Learning Goal: To understand that centripetal acceleration is the acceleration that causes motion in a circle. Acceleration is the time derivative of velocity. Because velocity is a vector, it can change in two ways: the length (magnitude) can change and/or the direction can change. The latter type of change has a special name, the centripetal acceleration. In this problem we consider a mass moving in a circle of radius R with angular velocity ω, r⃗ (t)=R[cos(ωt)i^+sin(ωt)j^] =Rcos(ωt)i^+Rsin(ωt)j^. The main point of the problem is to compute the acceleration using geometric arguments. (Figure 1) Part A What is the velocity of the mass at a time t? You can work this out geometrically with the help of the hints, or by differentiating the expression for r⃗ (t) given in the introduction. (Figure 2) Express this velocity in terms of R, ω, t, and the unit vectors i^ and j^. V⃗ (t) = Part Assume that the mass has been moving along its circular path for some time. You start timing its motion with a stopwatch when it crosses the positive x axis, an instant that corresponds to t=0. [Notice that when t=0, r⃗ (t=0)=Ri^. ] For the remainder of this problem, assume that the time t is measured from the moment you start timing the motion. Then the time − t refers to the moment a time t before you start your stopwatch. Part B What is the velocity of the mass at a time − t? Express this velocity in terms of R, ω, t, and the unit vectors i^ and j^. V⃗ (−t) = SubmitMy AnswersGive Up Part C What is the average acceleration of the mass during the time interval from − t to t? (Figure 3) Express this acceleration in terms of R, ω, t, and the unit vectors i^ and j^.

Given, Initial speed of the helicopter = 25 m/s. Final speed of the helicopter = 60 m/s, Time taken by the helicopter to change its speed from 25 m/s to 60 m/s is 5 seconds. Therefore, the acceleration of the helicopter is 7 m/s².Hence, option (A) is correct.

To find the acceleration of the helicopter, we can use the formula:

Acceleration= (Final velocity - Initial velocity)/Time,

Acceleration = (60 m/s - 25 m/s)/5 s,

Acceleration = 35 m/s/5 s,

Acceleration = 7 m/s²

Therefore, the acceleration of the helicopter is 7 m/s².

We know that,

Acceleration = change in velocity / time taken,

Acceleration can be defined as the rate at which an object changes its velocity.

It can be measured in units such as m/s², cm/s², etc.

Here, initial velocity = 25 m/s, Final velocity = 60 m/s time = 5 s.

Hence,

Acceleration= (Final velocity - Initial velocity)/Time

Acceleration = (60 m/s - 25 m/s)/5 s

Acceleration= 35/5

Acceleration= 7 m/s²

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Answer:

Keyboard and Mobile devices

Explanation:

got it right on edge 2021

Answer:

Outlook allows users to insert symbols and characters not located on the  

keyboard

, but they can have shortcut keys.  

Inserting horizontal lines in the message body breaks up different sections and mostly benefits people using Outlook on  

mobile devices

.

Explanation:

Answer:

A malleable material is one in which a thin sheet can be easily formed by hammering or rolling. ... In contrast, ductility is the ability of a solid material to deform under tensile stress.

A 10 kg mass slides down a flat hill that makes an angle of 10 degrees with the horizontal. Therefore, the resultant force on the sled is option (c) 97N.

If friction is negligible, the resultant force on the sled will be calculated below:

We know that gravitational force can be broken into two components - force parallel to the slope and force perpendicular to the slope.

The parallel component is given by

Fg * sin θ = 10*9.8*sin10 = 16.87 N.

The perpendicular component is given by

Fg * cos θ = 10*9.8*cos10 = 96.94 N.

The total force acting on the sled is the vector sum of the two components: Resultant force = √(16.87² + 96.94²) = 97 N.

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n = 3, ℓ = 1, mℓ = 1, ms = -1 is an impossible set of quantum numbers.

What are quantum numbers?

Quantum numbers are a set of values used to describe the unique energy states and properties of electrons in an atom. They provide a way to distinguish and characterize the different electron orbitals within an atom.

Among the given options:

a. n = 3, ℓ = 2, mℓ = 1, ms = -½

b. n = 3, ℓ = 1, mℓ = 1, ms = -1

c. n = 2, ℓ = 0, mℓ = 0, ms = -½

d. n = 1, ℓ = 0, mℓ = 0, ms = -½

Option (a) represents a valid set of quantum numbers. However, options (b), (c), and (d) are impossible sets of quantum numbers.

