You are designing a sensor that can measure action potentials from the medial cutaneous nerve in your arm. a) Use the tables of Fourier transform pairs and Fourier properties to determine the Fourier 1. Transform X(jw) of an idealized action potential x(t 1002C->u(t 1) b) Calculate X(jw) in terms of frequency w. Plot it. c) If we can ignore the frequency component of IXCja) whose amplitude is less than 1% of its maximum value, what is the highest frequency of X(ja)l? What is the corresponding Nyquist frequency? Sketch and describe how the amplitude of the Fourier transform is distorted when x(t) is sampled at 30 Hz. Based on your result from (c), design an ideal anti-aliasing filter. Specify the type of the filter, its cutoff frequency and sketch how the spectrum of the filtered signal looks like before and after sampling at 30 Hz.

Answer:

135N will cause a 45 kg object to accelerate at a rate of 3.0 m/s2

Explanation:

Force = mass * acceleration

Force = 45 kg * 3 m/s2

Force = 135N

Force=F

Apply Newton's second law

\(\\ \tt\longmapsto F=ma\)

\(\\ \tt\longmapsto F=45(3)\)

\(\\ \tt\longmapsto F=135N\)

Answer:

0.62

Explanation:

Ff = mg = 52 N

Since the speed is constant, the friction

force equals the force exerted by the

girl, 36 N.

Ff = ukFn

so uK = Ff/Fn

= 36 N/52 N

= 0.69

Force is needed to pull the sled across the snow at constant speed is 0.62 N

A force in physics is an effect that has the power to alter an object's motion. An object with mass can change its velocity, or accelerate, as a result of a force. An obvious way to describe force is as a push or a pull. A force is a vector quantity since it has both magnitude and direction.

Given in the question a boy exerts an unknown horizontal force as he pulls a 52 N sled across packed snow. The coefficient of friction is 0.12. If a person weighing 650 N sits on the sled,

Ff = mg = 52 N

Since the speed is constant, the friction force equals the force exerted by the boy, 36 N.

Ff = (uk).(Fn)

so uK = Ff/Fn

= 36 N/52 N

= 0.69 N

Force is needed to pull the sled across the snow at constant speed is 0.62 N

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In order to convert a temperature from Kelvin to degrees Celsius, we must deduct 273.15 from it, and in order to transfer a temperature from degrees Celsius to Kelvin, we must add 273.15 to it.

The formula T (K) = T (°C) + 273.15 can be used to convert between Celsius and Kelvin. These two temperature units are extremely dissimilar to one another. The temperature is expressed in °C on the Celsius scale, which was created by Anders Celsius. Lord Kelvin created the Kelvin scale, and K stands for the Kelvin unit of temperature.

The unit of measurement for temperature on a Celsius scale or a centigrade scale is Celsius. It is expressed as °C and can represent a range of temperatures or the difference between two temperatures.

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Answer:

14m/s

Explanation:

Given data

M1=45kg

U1=2m/s

M2=90kg

U2=7m/s

V1=?

V2=1m/s

Applying the expression for elastic collision

M1U1+M2U2=M1V1+M2V2

Substitute

45*2+90*7=45*V1+90*1

90+630=45V1+90

720-90=45V1

630=45V1

Divide both sides by 46

V1=630/45

V1=7.5m/s

Hence Bruce's speed is 14m/s

Answer:

Radiation Released by Fission

When an atom is split, three types of radiation that can damage living tissues are released. ... Ionization is the transfer of energy to the molecules that make up tissue, breaking chemical bonds and causing damage to cells and to DNA.

Explanation:

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A molecule with 5 bonding domains and 2 lone pairs will have an electron-domain geometry of pentagonal bipyramidal and a molecular geometry of seesaw.

The electron-domain geometry of a molecule with 5 bonding domains and 2 lone pairs can be predicted using the Valence Shell Electron Pair Repulsion (VSEPR) theory. This theory states that electron pairs around a central atom arrange themselves to minimize repulsion, resulting in specific geometries based on the number of electron pairs.

