit the ball, it will move however fast you hit the ball. The harder you hit the ball, the faster it will move.

hope this helps and brainliest please

Answer:

Newton's second law states that .

The rate of change of linear momentum is directly proportional to the force applied.

Formulically

F=ma

F=Force

m=mass

a=acceleration

The best example is hitting a tennis ball.

Which of the following statements are true?
(a) An object can move even when no force acts on it.

(b) If an object isn't moving, no external forces act on it.

(c) If a single force acts on an object, the object accelerates.

(d) If an object accelerates, a force is acting on it.

(e) If an object isn't accelerating, no external force is acting on it.

(f) If the net force acting on an object is in the positive x-direction, the object moves only in the positive x-direction.

I can't understand how A is true.

Answers

Answer: a, c, d

Explanation: a is true because the object will continue to move even without any force because of inertia (so yh thats why a is true). c is true because an object can accelerate if a single force acts on it. to accelerate (not move), it needs a force to act on it

Would appreciate brainly <3

Explanation:

inertia ...

you push or throw something, and you apply some force to it at that moment, but then it moves and keeps moving even long after you have no more connection to it, and no more force is applied to it.

please consider : we are only talking about moving. not about acceleration.

so, yes, (a) is true.

(c) is true.

(d) is true.

Which quantity or quantities is/are increasing for the object represented by line B?

Answers

Answer:

C. Velocity and Position

Explanation:

The quantities that are increasing for the object represented by line B are velocity and position. The correct option is b.

What is velocity?

The direction of a body or object's movement is defined by its velocity. In its basic form, speed is a scalar quantity. In essence, velocity is a vector quantity. It is the speed at which distance changes. It is the displacement change rate.

Velocity is the pace and direction of an object's movement, whereas speed is the time rate at which an object is traveling along a path. In other words, velocity is a vector, whereas speed is a scalar value.

The graph is given which represents the velocity and time with terms A, B, and C. As opposed to the position-time graph, which describes an object's motion over time, the velocity-time graph reveals an object's speed.

Therefore, the correct option is b. velocity and position.

To learn more about velocity, refer to the link:

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#SPJ2

The question is incomplete. Your most probably complete question is given below:

Velocity and acceleration

velocity and position

velocity only

velocity, position, and acceleration

Which would most likely form a homogenous mixture?

Answers

Answer:

B) a pinch of sugar mixed with cup of water

Answer:

A pinch of sugar mixed with a cup of water

Explanation:

A homogeneous mixtures have a uniform appearance where the parts are pretty evenly spaced out throughout the mixture. Picking out the individual pieces should be hard, ,which is why sugar in water is the best choice.  

what is a capacitor and capacitance.​

Answers

Answer:

Capacitor is a device which used to store energy .

Capacitance is the ratio of the change in an electric charge in a system to the corresponding change in its electric potential.

Explanation:

when a torque is acting on a fly wheel the angular velocity of the fly wheel changes from 10rad/sec to 25rad/sec in 5sec.what will be the magnitudes of the angular acceleration of the fly wheel?​

Answers

Hello!

We can use the following angular kinematic equation to solve:

α = Δω/Δt, or (ωf-ωi)/t

Plug in the given values:

α = (25 - 10)/5 = 15/5 = 3 rad/sec²

The coefficient of kinetic friction for a 22 kg bobsled on a track is 0.10. What force is required to push it down a 5.0 degree incline and achieve a speed of 62 km/h at the end of 75 m

Answers

Hi there!

We must begin by converting km/h to m/s using dimensional analysis:

[tex]\frac{62km}{1hr} * \frac{1hr}{3600sec}*\frac{1000m}{1km} = 17.22 m/s[/tex]

Now, we can use the kinematic equation below to find the required acceleration:

vf² = vi² + 2ad

We can assume the object starts from rest, so:

vf² = 2ad

(17.22)²/(2 · 75) = a

a = 1.978 m/s²

Now, we can begin looking at forces.