For option (b), the value of mℓ is not within the allowed range for the given ℓ value. In this case, ℓ = 1, which means that mℓ can have values -1, 0, or 1. The value of mℓ = 1 is outside this range.

For options (c) and (d), the values of n and ℓ are not consistent. According to the rules of quantum numbers, the principal quantum number (n) should be greater than or equal to the azimuthal quantum number (ℓ). However, in both options (c) and (d), the value of n is lower than ℓ, which is not possible.

Therefore, the correct answer is option (b), as it represents an impossible set of quantum numbers.

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Answer:

a. 0.947 m/s^2

b. 1304.54 N

c. 0.0966

Explanation:

mass of car = 13500 N = 13500/9.8 = 1377.55 kg

velocity = 50 km/h = 50,000 m/h = 13.9 m/s

raidus = 204 m

a. centripetal acceleartion = v^2/r = 13.9^2/204 = 0.947 m/s^2

b. centripetal force = m*centripetal acceleration = 1377.55 * 0.947 = 1304.54 N

c. In order for the car to round the curve safely, static friction = centripetal force

static friction = coefficient of friction (mu) * mg = mu* 1377.55*9.8 = 13500mu

13500mu = 1304.54

mu = 1304.54/13500 = 0.0966

The acceleration, force and coefficient of friction is required.

Centripetal acceleration is [tex]0.965\ \text{m/s}^2[/tex]

Centripetal force is [tex]1328\ \text{N}[/tex]

Coefficient of friction is [tex]0.1[/tex]

N = Weight of car = 13500 N

v = Velocity = [tex]50=\dfrac{50}{3.6}=13.89\ \text{m/s}[/tex]

r = Radius = [tex]2\times 10^2\ \text{m}[/tex]

m = Mass of car = [tex]\dfrac{N}{g}[/tex]

g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]

Centripetal acceleration is

[tex]a_c=\dfrac{v^2}{r}\\\Rightarrow a_c=\dfrac{13.89^2}{2\times 10^2}\\\Rightarrow a_c=0.965\ \text{m/s}^2[/tex]

Force is given by

[tex]F_c=ma_c\\\Rightarrow F_c=\dfrac{N}{g}a_c\\\Rightarrow F_c=\dfrac{13500}{9.81}\times 0.965\\\Rightarrow F_c=1328\ \text{N}[/tex]

Coefficient of friction is given by

[tex]\mu=\dfrac{F_c}{N}\\\Rightarrow \mu=\dfrac{1328}{13500}\\\Rightarrow \mu=0.098\approx 0.1[/tex]

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The linear homogeneous recurrence relation with constant coefficients that the sequence is: An = -(-1)n + (3)n or An = (1)n + (3)n

The sequence An = (n-3)(2n) + n(2n) satisfies the linear homogeneous recurrence relation with constant coefficients. In order to prove this, we need to first find the general formula of the sequence.

The formula can be found by replacing n with n+1 as follows:

An+1 = (n+1-3)(2n+2) + (n+1)(2n+2)

An+1 = n(2n+4) + (n+1)(2n+2)

An+1 = 2n² + 4n + 2n² + 4n + n + 2n + 2

An+1 = 4n² + 7n + 2

The characteristic equation of the linear homogeneous recurrence relation is given by:

r² - 2r - 3 = 0

Solving this quadratic equation, we get:r = -1 or r = 3

Hence, the general formula of the sequence is given by:An = A(-1)n + B(3)n

Now, we need to find the values of A and B using the initial values of the sequence. The first two terms of the sequence are given by:

A0 = -6 and A1 = 2

Substituting these values in the general formula, we get:

-6 = A + B2 = -A + 3B

From the above two equations, we can solve for A and B to get:

A = -1 and B = 1

Hence, the linear homogeneous recurrence relation with constant coefficients that the sequence An = (n-3)(2n) + n(2n) satisfies is given by:

An = -(-1)n + (3)n or An = (1)n + (3)n

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Answer:

For which the reasoning of the boy is correct for small heights, but as height increases his analysis is not correct.

Explanation:

The force of gravity comes from Newton's second law with the force the universal attraction

         F = ma

         F = [tex]G \frac{m_1 M}{(R_e +h)^2}[/tex]

we substitute

          [tex]G \frac{m_1 M}{ (R_e+ h)^2}[/tex] = m₁ a

where Re is the radius of the Earth 6.37 106 m

          a = [tex]G\frac{M}{R_e^2} \ ( 1 + \frac{h}{R_e})^{-2}[/tex]

In general, the height is much less than the radius of the earth, therefore the term ha / Re is very small and we can use a series expansion leaving only the first fears.

