Frequency more than 20,000 HZ​

Answers

Answer 1

Answer:

dvhd

Explanation:

xhxjjdvcbxhjddvifidid

Answer 2

Answer:

The units of frequency are called hertz (Hz). Humans with normal hearing can hear sounds between 20 Hz and 20,000 Hz. Frequencies above 20,000 Hz are known as ultrasound


Related Questions

what do solar winds and the earths magnetic field create

Answers

Answer:

bc earth rotates

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Explanation:

Answer:

The interaction between the solar wind and Earth's magnetic field, and the influence of the underlying atmosphere and ionosphere, creates various regions of fields, plasmas, and currents inside the magnetosphere such as the plasmasphere, the ring current, and radiation belts.

Explanation:

At the end of the passage, Sarah says to Emma, “You’re not the only one with tricks up your sleeves.” Explain what Sarah means by this. Use information from the passage to support your answer.

Answers

Answer:

Sarah plans to use trickery and cunning to get back at Emma. Sarah thinks she is smarter than Emma.

Explanation:

i think this is right i dont know tho

A certain type of laser emits light that has a frequency of 4.9 x 1014 Hz. The light, however, occurs as a series of short pulses, each lasting for a time of 2.9 x 10-11 s. The light enters a pool of water. The frequency of the light remains the same, but the speed of light slows down to 2.3 x 108 m/s. In the water, how many wavelengths are in one pulse

Answers

Answer:

N = 1.42 × 10⁴ cycles

Explanation:

Given that:

frequency f = 4.9 × 10¹⁴ Hz

Time = 2.9 × 10⁻¹¹ s

Speed = 2.3 × 10⁸ m/s

Recall that:

wavelength [tex]\lambda = \dfrac{c}{f} \\ \\[/tex]

Horizontal distance [tex]\Delta x = ct[/tex]

Number of wavelengths [tex](N) = \dfrac{\Delta x}{\lambda}[/tex]

[tex]N = \dfrac{ct}{c/f} \\ \\ N= ft[/tex]

N = (4.9 × 10¹⁴ cycles/s) (2.9 × 10⁻¹¹ s)

N = 14210

N = 1.42 × 10⁴ cycles

A 0.545-kg ball is hung vertically from a spring. The spring stretches by 3.56 cm from its natural length when the ball is hanging at equilibrium. A child comes along and pulls the ball down an additional 5cm, then lets go. How long (in seconds) will it take the ball to swing up and down exactly 4 times, making 4 complete oscillations before again hitting its lowest position

Answers

Answer:

t = 9.52 s

Explanation:

This is an oscillatory motion exercise, in which the angular velocity is

         w = [tex]\sqrt{ \frac{k}{m} }[/tex]

Let's use hooke's law to find the spring constant, let's write the equilibrium equation

        F_e - W = 0

        F_e = W

        k x = m g

        k = [tex]\frac{m g}{x}[/tex]

        k = 0.545 9.8 /0.0356

        k = 150 N / m

now the angular velocity is related to the period

          W = 2π / T

we substitute

          4π² T² = k /m

          T = 4pi² [tex]\sqrt{ \frac{m}{k} }[/tex]

we substitute

           T = 4 pi² [tex]\sqrt{ \frac{0.545}{150} }[/tex]

            T = 2.38 s

therefore for the spring to oscillate 4 complete periods the time is

            t = 4 T

            t = 4 2.38

            t = 9.52 s

a stationary object experiences two forces as shown in the diagram below

Answers

Answer: the answer is B

Explanation: 80 is not the same as 150 so it will go the way 150 units of force is pulling.

Time Vs Position of Battery Operated Car what type of relationship is shown in the graph?

Answers

A linear, positive relationship.

what must be the mass of a rock if a boy applies a 64N force and causes it to accelerate at 4.51m/s2

Answers

first of all the formula of force is F=ma,so we are searching for m,so we can divide a on both sides F/a=m, after this substitute the values given above 64N/4.51=14.2°Kg

an arrow is shot horizontally from the top of a tower at a speed of 15m/s and hits the ground with a speed of 25m/s. calculate the height of the tower

Answers

The height of the tower is 20.41 m.

