HURRY PLEASE!!

What are dichotomous keys used for?

Science

(a) The impulse reaction of the gadget is h(n) = δ(n+1) + δ(n-1). (b) The system is causal due to the fact the output simplest relies upon present and beyond inputs, x(n+1) and x(n-1), respectively. (c) The system is linear because it satisfies the homes of superposition and homogeneity.

(d) This is an IIR (Infinite Impulse Response) filter out because the output relies upon beyond inputs. (e) The frequency response of the machine is H(z) = \(z^(-1) + z\). (f) The device is BIBO strong considering its miles are linear and causal, and its impulse response is finite, ensuring bounded output for bounded enter.

(a) To discover the impulse reaction of the gadget, we want to enter an impulse signal. An impulse signal is represented as δ(n), that is 1 at n = 0 and 0 for all different values of n.

Let's alternative δ(n) into the given equation:

y(n) = x(n+1) + x(n-1)

For n = 0:

y(0) = x(1) + x(-1)

For n = 1:

y(1) = x(2) + x(0)

For n = -1:

y(-1) = x(0) + x(-2)

For n ≠ 0, 1, -1, y(n) = 0 considering the fact that x(n) = 0.

So, the impulse reaction of the gadget is:

h(n) = δ(n+1) + δ(n-1)

(b) The machine is causal if the output y(n) relies upon only on the present and beyond inputs x(n), x(n-1), x(n-2), and so forth. In this case, due to the fact that y(n) simplest depends on x(n+1) and x(n-1), it does not depend on future inputs. Therefore, the machine is causal.

(c) The system is linear if it satisfies the properties of superposition and homogeneity. Let's confirm these houses:

Superposition:

For two inputs \(x1(n)\)and \(x2(n)\) their corresponding outputs \(y1(n)\) , \(y2(n)\), we've got:

\(y1(n) = x1(n+1) + x1(n-1)\)

\(y2(n) = x2(n+1) + x2(n-1)\)

Now, allow's apply the machine to the linear aggregate of inputs:

\(y(n) = ax1(n) + bx2(n)\)

=\(a(x1(n+1) + x1(n-1)) + b(x2(n+1) + x2(n-1))\)

= \(ax1(n+1) + ax1(n-1) + bx2(n+1) + bx2(n-1)\)

This fits the output received when the system is carried out to the linear mixture of the person inputs. Therefore, the device is linear.

(d) This is an IIR (Infinite Impulse Response) clear out due to the fact the output depends on each contemporary and past input, as visible inside the equation y(n) = x(n+1) + x(n-1).

(e) To locate the frequency response, we will take the Z-remodel of the gadget's impulse response. Applying the Z-transform to the impulse response h(n) = δ(n+1) + δ(n-1), we get:

H(z) =\(Zh(n) = z^(-1) + z\)

The frequency response of the device is the Z-transform of the impulse reaction.

(f) BIBO (Bounded-Input Bounded-Output) stability refers to the belongings of the device to produce bounded output for any bounded input. In this example, because the system is linear and causal, and the impulse reaction is finite, the device is BIBO stable.

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The correct question is:

"A system with input x(n) and output y(n) is characterized by the following equation: y(n) = x(n+1) + x(n-1) (a) Find the impulse response of this system. (b) Is the system causal? Why or why not? (c) Is the system linear? Why or why not? (d) Is this an FIR or IIR filter? (e) Find the frequency response of the system. (f)Is this system BIBO stable? Why or why not?"

As the atomic number of group 15 elements increases, the first ionization energy drops/decrease. (option 1).

The energy needed to remove one electron from an atom is known as the ionization energy.

The members of the group 15 are N, P, As, Sb, Bi, and Mc.

The initial ionization energy is the force required to expel an electron from a neutral atom in the gaseous state. The general electron configuration for the valence shell of the elements in group 15 is ns² np³, where n is the primary quantum number (the same number of the row in which the element is).

As you move lower in the group, n rises, the valence electrons are farther from the nucleus, which means that their attraction to the nucleus is diminished. As a result, as you move down in the group, the electrons will be more readily removed, requiring less energy to do so.