             (1 + x)⁻² = 1 -2x + [tex]\frac{2 \ 1}{2!}[/tex]  x²

we substitute

          a = g₀ ([tex]1 - 2 \frac{h}{R_e}[/tex] )

with

         g₀ = [tex]G \frac{M}{R_e^2}[/tex]

let's launch the expression.

* For small height compared to the radius of the earth we can neglect the last term

          g = g₀

* For height comparable to the radius of the Earth

          g = g₀  [tex](1 - \frac{2h}{Re} )[/tex]

We see that the acceleration of gravity is decreasing.

For which the reasoning of the boy is correct for small heights, but as height increases his analysis is not correct.

The student's reasoning gone wrong when the analysis is undertaken for the increasing heights, to drop the object.

The given problem is based on the concept of gravity and gravitational force. The force of gravity comes from Newton's second law with the force the universal attraction as,

F = ma

[tex]F=G\dfrac{mM}{(R+h)^{2}}\\\\\\ma = G\dfrac{mM}{(R+h)^{2}}[/tex]

Here, a is the linear acceleration, m is the mass of object, M is the mass of Earth, R is the radius of Earth and h is the height from where the objects will be dropped. Then,

[tex]a = \dfrac{GM}{R^{2}} \times(1+h/R)^{-2}[/tex]

In general, the height is much less than the radius of the earth, therefore the term h/ R is very small, hence can be neglected.

[tex]a = \dfrac{GM}{R^{2}}\\\\a=g = \dfrac{GM}{R^{2}}[/tex]

g is the gravitational acceleration.

For small height compared to the radius of the earth we can neglect the last term as,

a = g

And for the height comparable to radius of Earth,

[tex]a = \dfrac{GM}{R^{2}} \times(1+h/R)^{-2}\\\\a=g \times(1+h/R)^{-2}[/tex]

Clearly, the acceleration of gravity is decreasing, for which the reasoning of the boy is correct for small heights, but as height increases his analysis is not correct.

Thus, we can conclude that the student's reasoning gone wrong when the analysis is undertaken for the increasing heights, to drop the object.

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The speed οf the ball when it reaches the lοwest pοint οf the track, rοlling withοut slipping is √10/7g(R-r).

Speed is a scalar quantity that measures hοw fast an οbject is mοving, withοut cοnsidering its directiοn. Speed is typically expressed in units such as meters per secοnd (m/s), kilοmeters per hοur (km/h), οr miles per hοur (mph).

Given:

The radius οf the ball is r.

The radius οf the track is R.

The acceleratiοn due tο gravity is 9.18 m/s².

The mοment οf inertia οf the spherical ball can be expressed as:

I=2/5m/r²

It is given that the ball is rοlling withοut slipping. The speed οf the ball can be expressed as:

v=rω

At the lοwest pοsitiοn οf the track, the ball has bοth types οf speed, namely angular and linear speed.

The tοtal energy οf the ball in the vertical circle can be expressed as:

cEₜ= Eᵦ+ K.Eₜ+ K.Eᵣ

mgR= mgr+ (1/2)mv²+ (1/2)Iω²

mg(R-r)=  (1/2)mv²+ (1/2)* (2/5) mr²ω²

g(R-r)= (1/2)v²+ (1/5)v²

Here,

Eₜ is the tοtal energy οf the ball οn the track,

Eᵦ is the ball's energy in the vertical circle at the highest pοint,

K.Eₜ is the translatiοnal kinetic energy οf the ball,

K.Eᵣ is the rοtatiοnal kinetic energy οf the ball, and g is the acceleratiοn due tο gravity.

The abοve equatiοn can be further sοlved as:

cg(R-r)= (7/10)v²

v²= (10/7)g (R-r)

v= √(10/7)g (R-r)

Therefοre, the speed οf the ball when it reaches the lοwest pοint οf the track is √10/7g(R-r).

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Some diversification benefits can be achieved by combining securities in a portfolio as long as the correlation between the securities isThe correct answer is C. Between 0 and 1.

Diversification benefits can be achieved by combining securities in a portfolio as long as the correlation between the securities is between 0 and 1. Correlation measures the degree to which two securities move in relation to each other. A correlation of 1 indicates a perfect positive correlation, where the securities move in perfect tandem. A correlation of less than 1 indicates a less than perfect positive correlation, where the securities move somewhat together but not completely. When the correlation is between 0 and 1, it means that the securities have some degree of independence and tend to move differently from each other. This allows for diversification benefits as losses in one security may be offset by gains in another, reducing overall portfolio risk.