To determine the height of the tower, we need to understand the concept of the energy conservation principle since the speed and acceleration due to gravity are involved in the system.

What is the energy conservation principle?

The principle of energy conservation lets us know that in an isolated system, energy can neither be created nor destroyed.

It can be expressed using the formula:

[tex]\mathbf{mgh = \dfrac{1}{2}mv_1^2 = \dfrac{1}{2}mv_2^2}[/tex]

From the parameters given:The initial speed [tex]v_1[/tex] = 15 m/sThe final speed [tex]v_2[/tex]  = 25 m/s

By applying the energy conservation principle, we have:

[tex]\mathbf{gh +\dfrac{1}{2}v_1^2 = \dfrac{1}{2}v_2^2}[/tex]

[tex]\mathbf{h = \dfrac{v_2^2- v_1^2 }{2 \times g}}[/tex]

[tex]\mathbf{h = \dfrac{25^2-15^2 }{2 \times 9.8}}[/tex]

h = 20.41 m

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A textbook of mass 2.10 kg rests on a frictionless, horizontal surface. A cord attached to the book passes over a pulley whose diameter is 0.100 m , to a hanging book with mass 3.10 kg . The system is released from rest, and the books are observed to move a distance 1.29 m over a time interval of 0.850 s . Part A What is the tension in the part of the cord attached to the textbook

Answers

Answer:

the tension in the part of the cord attached to the textbook is 7.4989 N

Explanation:

Given the data in the question;

As illustrated in the image below;

first we determine the value of the acceleration,

along vertical direction; we use the second equation of motion;

y = ut + [tex]\frac{1}{2}[/tex]a[tex]_y[/tex]t²

we substitute;

0 m/s for u, 1.29 m for y, 0.850 s for t,

1.29 = 0×0.850  + [tex]\frac{1}{2}[/tex]×a[tex]_y[/tex]×(0.850)²

1.29 = 0.36125a[tex]_y[/tex]

a[tex]_y[/tex] = 1.29 / 0.36125

a[tex]_y[/tex] = 3.5709 m/s²

Now when the text book is moving with acceleration , the dynamic equation will be;

T₁ = m₁a[tex]_y[/tex]

where m₁ is the mass of the text book ( 2.10 kg )

a[tex]_y[/tex] is the vertical acceleration ( 3.5709 m/s² )

so we substitute

T₁ = 2.10 × 3.5709

T₁ = 7.4989 N

Therefore, the tension in the part of the cord attached to the textbook is 7.4989 N

What is the work done by the 200.-N tension shown if it is used to drag the 150-N crate 25 m across the floor at a constant speed?

Answers

Answer:

0 J

Explanation:

Work equals force times distance, but the force is zero because the crate being dragged will have zero acceleration. Force equals mass times acceleration and since acceleration is zero, force has to equal zero as well. Since the force is zero, the work required also has to be zero.

Boxes A and B are in contact on a horizontal, frictionless surface. Box A has mass 21.0 kg and box B has mass 8.0 kg. A horizontal force of 100N is exerted on box A. What is the magnitude of the force that box A exerts on box B?

Answers

Answer:

2.75 N

Explanation:

Given that,

Box A has a mass 21.0 kg and box B has a mass 8.0 kg.

A horizontal force of 100N is exerted on box A.

Let a be the acceleration of the system. Using second law of motion,

[tex]F=(m_A+m_B)a\\\\a=\dfrac{F}{(m_A+m_B)}\\\\a=\dfrac{10}{(21+8)}\\\\a=0.344\ m/s^2[/tex]

Now applying Newton's second law to box B. So,

[tex]F_A=m_Ba\\\\=8\times 0.344\\\\=2.75\ N[/tex]

So, 2.75 N is the force that box A exerts on box B.