For an object moving down a ramp experiencing friction and an applied force, we have the forces:

Fκ  = μMgcosθ  = Force due to kinetic friction

Mgsinθ = Force due to gravity

A = Applied Force

We can write out the summation. Let down the incline be positive.

ΣF = A + Mgsinθ - μMgcosθ

Or:

ma = A + Mgsinθ - μMgcosθ

We can plug in the given values:

22(1.978) = A + 22(9.8sin(5)) - 0.10(22 · 9.8cos(5))

A = 46.203 N

7. A 1.0 kg metal head of a geology hammer strikes a solid rock with a velocity of 5.0 m/s. Assuming all the energy is retained by the hammer head, how much will its temperature increase

Answers

The increase in temperature of the metal hammer is 0.028 ⁰C.

The given parameters:

mass of the metal hammer, m = 1.0 kgspeed of the hammer, v = 5.0 m/sspecific heat capacity of iron, 450 J/kg⁰C

The increase in temperature of the metal hammer is calculated as follows;

[tex]Q = K.E\\\\mc \Delta T = \frac{1}{2} mv^2\\\\\Delta T = \frac{v^2}{2 c}[/tex]

where;

c is the specific heat capacity of the metal hammer

Assuming the metal hammer is iron, c = 450 J/kg⁰C

[tex]\Delta T = \frac{5^2}{2 \times 450} \\\\\Delta T = 0.028 \ ^0C[/tex]

Thus, the increase in temperature of the metal hammer is 0.028 ⁰C.

Learn more about heat capacity here: https://brainly.com/question/16559442

builder places a 3kg hammer on the top of a ladder, which is 4m above the ground. Calculate the gravitational potential energy of the hammer while on the ladder.

Answers

Answer:

[tex]E=mgh[/tex]

[tex]m=3kg[/tex]

[tex]h=4m[/tex]

[tex]g=9.8m/s^{2}[/tex]

[tex]E= 3*4*9.8=117.6J[/tex]

Explanation:

Only substitute amounts to formula.

Hope this helps ;)

Cheers :D

Students perform a set of experiments by placing a block of mass m against a spring, compressing the spring a distance x along a horizontal surface of negligible friction, releasing the block, and measuring the velocity v of the block as it leaves the spring, as shown in Figure 1. The experiments indicate that as x increases, so does v in a linear relationship. The surface is now lifted so that the surface is at an angle θ above the horizontal. Which of the following indicates how the relationship between v and x changes?

Answers

Increasing the angle of inclination of the plane decreases the velocity of the block as it leaves the spring.

The statement that indicates how the relationship between v and x changes is; As x increases, v increases, but the relationship is no longer linear and the values of v will be less for the same value of x.

Reasons:

The energy given  to the block by the spring = [tex]\mathbf{0.5 \cdot k \cdot x^2}[/tex]

According to the principle of conservation of energy, we have;

On a flat plane, energy given to the block = [tex]0.5 \cdot k \cdot x^2[/tex] = kinetic energy of

block = [tex]0.5 \cdot m \cdot v^2[/tex]

Therefore;

0.5·k·x² = 0.5·m·v²

Which gives;

x² ∝ v²

x ∝ v

On a plane inclined at an angle θ, we have;

The energy of the spring = [tex]\mathbf{0.5 \cdot k \cdot x^2}[/tex]

The force of the weight of the block on the string, [tex]F = m \cdot g \cdot sin(\theta)[/tex]

The energy given to the block = [tex]0.5 \cdot k \cdot x^2 - m \cdot g \cdot sin(\theta)[/tex] = The kinetic energy of block as it leaves the spring = [tex]\mathbf{0.5 \cdot m \cdot v^2}[/tex]

Which gives;

[tex]0.5 \cdot k \cdot x^2 - m \cdot g \cdot sin(\theta) = 0.5 \cdot m \cdot v^2[/tex]

Which is of the form;

a·x² - b = c·v²

a·x² + c·v² = b

Where;

a, b, and c are constants

The graph of the equation a·x² + c·v² = b  is an ellipse

Therefore;

As x increases, v increases, however, the value of v obtained will be lesser than the same value of x as when the block is on a flat plane.