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Based on the given information of a lattice parameter and powder X-ray wavelength, the crystal structure can be determined by considering only the FCC and BCC structures as possibilities.

The lattice parameter, denoted as 'a,' represents the distance between the lattice points in a crystal structure. In this case, the given value of 'a' is 0.4997 nm. To determine the crystal structure, we need to compare this lattice parameter with the characteristic values of the FCC (face-centered cubic) and BCC (body-centered cubic) structures.

For the FCC structure, the relationship between the lattice parameter 'a' and the radius of the atoms or ions in the structure is given by a = 4√2r, where 'r' represents the atomic or ionic radius. Similarly, for the BCC structure, the relationship is a = 4√3r.

By rearranging the equations, we can solve for the radius 'r.' For the FCC structure, r = a/(4√2), and for the BCC structure, r = a/(4√3). Substituting the given lattice parameter 'a' into these equations, we can calculate the corresponding radii for each structure.

Next, we compare the calculated radii with the typical atomic or ionic radii for different elements. If the calculated radius matches closely with the known radius of an element, then that element is likely to form the crystal structure.

Lastly, to confirm the crystal structure, we can consider the powder X-ray wavelength (λ) provided. The X-ray diffraction pattern obtained from the powder X-ray experiment can help identify the characteristic peaks for different crystal structures. By comparing the observed diffraction pattern with the known patterns for FCC and BCC structures, we can determine the crystal structure based on the closest match.

In conclusion, by calculating the radii for FCC and BCC structures using the given lattice parameter, and by analyzing the X-ray diffraction pattern obtained from the powder X-ray experiment, the crystal structure can be determined as either FCC or BCC.

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Answer:

C hope it helps

call the client and inform her that she was incorrectly charged.

Answer:

The answer is C. call the client and inform her that she was incorrectky charged

Explanation:

The total work performed around the thermodynamic cycle in Figure 7 is 350 Joules.
In a thermodynamic cycle, the total work performed is equal to the net area enclosed by the cycle on a pressure-volume (PV) diagram. From the given figure, we can observe that the cycle consists of two quasi-static processes: Process 1-2 and Process 2-3.

In Process 1-2, the system undergoes an expansion at constant pressure, represented by the horizontal line between points 1 and 2 on the PV diagram. As the volume increases, work is done by the system, and the work done in this process is given by the equation W_1-2 = PΔV = 2 * 50 = 100 Joules (where P is the constant pressure and ΔV is the change in volume).

In Process 2-3, the system undergoes compression at constant volume, represented by the vertical line between points 2 and 3 on the PV diagram. Work is done on the system during this process, and the work done in this process is given by the equation W_2-3 = -PΔV = -3 * 50 = -150 Joules (where P is the constant pressure and ΔV is the change in volume).

The total work performed around the cycle is the sum of the work done in each process, i.e., 100 Joules + (-150 Joules) = -50 Joules.
The total work performed around the thermodynamic cycle in Figure 7 is -50 Joules.

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Answer:

Batteries are intended to store chemical energy and then it converts into electricidal energy. So overall, batteries produce energy.

Explanation:

Planet A exerts a force on planet B, the magnitude and direction of the gravitational force planet B exerts on planet A the gravitational force exerted by planet B on planet A is the same as the magnitude of the gravitational force exerted by planet A on planet B

Newton's third law states that if object A exerts a force on object B, then object B exerts an equal and opposite force on object A. Hence, if planet A exerts a gravitational force on planet B, then planet B exerts an equal and opposite gravitational force on planet A.The magnitude of the gravitational force exerted by planet B on planet A is the same as the magnitude of the gravitational force exerted by planet A on planet B, this is according to the law of universal gravitation,

This law states that the force of gravity between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers. The direction of the gravitational force exerted by planet B on planet A is towards planet B's center, just as the direction of the gravitational force exerted by planet A on planet B is towards planet A's center. Therefore, we can say that the magnitude and direction of the gravitational force planet B exerts on planet A is equal and opposite to the gravitational force planet A exerts on planet B

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Kinetic energy is the energy that an object possesses due to its motion. Hence, the correct option is [109.5 J].