The key concept here is that when two objects interact, the forces they exert on each other are equal in magnitude but opposite in direction. The magnitude of the force that box A exerts on box B is also 100N.

Newton's third law of motion states that for every action, there is an equal and opposite reaction. It is one of the three fundamental principles described by Sir Isaac Newton in his Laws of Motion.

According to Newton's third law, when an object exerts a force on another object, the second object simultaneously exerts a force of equal magnitude but in the opposite direction on the first object. In simpler terms, if object A applies a force on object B, then object B applies an equal force in the opposite direction on object A.

Since boxes A and B are in contact on a frictionless surface, the force exerted on box A will be transmitted to box B. According to Newton's third law of motion, the magnitude of the force that box A exerts on box B will be equal in magnitude but opposite in direction to the force exerted on box A.

Therefore, the magnitude of the force that box A exerts on box B is also 100N.

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A dump truck contains a load of soil. Which action will leave the dump truck's
inertia unchanged?
A. Dump out some of the soil.
B. Add gas to its fuel tank.
C. Add more soil.
D. Increase the force applied by the engine.

Answers

Answer:

D

Explanation:

This will not change the weight and therefore not change the inertia

A dump truck contains a load of soil. The action that will leave the dump truck's inertia unchanged is that increase the force applied by the engine. Hence, option D is correct.

What is inertia?

A body's ability to resist being propelled into motion or, if already moving, to modify the direction or magnitude of its velocity is known as inertia.

An object's lethargy is a passive quality that prevents it from doing anything other than obstructing active forces and torques. The only reason a moving body continues to move is the lack of a force that might slow it down, alter its trajectory, or accelerate it. This is not due to inertia.

A body's inertia moment about a certain axis and its mass, which determine how resistant it is to the application of forces to that axis, respectively, are two statistical measures of inertia.

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Which is true regarding a child standing up for their own rights?

Answers

Answer:

hey mate......looks like the question is incomplete

Plz help

What factors determine
how the speed of the marbles changes in a
collision?

Answers

Answer:

Force,friction,inertia and momentum

Explanation:

The speed that the marble is moving at can be determined by the amount of force used when pushed or pulled and what kind of surface it's on.Momentum is also a factor because of the mass of the marbles.

The work done is a vector quantity and SI base unit is J

Answers

Answer:

Is this your question? Also I think work done is a scalar quantity.

Explanation:

PLEASE HELP! I'LL GIVE BRAINLEST​

Answers

Answer:

1.62 m/s²

Explanation:

Please help due today

Answers

Answer:

8

Explanation:

(8√2)² = x² + x²

8² × √2² = 2x²

64 × 2 = 2x²

128 = 2x²

64 = x²

x = 8

give me brainliest please

Which cloud types would most likely indicate that a thunderstorm is on the way?

cirrus clouds
dull, gray, stratus clouds
cumulus clouds that are small and round
cumulus clouds that are tall with a flat top

Answers

Answer:

The aswer is D

Explanation:

Answer:

d

Explanation:

cumulus clouds that are tall with a flat top

If a weather service map has a circle that
is shaded completely in, what does that
mean about the cloud cover in that area?
A. There is 100% cloud cover in that area.
B. There is 0% cloud cover in that area.
C. There is a good chance of rain.
D. There are sunny skies.

Answers

Answer:

A. There is 100% cloud cover in that area.

Explanation:

Cloud cover is recorded on weather charts by shading in parts of the circle.

If there are no clouds, the circle is left white and if the sky is completely covered in cloud, the circle is shaded completely in which means 100% cloud cover in that area.

A uniform, 4.5 kg, square, solid wooden gate 2.0 m on each side hangs vertically from a frictionless pivot at the center of its upper edge. A 1.3 kg raven flying horizontally at 5.0 m/s flies into this door at its center and bounces back at 2.0 m/s in the opposite direction.