In this case, there are a total of 7 electron pairs: 5 bonding domains and 2 lone pairs. This corresponds to an electron-domain geometry of pentagonal bipyramidal. However, molecular geometry is determined by considering only the bonding domains and ignoring the lone pairs.

To determine the molecular geometry, we must identify the positions of the lone pairs within the pentagonal bipyramidal structure. Lone pairs are typically located in equatorial positions, as these provide more space and minimize repulsion. With 2 lone pairs occupying 2 of the 5 equatorial positions, there will be 3 equatorial bonding domains left.

The molecular geometry, taking into account the 3 equatorial bonding domains and the 2 axial bonding domains, is called seesaw (or disphenoidal). This geometry is characterized by a central atom bonded to two axial atoms and three equatorial atoms, with the axial bonds in a linear arrangement and the equatorial atoms forming a trigonal planar configuration around the central atom.

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Answer:

The net displacement is

\(R= \sqrt{A^2+B^2+2AB \cos \theta}\)

\(=\sqrt{15^2+25^2+2AB \cos 30^\circ} \\\\=\sqrt{225+625+ \cos30^0} \\\\=38.7m\)

Explanation:

Suppose that you walk 15 meters at 30 degrees as measured from the East. Then you walk another 25 meters at 60 degrees from the East what is your net displacement

Given data

A = 15 m

B = 25 m

Angle between the vectors A and B is θ = 30°

The net displacement is

\(R= \sqrt{A^2+B^2+2AB \cos \theta}\)

\(=\sqrt{15^2+25^2+2AB \cos 30^\circ} \\\\=\sqrt{225+625+ \cos30^0} \\\\=38.7m\)

Answer:

The time for both of the balls to reach the ground would be the same.

Explanation:

If no air resistance is present, the rate of descent depends only on how far the object has fallen, no matter how heavy the object is. This means that two objects will reach the ground at the same time if they are dropped simultaneously from the same height. ... In air, a feather and a ball do not fall at the same rate.

The magnitude of the force exerted on the 0.02 C point charge in a 400 N/C electric field is 8 N. It is important to note that the force is a vector quantity, meaning it has both magnitude and direction. In this case, the direction of the force would depend on the polarity of the charge and the direction of the electric field.

The magnitude of the force exerted on a point charge in an electric field can be determined using the formula:

F = q * E

Where: F is the force exerted on the charge, q is the magnitude of the charge, and E is the magnitude of the electric field.

In this case, the magnitude of the electric field is given as 400 N/C, and the magnitude of the point charge is 0.02 C.

Substituting these values into the formula, we get:

F = (0.02 C) * (400 N/C)

Calculating this, we find:

F = 8 N

Therefore, the magnitude of the force exerted on the 0.02 C point charge in a 400 N/C electric field is 8 N. It is important to note that the force is a vector quantity, meaning it has both magnitude and direction. In this case, the direction of the force would depend on the polarity of the charge and the direction of the electric field.

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The effect of the force of the fifth lumbar vertebra can be resolved into two forces which produce the same effect

Reason:

Given parameters in a similar question are;

Weight of head, w₁ = 0.07·w, location (distance from weight) = 0.72 m, angle formed with vertebra = 60°

Weight of arms, w₂ = 0.12·w, location = 0.48 m, angle = 60°

Weight of trunk, w₃ = 0.46·w, location, 0.36 m, angle = 60°

Force of muscle = \(F_M\), location = 0.48 m, angle = 12°

At equilibrium, we have, ∑M = 0, therefore;

0.48×sin(30°)×cos(18°) ×\(F_M\) - 0.48×cos(30°)×sin(18°) ×\(F_M\) = 0.72×sin(60°)×w₁ + 0.48×sin(60°)×w₂ + 0.36×sin(60°)×w₃

Where;

cos(18°) ×\(F_M\) = \(F_{Mx}\)

sin(18°) ×\(F_M\) = \(F_{My}\)

Which gives;