The formula to calculate kinetic energy is KE = 1/2 mv², where m is the mass of the object and v is its velocity or speed. In this question, the wagon wheel is a ring of mass 77.1 kg and radius 0.630 m. The wagon is moving at 1.22 m/s. We have to calculate the total kinetic energy of the wheel.The velocity of the wagon can be converted into the angular velocity of the wheel by using the formula v = wr. The angular velocity w is calculated as:w = v/rw = 1.22 m/s ÷ 0.630 m ≈ 1.936 rad/sNow that we know the angular velocity of the wheel, we can calculate its total kinetic energy using the formula KE = 1/2 Iw², where I is the moment of inertia of the wheel. For a ring-shaped object, the moment of inertia is given by I = mr².KE = 1/2 Iw²KE = 1/2 (mr²) (w²)KE = 1/2 (77.1 kg) (0.630 m)² (1.936 rad/s)²KE ≈ 109.5 J. Therefore, the total kinetic energy of the wagon wheel is approximately 109.5 J. Note: As per the question, the total kinetic energy of the wheel is required, not of the wagon.

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Answer:

The correct option is (C).

Explanation:

The distance between two charges, [tex]d=10^{-8}\ m[/tex]

We need to find the magnitude of the electrostatic force between two electrons. The formula for the electric force between charges is given by :

[tex]F=\dfrac{kq^2}{r^2}\\\\=\dfrac{9\times 10^9\times (1.6\times 10^{-19})^2}{(10^{-8})^2}\\\\=2.304\times 10^{-12}\ N[/tex]

So, the magnitude of force is equal to [tex]2.30\times 10^{-12}\ N[/tex].

The figure skater spinning with her arms held straight out will have greater rotational speed at her fingertips compared to her shoulders.

When the skater extends her arms straight out, the moment of inertia increases as the mass is distributed farther from the axis of rotation (her body).

According to the conservation of angular momentum, the product of moment of inertia and angular velocity remains constant unless an external torque acts on the system.

Since the moment of inertia increases when her arms are extended, the angular velocity must decrease to maintain the constant angular momentum.

As a result, the rotational speed is higher at her fingertips because they have a larger distance from the axis of rotation compared to her shoulders.

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Answer:

the final velocity of the skater after throwing the snowball is 3.17 m/s.

Explanation:

Given;

mass of the ice skater, m₁ = 72 kg

initial velocity of the ice skater, u₁ = 3.1 m/s

mass of the snowball, m₂ = 0.21 kg

initial speed of the snowball, u₂ = 28.0 m/s

Let the final velocity of the skater after throwing the snowball = v

Apply the principle of conservation of linear momentum to determine v;

m₁u₁ + m₂u₂ = v(m₁ + m₂)

72 x 3.1   +  0.21 x 28  = v(72 + 0.21)

229.08 = v(72.21)

v = 229.08 / 72.21

v = 3.17 m/s

Therefore, the final velocity of the skater after throwing the snowball is 3.17 m/s.

Answer:

0.05 m/s

Explanation:

We start by finding the average velocity of water in the pipe. This is done by saying

R(e) = ρv(avg)d/μ

Where,

R(e) = Reynolds number, and that's 2000

ρ = Density of water, 1000 kg/m³

μ = Viscosity of water, 10^-3

d = diameter of pipe

v(avg) = average velocity

Since we're interested in average velocity, we make v(avg) the subject of formula. So that

V(avg) = R(e).μ/ρ.d

V(avg) = 2000 * 10^-3 / 1000 * 0.08

V(avg) = 2 / 80

V(avg) = 0.025 m/s

The maximum flow rate of water in the pipe usually is twice the average velocity, and as such

V(max) = 2 * V(avg)

V(max) = 2 * 0.025

V(max) = 0.05 m/s

Answer:

Yes

Explanation:

sorry if im wrong

it should be rising and converging

The global winds and moisture belts indicate that large amounts of rainfall occur at the Earth's equator because air is rising and converging.

The equator is a large circle that circles the planet Earth, lying in a plane perpendicular to the axis of the planet and being equally spaced from all four geographic poles.

Rainfall in equatorial regions averages 4000mm per year. Every other raining produces about 22 days of precipitation in a month. The equatorial regions have higher temperatures because solar radiation produces a lot of heat there.

The cold air filters down into the lower levels of the atmosphere because the hot air is less dense here than the cold air. The tropical regions become warmer as a result. In the tropical rain belt, the tropical climate predominates.

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An automobile engine slows down from 4,087 rpm to 1,830 rpm in 1,419 revolutions the magnitude of the angular acceleration of the automobile engine is approximately 77.75 rad/s².