Required:
a. What is the angular speed of the gate just after it is struck by the unfortunate raven?
b. During the collision, why is the angular momentum conserved but not the linear momentum?

Answers

Answer:

its a. and jusing that youl

49. \ A rectangular plate is rotating with a constant angular speed about an axis that passes perpendicularly through one corner, as the drawing shows. The centripetal acceleration measured at corner A is n times as great as that measured at corner B. What is the ratio L1/L2 of the lengths of the sides of the rectangle when n

Answers

Answer:

    [tex]\frac{L_1}{L_2} = \sqrt{(n^2 - 1)}[/tex]

Explanation:

For this interesting problem, we use the definition of centripetal acceleration  

      a = v² / r  

angular and linear velocity are related  

     v = w r  

we substitute  

    a = w² r

the rectangular body rotates at an angular velocity w  

We locate the points, unfortunately the diagram is not shown. In this case we have the axis of rotation in a corner, called O, in one of the adjacent corners we call it A and the opposite corner A  

the distance OB = L₂  

the distance AB = L₁

the sides of the rectangle  

It is indicated that the acceleration in in A and B are related  

      [tex]a_A = n \ a_B[/tex]  

we substitute the value of the acceleration  

    w² r_A = n r_B  

the distance from the each corner is  

    r_B = L₂  

    r_A = [tex]\sqrt{L_1^2 + L_2^2}[/tex]  

we substitute  

   \sqrt{L_1^2 + L_2^2} = n L₂  

    L₁² + L₂² = n² L₂²  

    L₁² = (n²-1) L₂²  

A 25.0kg girl pushes a 50.0kg boy so that he accelerates at 4.00m/s2. What is the force of the boy on the girl? A. 200N B. 100N C. 12.5 D. 400N

Answers

Answer:

a

Explanation:

so the answer is 200N

and I hope it is correct

If a virtual image is formed 10.0 cm along the principle axis from a convex mirror of focal length-15.0 cm, how far is the object from the mirror

Answers

Answer:

U=30cm

Explanation:

All you have to do is to put

Mirror formula , 1/f=1/u + 1/v

You should be careful in sign convention .

Virtual image is negative

we take focal length of convex lens negative even if its not given and so on...

Calculate the acceleration of a car if the force on the car is 450 Newtons and the mass is 1300 kilograms.

Answers

[tex] \Large {\underline { \sf {Required \; Solution :}}}[/tex]

We have

Force, F = 450 NMass of the car, m = 1300 kg

We have been asked to calculate the acceleration of the car.

[tex]\qquad \implies\boxed{\red{\sf{ F = ma }}}\\[/tex]

F denotes Forcem denotes massa denotes acceleration

[tex] \quad \twoheadrightarrow\sf { 450 = 1300a} \\ [/tex]

[tex] \quad \twoheadrightarrow\sf {\cancel{ \dfrac{450}{1300}} = 1300a} \\ [/tex]

[tex]\quad\twoheadrightarrow\boxed{\red{\sf{0.346 \; ms^{-2} = a }}}\\[/tex]

Therefore, acceleration of the car is 0.346 m/.

Sparks occur when the electric field in air exceeds 3 x 106 N/C. This is because free electrons normally present in air are accelerated to such high speeds that their kinetic energy will overcome the potential energy holding other electrons to atoms. When those electrons rearrange themselves after such a collision, a flash of light is emitted. Let us suppose that the work done on an electron must give it an energy of 3 x 10-19 J to cause this ionization. How far does an electron involved in making in a spark travel through the air before it collides with an atom

Answers

Answer:

h = 5.38 10¹⁶ m

Explanation:

Let's start this exercise by assuming that all the potential energy of the electron is converted into kinetic energy, let's use the conservation of energy

starting point. Just before ionization

          Em₀ = U = qE

final point. Right after ionization

           Em_f = K = ½ m v²

Energy is conserved

           Em₀ = Em_f

           q E = ½ m v²

           v² = 2qE / m

Now we can use the relationship between net work and kinetic energy

           W_net = ΔK

net work is the work done by the electron minus the binding energy with the atom, called the work function, Ф = 3 10-19 J