This enables you to forecast the following arrangement in the initial ionization energies:

N > P > As > Sb > Bi > Mc

The question is incomplete, it should be:

As the first five elements in group 15 are considered in order of increasing atomic number, first ionization energy

(1) decreases

(2) increases

(3) decreases, then increases

(4) increases, then decreases

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The halo of the sky has a large distribution of dark matter. The amount of radiation that dark matter emits is negligible or nonexistent. Surprisingly quickly, stars far from the galactic Centre orbit the sun.

indicating that invisible stuff in the halo is causing gravitational impacts on these stars. The amount of matter in the cosmos that is assumed to be made up of dark matter is thought to be around 85%. Since dark matter does not appear to interact with the electromagnetic radiation, it does not absorb, reflect, or emit electromagnetic radiation, hence the name "dark" matter. Radiation is the emission or transmission of energy via space or a material medium as waves or particles.

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Answer:

TEJ as this is a thing you wont get

As per the given scenario, Ram will cross the river first.

Ram chooses the quickest route and crosses the river without turning. Ram and Hari can both swim in still water at the same speed, hence their respective speeds when crossing the river will be equivalent. Ram will be able to get to the opposing bank the fastest as a result.

Hari, on the other hand, is swimming upstream at an angle and is being dragged downstream by the water.

Hari's overall travel distance will increase due to the downstream current, however his effective speed will drop in comparison to the other bank. Hari will therefore take longer than Ram to get to the opposing bank.

Thus, Ram will cross the river first in this scenario.

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Answer: The charges must be like charges (both positive or both negative).

Explanation:

The diameter should of the double-walled tube be big for an ice box​ is increase the thermal temperature.

Double wall pipes are an important safety tool when storing liquids such as those used in the chemical wastewater food and beverage industries. The majority of jacketed piping applications are for chemical wastewater-contaminated groundwater and chemical process safety.

Tube wall thickness in excess of that required for internal stresses and known loads not only provides a margin of safety against unexpected loads but also increases survival time when corrosion or fatigue mechanisms are in action. increase. Single-layer tubing is lightweight, flexible, and can be installed quickly with less personnel and equipment. The double wall is a smooth inner double wall with a corrugated outer tube.

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The combined abundance of elements heavier than hydrogen and helium in the Sun and the majority of its neighbours is between 1 and 4 percent of the star's mass.

When a star's core runs out of hydrogen, it starts to fuse helium to create ever heavier elements, such carbon and iron. Most open-cluster stars have 1-4% of their mass in the form of heavy elements, according to spectra.

In the centres of most stars, lighter elements like hydrogen and helium are fused to form the most prevalent elements, such as carbon and nitrogen. Yet, the heaviest elements, like iron, can only be created in the vast stars which end their lives in supernova explosions.

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Answer:

The wave takes 3.2 s at 0.75 m s⁻¹ to travel across a slinky. The length of slinky is 2.4 m.

What is a wave?

A wave is defined as a disturbance that travels from one spot to another through a medium. As an example of a wave, consider a slinky wave. When stretched from end to end and maintained at rest, the slinky assumes a natural state known as the equilibrium or rest position. The slinky's coils automatically assume this posture, equally spaced apart. The initial particle is pushed or shifted from its equilibrium or rest state to introduce a wave into the slinky.

Explanation:

As we know, \(speed=\frac{distance}{time}\)

Therefore, \(distance = speed\) × \(time\)

=> d = 0.75 x 3.2

       = 2.4 m

Hence, the distance the wave travels or the length of the slinky is 2.4 m.

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The block would accelerate down the inclined plane. The force applied is greater than the maximum force of static friction. The correct answer is (B).

Angle of inclination of plane, θ = 30, Coefficient of static friction, µs = 0.2, Coefficient of kinetic friction, µk = 0.2The block is stationary, A block (A) remains stationary on the inclined plane, which implies that the force of static friction fsmax acting upwards balances the force of gravity mgsinθ acting downwards.