(0.48×sin(30°)×cos(18°) - 0.48×cos(30°)×sin(18))×\(F_M\) = 0.72×sin(60°)×w₁ + 0.48×sin(60°)×w₂ + 0.36×sin(60°)×w₃

\(F_M = \dfrac{0.72 \times sin(60^{\circ}) \times w_1 + 0.48 \times sin(60^{\circ}) \times w_2 + 0.36 \times sin(60^{\circ}) \times w_3}{(0.48 \times sin(30^{\circ})\times cos(18^{\circ}) - 0.48\times cos(30^{\circ})\times sin(18^{\circ})) }\)

Therefore;

\(F_M = \dfrac{0.72 \times sin(60^{\circ}) \times0.07\cdot w + 0.48 \times sin(60^{\circ}) \times 0.12 \cdot w + 0.36 \times sin(60^{\circ}) \times 0.46\cdot w}{(0.48 \times sin(30^{\circ})\times cos(18^{\circ}) - 0.48\times cos(30^{\circ})\times sin(18^{\circ})) }\)

\(F_M = \dfrac{0.236944550476}{0.099797611593} \approx 2.374\)

At equilibrium sum of forces, ∑F = 0

∑Fₓ = \(F_{Mx}\) = cos(18°) ×\(F_M\)

∴ ∑Fₓ = 2.374 × cos(18°) ≈ 2.258·w

\(\sum F_y\) = \(F_{My}\) + w₁ + w₂ + w₃

∴ \(\sum F_y\) = sin(18°) ×\(F_M\)  + w₁ + w₂ + w₃

\(\sum F_y\) ≈ 0.734·w + 0.07·w + 0.12·w + 0.46·w ≈ 1.384·w

\(Force \ on \ vertebra, \ F_v = \sqrt{\left(\sum F_x \right)^2 + \left(\sum F_y \right)^2}\)

Therefore;

\(Force \ on \ vertebra, \ F_v = \sqrt{\left(2.258\right)^2 + \left(1.384 \right)^2} \approx 2.648\)

The force acting on the fifth vertebra, \(F_v\) ≈ 2.648·w

\(The \ direction \ of \ the \ force,\, \theta = tan^{-1} \left(\dfrac{F_{My}}{F_{Mx}} \right)\)

\(\theta = tan^{-1} \left(\dfrac{1.384}{2.258} \right) \approx 31.5^{\circ}\)

The direction of the force, θ ≈ 31.5°

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Answer:

(i) Half

(ii) 3 V

(iii) V₁

Explanation:

(i) The given parameters are;

The circuits have identical resistances

The number of resistors in circuit 1 = 1 resistor

The number of resistors in circuit 2 = 2 resistors

Let 'R' represent the value of each resistor, we have;

The total resistance of circuit 1 = R Ohm

The total resistance of circuit 3 = 2·R Ohm

∴ The total resistance of circuit 1  = (1/2) × The total resistance of circuit 3

∴ The resistance of circuit 1 is half the resistance of circuit 3

(ii) The potential difference of each cell, V = 1.5 volts

The number of cells in circuit 2 = 2 cells

The total potential difference of the cells of circuit 2 = 2 × 1.5 volts = 3 × volts = 3 V.

The voltmeter reading = The potential difference across the cell or cells it is applied

∴ The voltmeter reading on voltmeter, V₂, applied across the cells of circuit 2 = 3 V

(iii) The voltmeter reading V₁ = 1.5 V

The voltmeter reading V₂ = 3 V

The voltmeter reading V₃ = 4.5/(2·R) × R = 2.25 V

Therefore, the voltmeter reading with the smallest volt, is V₁ = 1.5 V

The example that is NOT an example of Simple Harmonic Motion is:

A ball rolling down a hill.

Simple Harmonic Motion is a type of periodic motion where the restoring force is proportional to the displacement from equilibrium and is directed towards the equilibrium point. The classic example of simple harmonic motion is the motion of a mass attached to a spring that is oscillating back and forth. Other examples of simple harmonic motion include a pendulum swinging back and forth and the vibration of a guitar string.