           W - Ф = K_f - K₀

we assume that the electron converts all its initial initial kinetic energy to be zero

           W -Ф = ½ m v² - 0

            W = ½ m v² +Ф

we substitute

           W = 1/2 m 2qE/m + E

           W =  qE +Ф

           W = 1.6 10⁻¹⁹ 3 10⁶ + 3 10⁻¹⁹

            W = 4.8 10⁻¹³ + 3 10⁻¹⁹

           W = 4.8 10⁻¹³ J

When the electron is in air, its kinetic energy can be transformed into gravitational potential energy

           

As the electron is in the air, all work is transformed into scientific energy

           W = K

starting point Em₀ = K = W

end point Em_F = U = m g h

energy conservation Em₀ = Em_f

                       W = m g h

                       h = [tex]\frac{W}{mg}[/tex]

let's calculate

                       h = [tex]\frac{4.8 \ 10^{-13} x}{9.1 \ 10^{-31} \ 9.8 }[/tex]

                       h = 5.38 10¹⁶ m

Electron involved in making in spark travel through the air before it collides with an atom will be at the distance of 5.38 10¹⁶ m.

What is an electric field?

An electric field is an electric property that is connected with any location in space where a charge exists in any form. The electric force per unit charge is another term for an electric field.

Let's begin this exercise by assuming that all of the electron's potential energy is turned into kinetic energy, and then we'll apply the law of conservation of energy.

Energy before ionization;

[tex]\rm Em_0 = U = qE[/tex]

Energy after ionization;

[tex]Em_f = K = \frac{1}{2} mv^2[/tex]

From the law of conservation of energy principle;

[tex]Em_0 = Em_f \\\\ q E =\frac{1}{2} m v^2\\\\ v^2 = \frac{2qE }{m}[/tex]

The relationship between net work and kinetic energy;

[tex]W_{net} = \triangle K[/tex]

The work function is defined as net work, which is the work done by the electron minus the binding energy with the atom.

[tex]W - \phi = K_f - K_0[/tex]

[tex]W = K_f+ \phi[/tex]

[tex]W = \frac{1}{2} m \times \frac{2qE}{m} + E\\ \\W = qE + \phi \\\\ \rm W = 1.6 \times 10^{-19}\times 3 \tims 10^6 3 10⁶ +3 \times 10^{-19} \\\\ W = 4.8 \times 10^{-13}+ 3 \times 10^{-19}\\\\ W = 4.8 \times 10^{-13} J[/tex]

EMF at starting point;

[tex]\rm Em_0 = K = W[/tex]

EMF at the endpoint;

[tex]\rm Em_F = U = m g h[/tex]

From the law of conservation of energy principle;

[tex]Em_0 = Em_f \\\\ W = m g \\\\ h = \frac{W}{mg}\\\\\ h = \frac{4.8 \timjes 10^{-13}}{9.1 \times 10^{-31} \times 9.81 }\\\\ \rm h= 5.38 \times 10^{16}[/tex]

Hence electron involved in making in spark travel through the air before it collides with an atom will be at a distance of 5.38 10¹⁶ m.

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What do we call the small changes that
could result in large future changes?
A. the "butterfly effect"
B. the "snowflake effect"
C. the "ripple effect"
D. the "trickle-down effect"

Answers

Answer:

The "butterfly Effect"

Explanation:

The "butterfly effect" will probably have big changes in the future.

The force of Earths gravity keeps earth in orbit true or false

Answers

Answer:

True

Explanation:

The force of gravity keeps all of the planets in orbit around the sun

True. The force of gravity keeps all of the planets in orbit around the sun.

What is Gravity?

The force that pulls items toward the center of a planet or other entity is called gravity. All of the planets are kept in orbit around the sun by gravity.