Using the formula of maximum force of static friction, we get; fsmax = µs x mg cosθ = 0.2 x mg x cos 30 ......(1)Also, the maximum force of static friction, in this case, is less than the force of gravity acting downwards. Hence, the block will slide down the incline.

On substituting the values in eq. (1), we get; fsmax = (0.2) (9.8) (0.866) ≈ 1.69 N. The force of gravity acting on the block will be; Fg = mg sinθ = 0.5mg N. Since the force applied, Fapp is greater than fsmax, the block will accelerate down the plane. So, the correct answer is (B).

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Answer:

A) Kinetic Energy

Explanation:

Therefore, the tension in the string is 2.3658 N.

The buoyant force on the ball is equal to the weight of the water displaced by the ball. Therefore, the buoyant force is given by:

buoyant force = density of water x volume of ball x acceleration due to gravity

buoyant force = 1000 kg/m³ x 60 cm³ x 9.81 m/s²

buoyant force = 0.5886 N

The weight of the ball is given by:

weight = mass x acceleration due to gravity

weight = density x volume x acceleration due to gravity

weight = 507 kg/m³ x 60 cm³ x 0.01 m/cm x 9.81 m/s²

weight = 2.9544 N

Since the ball is in equilibrium, the tension in the string is equal to the weight of the ball minus the buoyant force:

tension = weight - buoyant force

tension = 2.9544 N - 0.5886 N

tension = 2.3658 N

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The upper control limit (UCLp) for the control chart is 0.290, rounded to three decimal places. The lower control limit (LCLp) is 0, rounded to three decimal places.

Based on the developed control limits, we cannot determine whether the process has been in control or not without additional information.

Control charts are used to monitor and control processes by analyzing data and identifying variations. In this case, the control chart is being used to monitor the number of defectives in a process. The question provides the control limits for the chart, which are the upper control limit (UCLp) and the lower control limit (LCLp).

The UCLp is the highest acceptable value for the proportion of defectives in the process, while the LCLp is the lowest acceptable value. In this case, the UCLp is given as 0.290, which means that if the proportion of defectives exceeds this value, it would be considered out of control. The LCLp is given as 0, indicating that there is no lower limit for the proportion of defectives.

However, the question does not provide any data or information about the actual proportion of defectives observed in the process. Without this data, we cannot determine whether the process has been in control or not. To make a determination, we would need to compare the observed proportion of defectives to the control limits provided.

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The total momentum after collision is 143.466 kgm/s.

The momentum of a body is the product of the mass and the velocity of the body.

Assuming the right direction as positive and left direction as negative:

Momentum to the right = 46.3 * 5.24 = +242.612 kgm/s

Momentum to the left = 55.4 * -1.79 = -99.166 kgm/s

Total momentum before collision =  +242.612 kgm/s - 99.166 kgm/s

Total momentum before collision = 143.466 kgm/s

Thus, total momentum after collision is 143.466 kgm/s.

In conclusion, momentum is conserved in an isolated system of colliding bodies.

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Answer:

last is the answer.

increase the voltage in order to send energy faster.

Answer:

W= 20000j

Explanation:

W= fd = 1000N(20m) = 20000j

The third harmonic of a tone with a fundamental frequency of 150 Hz is 450 Hz.

The frequency of an event is the number of times it occurs or repeats in a specific amount of time. Frequency refers to the rate at which waves oscillate or vibrate in the context of waves, such as sound or light waves. It is expressed in hertz (Hz), or the number of cycles per second. A higher frequency denotes more cycles taking place in a specific amount of time, giving rise to higher pitches for sound waves or a bluer color for light waves. Understanding different phenomena relating to waves, communication networks, and the behavior of electromagnetic radiation depends fundamentally on frequency.

The extra frequencies that are generated in addition to a wave's basic frequency are referred to as harmonics. There are harmonics that are integer multiples of the fundamental frequency that are produced when a vibrating item or sound source makes a tone with that fundamental frequency. To find the third harmonic of a tone with a fundamental frequency of 150 Hz, we multiply the fundamental frequency by 3.