A ball rolling down a hill does not exhibit simple harmonic motion because it does not have a restoring force that is proportional to the displacement from equilibrium. The motion of the ball is affected by the force of gravity, which is not directed towards the equilibrium point, and the frictional force between the ball and the surface of the hill, which is not proportional to the displacement from equilibrium. Therefore, it is not an example of simple harmonic motion.

Answer:

earth and venus

Explanation:

Answer:

Answer:

Speed of sound ways in railroad = 6,135.8 m/s

Explanation:

Given:

Distance cover by sound wave = 2,350 meter

Time taken by sound wave to cover distance = 0.383 seconds

Find:

Speed of sound ways in railroad

Computation:

Speed = Distance / Time

Speed of sound ways in railroad = Distance cover by sound wave / Time taken by sound wave to cover distance

Speed of sound ways in railroad = 2,350 / 0.383

Speed of sound ways in railroad = 6,135.77

Speed of sound ways in railroad = 6,135.8 m/s

The inductor's average induced emf during the course of the 10 ms time period is 100 V. This value signifies the opposing force generated by the inductor as the current rapidly decreases to zero when the circuit switch is opened.

To calculate the average induced electromotive force (emf) in the inductor during the given time interval, we can use the formula:

ε = -L * (ΔI / Δt)

where ΔI is the change in current, Δt  is the change in time, L is the inductance, and ε denotesthe induced emf.

Given that the inductance is 2 H, the initial current is 0.5 A, and the final current is effectively zero (0 A), and the time interval is 10 ms (0.01 s), we can calculate the average induced emf:

ΔI = (0 A - 0.5 A) = -0.5 A

Δt = 0.01 s

ε = -(2 H) * (-0.5 A / 0.01 s) = 100 V

Therefore, the average induced emf in the inductor during the 10 ms time interval is 100 V.

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The angle of corresponding reflected ray is 30.88 degree.

To solve this problem, we can use Snell's law, which states that the ratio of the sines of the angles of incidence and refraction is equal to the ratio of the refractive indices of the two media involved.

Let's denote the angle of incidence in glass as θ₁ and the angle of refraction in carbon tetrachloride as θ₂. The refractive indices of glass and carbon tetrachloride are given as follows:

Refractive index of glass (n₁) = 1.52

Refractive index of carbon tetrachloride (n₂) = 1.46

Using Snell's law, we have:

n₁ * sin(θ₁) = n₂ * sin(θ₂)

Plugging in the given values:

1.52 * sin(30°) = 1.46 * sin(θ₂)

Now we can solve for θ₂:

sin(θ₂) = (1.52 * sin(30°)) / 1.46

θ₂ = arcsin((1.52 * sin(30°)) / 1.46)

Using a calculator, we can evaluate the expression:

θ₂ ≈ 30.88°

Therefore, the angle of the refracted ray in carbon tetrachloride is approximately 30.88 degrees.

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An object's mass rather than its weight is used to indicate the amount of matter it contains because weight is defined as the amount of force experienced by a body. The weight of a body can be different as it depends on where and under what conditions it is kept.

However mass is the measure of the amount of matter contained in a body and it is always constant.

Weight is defined as the amount of force exerted on a body by a moving gravitational field or any other field at the surface of the earth. It is measured in Newton (N)

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The jumper's takeoff speed and hang time are equal to 5.50 m/s and 1.12 seconds respectively.

In this scenario, the takeoff speed is the same as the initial velocity of the jumper and it can be calculated by using the third equation of motion. Also, since the motion is against gravity, the acceleration due to gravity will be negative (g = -9.81 m/s²)

V² = U² - 2gS

U² = V² + 2gS

U² = 0² + 2(9.81)(1.54)

U = √30.22

U = 5.50 m/s.

Next, we would calculate the peak time by using the first equation of motion:

V = U - at

0 = 5.5 - 9.81t

9.81t = 5.5

t = 5.5/9.81

t = 0.56 seconds.

Mathematically, the hang time is double the peak time:

Hang time = 2 × 0.56

Hang time = 1.12 seconds.