Gravity applies to everything that has mass. Gravity is stronger for objects with higher mass. Along with distance, gravity weakens as well. Therefore, the gravitational pull of two things becomes stronger the closer they are to one another.

The mass of the Earth is what creates gravity. The combined gravitational force of all of its mass acts on the mass in your body.

Therefore, True. The force of gravity keeps all of the planets in orbit around the sun.

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Calculate the induced electric field (in V/m) in a 52-turn coil with a diameter of 17 cm that is placed in a spatially uniform magnetic field of magnitude 0.45 T so that the face of the coil and the magnetic field are perpendicular. This magnetic field is reduced to zero in 0.10 seconds. Assume that the magnetic field is cylindrically symmetric with respect to the central axis of the coil. (Enter the magnitude.)

Answers

Answer:

the induced electric field is 9.95 V/m

Explanation:

Given the data in the question;

Number of turns N = 52

Diameter of coil D = 17 cm = 0.17 m

Radius r = D/2 = 0.17/2 = 0.085 m

Now,

cross-section area A of the coil  = πr²

A = π × ( 0.085 m )²

A = 0.0227 m²

Also given that;

Initial magnetic field B₁ = 0.45 T

Final magnetic field B₂ = 0

∴ change in magnetic field ΔB = B₁ - B₂ = 0.45 T - 0 = 0.45 T

Time taken dT = 0.10 seconds

Now, we know that;

Induced emf ∈ = N[tex](\frac{d\eta }{dt} )[/tex]

where η = BAcosθ

We know that, magnetic field is cylindrically symmetric, coil is also perpendicular to magnetic field.

Hence, the angle between B & A is 0°

∴ θ = 0°

Induced emf ε = N[tex](\frac{d }{dt} )BAcos\theta[/tex]

we substitute

ε = N[tex](\frac{d }{dt} )[/tex] (BAcos0°)

A is constant and cos0° = 1

so

ε = NA[tex](\frac{dB }{dt} )[/tex]  

We now substitute in our values;

ε = 52 × 0.0227 m² ×  [tex](\frac{0.45T }{0.10s} )[/tex]

ε  =  5.3118 V

we know that, from the relation between electric and emf

ε = ∫∈.dl or ε = ∈∫dl         { for coil; ∫dl = πD }

so we have;

ε = ∈πD

solve for ∈

∈ = ε/πD

we substitute

∈ = 5.3118 V / ( π × 0.17 m )

∈ = 9.95 V/m

Therefore, the induced electric field is 9.95 V/m

The aircraft wing from problem 6 experiences temperature extremes that span 210 degrees Celsius. The component for the wing will have a length of exactly 3 meters. Testing indicates that the aircraft wing will remain stable only if the component never expands to a length larger than 3.017 meters. If the component is made from the metal alloy in question, will it meet this requirement. An unknown metal alloy is being tested to discover its thermal properties to see if it is suitable for use as a component in an aircraft wing. The alloy is formed into a bar measuring 1 meter in length, and is then heated from its starting temp. of 30°C to a final temperature of 100°C. The length of the heated bar is measured to be exactly 1.002 meters in length.

Required:
What is the coefficient of thermal expansion of the alloy?

Answers

Answer:

α = 2,857 10⁻⁵ ºC⁻¹

Explanation:

The thermal expansion of materials is described by the expression

          ΔL = α Lo ΔT

          α = [tex]\frac{\Delta L}{L_o \ \Delta T}[/tex]

in the case of the bar the expansion is

         ΔL = L_f - L₀

         ΔL=   1.002 -1

         ΔL = 0.002 m

the temperature variation is

         ΔT = 100 - 30

         ΔT = 70º C

we calculate

         α = 0.002 / 1 70

         α = 2,857 10⁻⁵ ºC⁻¹

True or false? Pls help

Answers

False.

Tripling the height will triple the potential energy.

Speed has nothing to do with potential energy.

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