Third harmonic frequency = Fundamental frequency × 3

Third harmonic frequency = 150 Hz × 3

Third harmonic frequency = 450 Hz

Therefore, the third harmonic of a tone with a fundamental frequency of 150 Hz is 450 Hz.

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Answer:

B

Explanation:

what could i say except ur welcome

Dear student Concept:

The acceleration due to gravity near the earth surface is almost constant and equal to g=9.8m/s2 Here the option A and D ar... More

No option , The curler apply to the stone to bring it up to speed is  42.85 N.

F = 20*3/1.4 = 42.85 N

3 *3 = 2*u*9.8*40

u = 0.0114

Despite their similarities, distance and displacement have very different meanings, as do speed and velocity. Speed, a scalar quantity, describes "how quickly an object is travelling." You can think of speed as the rate at which an object travels a distance. An object moving quickly has a high speed and travels a fair distance in a brief period of time. In contrast, a slow-moving object travels a comparatively short distance in the same amount of time because of its low speed. Zero speed refers to an object that is completely stationary.

[Speed = Distance Time] is the general formula for calculating an object's speed. The SI speed unit is m/s.

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Answer:A

Explanation:I’m taking the test rn and I got it correct

It is a conductor that allows charge to move freely through it.

Resistivity is a distinctive attribute of any material that may be used to compare different materials based on how well they conduct electric currents. Low conductivity is indicated by high resistance.

The Greek letter rho is frequently used to represent resistance, which is numerically equivalent to the resistance R of a wire-like specimen, multiplied by its cross-sectional area A, and divided by its length.

The ohm is the measurement of resistance. The ratio of area in square meters to length in meters is reduced to merely meters in the metre-kilogram-second (mks) method.

Therefore, It is a conductor that allows charge to move freely through it.

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As the experiment of the motion of the pendulum is performed. The motion of the pendulum is periodic.

This periodic nature can be represented the sinusoidal functions.

In the given graph, the position of the pendulum is repeated after a regular interval of time which can be represented by the sinusoidal function whose value oscillates.

Thus, the sinusoidal function can best fit the data represented in the figure.

Hence, the second option is the correct answer.

The given problem involves calculating the definite integral of a function f(x) over a specific range. The function f(x) is defined differently for different values of x, and the final result of the definite integral \(1^2\)₁ f(x) dx, where x ≤ n, is -cos(n) - (-cos(1)) + 3cos(T) - 3cos(n) + infinity.

To calculate the definite integral 1²₁ f(x) dx, where x ≤ n, we need to evaluate the integral of the given function f(x) over the specified range. The function f(x) has different definitions depending on the value of x. For x ≤ n, the function is sin(x), and for x > n, the function is -3sin(x). Additionally, the function is defined as 2x for values of x greater than a certain threshold T.

To solve this problem, we need to consider the different intervals of the range separately. First, we integrate sin(x) over the interval 1 to n. The integral of sin(x) is -cos(x), so the value of this part of the integral becomes -cos(n) - (-cos(1)).

Next, we need to integrate -3sin(x) over the interval n to T. The integral of -3sin(x) is 3cos(x), so this part of the integral becomes 3cos(T) - 3cos(n).

Lastly, we integrate 2x over the interval T to infinity. The integral of 2x is \(x^2\), so this part of the integral becomes infinity.

Combining these three parts, the final result of the definite integral \(1^2\)₁ f(x) dx, where x ≤ n, is -cos(n) - (-cos(1)) + 3cos(T) - 3cos(n) + infinity.

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Answer:

375m per minute

Explanation:

Just divide 1,500 by 4 and you will get your answer!

Answer:

375 meters per minute

Explanation:

it is that much

Answer:

D.20 W

Explanation:

P=W/t

P=2000 J / 10 sec

p=200 Watts

It is easier to stop a lightly loaded truck than a heavier one with equal speed due to the concept of inertia and the relationship between mass and momentum.

Inertia is the tendency of an object to resist changes in its state of motion. The greater the mass of an object, the greater its inertia. When a truck is in motion, it possesses kinetic energy and momentum.