To calculate the magnitude of the angular acceleration, we can use the formula:

Angular acceleration (α) = (ω2 - ω1) / (t2 - t1)

where:

ω1 and ω2 are the initial and final angular velocities, respectively, and

t1 and t2 are the initial and final times, respectively.

Initial angular velocity ω1 = 4087 rpm

Final angular velocity ω2 = 1830 rpm

Number of revolutions (n) = 1419

First, we need to convert the angular velocities from rpm to radians per second (rad/s):

ω1 = (4087 rpm) * (2π rad/1 min) * (1 min/60 s) ≈ 426.97 rad/s

ω2 = (1830 rpm) * (2π rad/1 min) * (1 min/60 s) ≈ 191.46 rad/s

Next, we can calculate the time interval (t2 - t1) using the number of revolutions and the initial and final angular velocities:

t2 - t1 = (n / ω2) - (n / ω1)

t2 - t1 = (1419 / 191.46) - (1419 / 426.97) ≈ 3.3 s

Finally, we can calculate the magnitude of the angular acceleration:

α = (ω2 - ω1) / (t2 - t1)

α = (191.46 rad/s - 426.97 rad/s) / (3.3 s)

α ≈ -77.75 rad/s²

Therefore, the magnitude of the angular acceleration of the automobile engine is approximately 77.75 rad/s².

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Answer:

V = f x λ

Explanation:

The options are confusing

The resistance (R) of the copper wire in the circuit is 1600 ohms, and the potential (V) at the other end of the wire is 21 volts.

To calculate the resistance of the copper wire (R), we use the formula R = (ρ * L) / A, where ρ represents the resistivity of the material, L is the length of the wire, and A is the cross-sectional area. In this case, the length of the copper wire is 8 mm (0.008 m), and the cross-sectional area is 0.3 mm^2 (0.3 * 10^(-6) m^2). With the resistivity of copper being 6 * 10^7 ohm^(-1) m^(-1), we can calculate the resistance as follows: R = (6 * 10^7 ohm^(-1) m^(-1) * 0.008 m) / (0.3 * 10^(-6) m^2), which gives us 1600 ohms. Since the copper wire and the carbon resistor are connected in series in the circuit, the potential difference (V) across each component is the same. In the case of a thick, high-conductivity copper wire, the small drift speed of electrons requires only a very small electric field, resulting in a negligible potential difference along the wire. As a result, the potential throughout the thick copper wire remains practically uniform.

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chemical reaction for fossil fuels:combustion of fuels.

Difference between biomass and fossil fuels:how much carbon dioxide is produced

Comparing biomass with other renewable energy sources:does not have as much energy potential as fossil fuels.

Making energy choices locally: biomass

Answer:

The bright blue and black colors represent the following:

Constructive and destructive interference are represented by the blue and black colors, respectively. When two water waves collide constructively, the resultant wave is bright blue, while when they collide destructively, the resultant wave is black in color.

Explanation:

   When two propagating waves with the same frequency (say, [tex]\nu[/tex]) and wavelength (say,[tex]\lambda[/tex]) but slightly different amplitudes (say, A 1 and A 2) traveling in the same direction interfere or are superimposed on each other (that is incident at the same point or object), a third resultant wave with a different amplitude (increased or decreased) but same wavelength and frequency is generated.  

The direction difference between the two waves determines whether they intervene constructively (increasing the amplitude of the resultant wave) or destructively (increasing the amplitude of the resultant wave) (decreased amplitude of the resultant wave). To put it another way, when the difference in direction between the two waves is of the form -

      [tex]\Delta x = n\lambda , n=0,1,2,.....[/tex] is the order of interference.

The two waves are then assumed to be in phase, and the interference is constructive, resulting in the resultant wave having a larger amplitude (which is the sum of the two amplitudes [tex]A_1 +A_2[/tex]  also known as a maxima). When the difference in direction between the two waves is in the form -

  [tex]\Delta x = (2n-1)\frac{\lambda}{2} , n=0,1,2,.....[/tex] is the order of interference.

The two intervening waves are then said to be out of phase, and the interference is disruptive, resulting in the resultant wave having a lower amplitude (which is the difference between the two amplitudes [tex]A_1+A_2[/tex], also known as a minima).

Hence , the graphical representation of constructive (blue )and destructive (black) is attached.

Answer:

B. blocks 2 & 3.

Explanation:

Block 1 has equal & opposite forces acting on it.

Block 2 has 5N on one side, 3N on the other. It will move in the direction the 5N of force is pushing.

Block 3 has no opposing force.