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Answer:

B

Explanation:

Answer:

Convection

Explanation:

The method of heat transfer that occurs when particles move through a fluid is called convection.

Conduction is where heat is transferred from one substance to another through direct contact.

Thermal energy is the amount of heat that a substance has.

I hope this helps!

Answer:

333 mAmps      .33 A

Explanation:

v = ir

v/r = i

120 v / 360 ohm = .3333 A

Answer:

A). 0.33A

Explanation:

The average force applied to the ball is 200 N.

The average force applied to the ball is:

F = Δp/Δt

Where F is the average force applied to the ball

Δp is the change in momentum

Δt is the change in time

Change in momentum is given by the formula:

Δp = m * Δv

Where Δp is the change in momentum

m is the mass of the ball Δv is the change in velocity

Change in time is given as Δt = 0.01 s

The mass of the ball is 0.1 kg

The initial velocity of the ball is 20 m/s

The final velocity of the ball is zero because the ball has stopped.

Δv = -20 m/s

Substitute the values in the formula,

Δp = m * ΔvΔp = 0.1 * (-20)Δp = -2 Ns

F = Δp/ΔtF

= (-2 Ns) / (0.01 s)

F = -200 N

The negative sign in the result indicates that the direction of force is opposite to the direction of motion.

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Answer:

The description of the energy of the skier at point W is,

At point W, potential energy is greatest and kinetic energy is least

Explanation:

The given parameters are;

The highest point of the skier's path = Point W

The lowest point of the skier's path = Point X

The height of the peak after point X = Intermediate between point W and point X

The potential energy of the skier, P.E. = m·g·h

The kinetic energy of the skier, K.E. = 1/2·m·v²

The total mechanical energy of the skier, M.E. = P.E. + K.E.= Constant

Where;

m = The mass of the skier

g = The acceleration due to gravity = 9.8 m/s²

h = The height of the skier

v = The velocity of the skier

Therefore, the P.E. of the skier is highest at the highest point of the skier's path which is the point W, where h = Maximum

Similarly, the potential energy of the skier will be lowest at point X which is the lowest point on the skier's path

From P.E. + K.E. = constant, the kinetic energy will be least at point W, where the potential energy is highest.

Therefore, the description of the energy of the skier at point W is that the potential energy is greatest and kinetic energy is least.

Answer:

Dog2 has most kinetic energy .

B:324j

Answer:

15

Explanation:

speed = distance/time

distance = 60 meters

time = 4 seconds

speed = 60/4= 15 ms-¹

Speed:-

Answer:

The tiny object's acceleration will be much greater.

Explanation:

Because of Newton's second law of motion, (which is F=ma or Force= Mass*acceleration), then if the force maintains the same (which in this case it does, because it says 50 N to both obejcts) and  one mass was much greater than the other, then the ould be less for the more massive object and much greater for the lighter object.

For example:  If 50 N were applied to a 500 kg object and a 50 kg object, then theformulas for each (respectively) are:

50 = 500*acceleration

and

50 = 50*acceleration

(Because of Newtons Second Law of Motion)

Then, solving for the equations, we get for equation 1:

Acceleration = .1 m/s^2

And for equation 2:

Accleration = 1 m/s^2

Thus, you can see that more massive objects (when applied he same amount of force as the smaller object) clearly have less accleration than the smaller objects.

The speed v of the piece of ice as it passes over the top of the hill if the normal force exerted on it at this point is half its weight is \(v= \sqrt{\dfrac{gr}{2}}\).

During the sliding over the top of the hill, the forces acting upon the piece of ice are gravitational force, \(F_g\) = mg toward the ground,

and centrifugal force is given by \(F_c = \dfrac{mv^2}{r}\)  acting from the ground,

so the total normal force exerted on it at this point (toward the ground) is half its weight, \(F_n = \dfrac{1}{2} mg\).

Using the Newton's second law of motion,

\(\dfrac{1}{2} mg = mg - \dfrac{mv^2}{r}\)

Solving this,

\(v= \sqrt{\dfrac{gr}{2}}\)

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