Please find attached a drawing related to the question obtained from a similar question online

The possible question options are;

As x increases, v increases, but the relationship is no longer linear and the values of v will be less for the same value of xThe relationship is no longer linear and v will be more for the same value of xThe relationship is still linear, with lesser value of vThe relationship is still linear, with higher value of vThe relationship is still linear, but vary inversely, such that as x increases, v decreases

Learn more here:

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A fat wire, radius a, carries a constant current I, uniformly distributed over its cross section. A narrow gap in the wire, of width w << a, forms a parallel-plate capacitor. Find the magnetic field in the gap, at a distance s < a from the axis.

Answers

The magnetic field in the gap is [tex]\mathbf{B = \dfrac{\mu_oI s}{2 \pi a^2}}[/tex]

In a circuit, a magnetic flux is circulated or followed via a confined environment or passage called a magnetic field gap. The narrow air gap is a non-magnetic component of a magnetic circuit that is normally connected to the remainder of the circuit magnetically in series. This enables a significant amount of magnetic flux to pass via the gap.

The magnetic field in the gap at a distance s < a can be computed by using the formula:

[tex]\mathbf{ \oint Bdl = \mu_oI_{enclosed}}[/tex]

where;

Magnetic flux density = Bdistance = d

[tex]\mathbf{B( 2 \pi d) = \mu _o \oint _s J_d da }[/tex]

where;

[tex]\mathbf{J_d}[/tex] = drift current density

[tex]\mathbf{B( 2 \pi d) = \mu _o J_d \oint _sda }[/tex]

[tex]\mathbf{B( 2 \pi d) = \mu _o J_d (\pi d^2) }[/tex]

Making the magnetic flux density the subject, we have:

[tex]\mathbf{B =\dfrac{ \mu _o J_d (\pi d^2) }{( 2 \pi d)}}[/tex]

[tex]\mathbf{B =\dfrac{ \mu _o J_dd}{ 2 }}[/tex]

Recall that, the drift current density [tex]\mathbf{J_d = \dfrac{I}{\pi a^2}}[/tex]

[tex]\mathbf{B = \dfrac{\mu_o d}{2}(\dfrac{I}{\pi a^2})}[/tex]

Recall that distance in question is said to be (s);

[tex]\mathbf{B = \dfrac{\mu_oI s}{2 \pi a^2}}[/tex]

Learn more about the magnetic field here:

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Question below...............

Answers

Answer:

friction force

Explanation:

force of friction is opposite to the force applied it resist the motion

Danny is competing in the high jump. When he is in the air, his body has _______ energy due to its height, and it has _______ energy due to its motion.

Answers

Answer:

Gravitational potential energy

kinetic energy

A solenoid 91.0 cm long has a radius of 1.50 cm and a winding of 1300 turns; it carries a current of 3.60 A. Calculate the magnitude of the magnetic field inside the solenoid.

Answers

The magnitude of the magnetic field inside the solenoid is [tex]6.46 \times 10^{-3} \ T[/tex].

The given parameters;

length of the solenoid, L = 91 cm = 0.91 mradius of the solenoid, r = 1.5 cm = 0.015 mnumber of turns of the solenoid, N = 1300 current in the solenoid, I = 3.6 A

The magnitude of the magnetic field inside the solenoid is calculated as;

[tex]B = \mu_0 nI\\\\B = \mu_o(\frac{ N}{L} )I\\\\[/tex]

where;

[tex]\mu_o[/tex] is the permeability of frees space = 4π x 10⁻⁷ T.m/A

[tex]B = (4\pi \times 10^{-7}) \times (\frac{1300}{0.91} ) \times 3.6\\\\B = 6.46 \times 10^{-3} \ T[/tex]

Thus, the magnitude of the magnetic field inside the solenoid is [tex]6.46 \times 10^{-3} \ T[/tex].

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