When we apply brakes to stop a moving truck, we need to counteract its momentum. Momentum is the product of an object's mass and velocity and is a measure of how difficult it is to change the object's motion. The momentum of an object is directly proportional to its mass.

In the case of a heavily loaded truck, it has a greater mass compared to a lightly loaded truck. Consequently, it possesses a greater amount of momentum at the same speed. The greater momentum requires a greater force to stop the truck.

When braking is applied, the force of friction between the truck's tires and the road surface acts as the decelerating force. The force of friction is the same for both the lightly loaded and heavily loaded truck, assuming all other factors remain constant. However, the heavier truck has a greater resistance to changes in motion due to its higher mass and momentum. As a result, it requires a stronger and longer-lasting braking force to overcome its inertia and bring it to a stop.

In summary, the greater mass and momentum of a heavily loaded truck make it more challenging to stop compared to a lightly loaded truck with the same speed. The heavier truck's increased inertia necessitates a greater force to counteract its momentum and bring it to a halt.

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The ball is in the air for 0.05 seconds after it leaves the launcher. By analyzing the impulse experienced by the ball, we can determine the final velocity of the ball

To determine the time the ball is in the air after it leaves the launcher, we can use the concept of impulse. Impulse is defined as the product of force and time, and it is equal to the change in momentum of an object.

Impulse (J) = Force (F) * Time (Δt)

We know that the net force acting on the ball is 60 N and the time it is applied is 0.05 seconds. The mass of the ball is 0.3 kg.

The impulse can also be expressed as the change in momentum (Δp) of the ball:

Δp = m * Δv

Where:

m is the mass of the ball,

Δv is the change in velocity.

Since the ball starts from rest, the change in velocity (Δv) is the final velocity (v) of the ball after it leaves the launcher.

By rearranging the equation for impulse, we can solve for the final velocity:

Δv = J / m = F * Δt / m

= 60 N * 0.05 s / 0.3 kg

Simplifying the calculation:

Δv = 10 m/s

Now, to find the time the ball is in the air after it leaves the launcher, we divide the displacement of the ball by the final velocity.

The displacement of the ball is given by:

Δs = v * t

Since the initial velocity is 0 m/s and the final velocity is 10 m/s, the displacement is:

Δs = 10 m/s * t

We want to find the time (t) when the displacement is zero, as the ball reaches the highest point in its trajectory. Therefore, we set Δs = 0 and solve for t:

10 m/s * t = 0

This implies that the ball is in the air for 0.05 seconds after it leaves the launcher.

The ball remains in the air for 0.05 seconds after it leaves the launcher. By analyzing the impulse experienced by the ball, we can determine the final velocity of the ball and use it to find the time it takes for the ball to reach its maximum height. This calculation helps us understand the motion of objects under the influence of external forces and provides insights into projectile motion.

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If, after focusing in low power, you need to use only the fine adjustment to focus the specimen at higher powers, the microscope is said to be parfocal.

When a microscope is said to be parfocal, it means that once you have achieved a sharp focus on the specimen using the low-power objective lens, you can switch to a higher-power objective lens and only need to use the fine adjustment knob to fine-tune the focus. The concept of parfocality is based on the principle of maintaining the relative distance between the objective lens and the specimen when switching between different magnifications.

In a parfocal microscope, the optical system is designed in such a way that the focal point of each objective lens is nearly the same. This means that when you switch from a lower-power objective to a higher-power one, the focal plane of the specimen remains relatively unchanged. As a result, the specimen will still be relatively in focus, requiring only minor adjustments using the fine adjustment knob to achieve a clear image.

The advantage of a parfocal microscope is that it saves time and effort in refocusing the specimen each time you change the objective lens. Without the parfocal feature, you would need to use the coarse adjustment knob extensively to reestablish focus, which can be time-consuming and increase the risk of losing the specimen's position.

It is important to note that while parfocality simplifies the process of switching between objective lenses, fine adjustments are still necessary to achieve the sharpest focus when moving to higher magnifications. The fine adjustment knob allows for precise control and minor corrections to obtain optimal clarity and detail in the specimen at higher magnifications.

Therefore, the correct answer is parfocal.

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