The most interpersonal constructive passion response to relational conflict is..

Answers

Answer 1
Answer:

Answer:

loyalty

Explanation:


Related Questions

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An engineer is designing a small toy car that a spring will launch from rest along a racetrack. She wants to maximize the kinetic energy of the toy car when it launches from the end of a compressed spring onto the track, but she can make only a slight adjustment to the initial conditions of the car. The speed of the car just as it moves away from the spring onto the track is called the launch speed. Which of the following modifications to the car design would have the greatest effect on increasing the kinetic energy of the car? Explain your reasoning.Decrease the mass of the car slightly.Increase the mass of the car slightly.Decrease the launch speed of the car slightly.Increase the launch speed of the car slightly.
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Two loudspeakers, A and B, are driven by the same amplifier and emit sinusoidal waves in phase. The frequency of the waves emitted by each speaker is 641Hz . You are standing between the speakers, along the line connecting them and are at a point of constructive interference.How far must you walk toward speaker B to move to reach the first point of destructive interference?
Take the speed of sound to be 344 .

Answers

Final answer:

You must walk approximately 0.2685 m, or 26.85 cm, towards speaker B to encounter the first point of destructive interference. This calculation is arrived at by determining the half-wavelength of the sound wave.

Explanation:

Interference occurs when two sound waves from the same source meet. When they constructively interfere, their amplitudes add together creating a louder sound, while when they destructively interfere, they cancel each other out creating a point of silence. Since you are initially in a position of constructive interference, you need to move towards speaker B at a distance that would change the path length difference to be equivalent to a half wavelength.

To find this distance, we first need to find the wavelength from the frequency. The formula for this is:

  • Wavelength = Speed of sound / Frequency

Given the speed of sound is 344 m/s and the frequency is 641 Hz, we find the wavelength to be roughly 0.537 m. A half wavelength, which characterizes the distance needed for destructive interference from total constructive interference, would then be 0.2685 m.

You must walk approximately 0.2685 m, or 26.85 cm, towards speaker B to encounter the first point of destructive interference.

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Final answer:

To find the distance at which the first point of destructive interference occurs, divide the wavelength by 2. In this case, the distance is approximately 0.268 meters or 26.8 centimeters. Therefore, you would need to walk about 26.8 centimeters toward speaker B to reach the first point of destructive interference.

Explanation:

To determine the distance at which the first point of destructive interference occurs, we need to understand the concept of interference and the conditions for constructive and destructive interference. Constructive interference occurs when the waves from both speakers are in phase and add up to create a larger amplitude. Destructive interference occurs when the waves from both speakers are out of phase and cancel each other out, resulting in a smaller amplitude. In this case, since the speakers are emitting waves in phase, the distance at which destructive interference occurs is equal to half the wavelength of the waves.

The wavelength of a wave can be calculated using the formula:
Wavelength = Speed of sound / Frequency

In this case, the frequency is given as 641 Hz and the speed of sound is given as 344 m/s. Plugging in these values into the formula, we get:
Wavelength = 344 m/s / 641 Hz

Solving this, we find that the wavelength is approximately 0.536 meters. To find the distance to the first point of destructive interference, we divide the wavelength by 2:
Distance to first point of destructive interference = Wavelength / 2

Plugging in the calculated wavelength, we get:
Distance to first point of destructive interference = 0.536 meters / 2

Simplifying, we find that the distance is approximately 0.268 meters or 26.8 centimeters. Therefore, you would need to walk about 26.8 centimeters toward speaker B to reach the first point of destructive interference.

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The girlfriend of a PHS1282 student proposes to him and offers him a "golden" ring in the process. However, his girlfriend is a student in the Department of Accountancy and Finance, and would struggle to afford a gold ring. Additionally she is known to have a slightly dodgy character. After initially being flattered by the marriage proposal he therefore gets suspicious about whether the ring is actually made of gold. He therefore bring the ring to the PHS「282 lab and measure its mass to be 2.80± 0.02 g and its volume to be 0.16 ± 0.03 cm3 . Gold has a mass density of 19.3 g/cm3. Could the ring be made of gold? (Explain your answer) . Brass has a mass density of 8.4 g/cm3 to 8.7 g/cm3 (depending on the composition of the alloy). Could the ring be made of brass?

Answers

Answer:

The density of the ring is:

\rho=17.5\pm 3 \, g/cm^3

This means the ring could very well be made of gold, but it is very unlikely that it is made of brass.

Explanation:

For a quantity f(x,y) that depends on other quantities (in this case two) x and y, the error is given by:

\sigma_f=\sqrt{\left((\partial f)/(\partial x)\right)^2\sigma_x^2+\left((\partial f)/(\partial y)\right)^2\sigma_y^2 }

where \sigma_x and \sigma_y are the standard deviations on errors of the variables x and y.

In our case \rho=f(m,V)=(m)/(V) where m is the mass and V is the volume.

Knowing that \sigma_m=0.02 and \sigma_V=0.03 we can estimate the error on the density

\sigma_(\rho)=\sqrt{\left((1)/(V)\left)^2\sigma_m^2+\left((m)/(V^2)\right)^2\sigma_V^2}\approx 3 (values were directly plugged)

The density is by using the given values

\rho=(m)/(V)=(2.80)/(0.16)=17.5 \, g/cm^3

The density with error is given by

\rho=(m)/(V)\pm \sigma_(\rho)=17.5\pm 3 \, g/cm^3

Which means it could go as high as 20.5 or as low as 14.5, Meaning that the ring could very well be made of gold, but it is very unlikely that it is made of brass.

An airplane with a speed of 92.3 m/s is climbing upward at an angle of 51.1 ° with respect to the horizontal. When the plane's altitude is 532 m, the pilot releases a package. (a) Calculate the distance along the ground, measured from a point directly beneath the point of release, to where the package hits the earth.
(b) Relative to the ground, determine the angle of the velocity vector of the package just before impact. (a) Number Units (b) Number Units

Answers

Answer:

a

D =  1162.7 \  m

b

\beta =- 65.55^o

Explanation:

From the question we are told that

  The speed of the airplane is  u  =  92.3 \ m/s

   The  angle is  \theta = 51.1^o

    The altitude of the plane is  d =  532 \  m

Generally the y-component of the airplanes velocity is  

       u_y  =  v *  sin (\theta )

=>     u_y  =   92.3 *  sin ( 51.1 )

=>     u_y  =  71.83  \ m/s

Generally the displacement  traveled by the package in the vertical direction is

       d =  (u_y)t +  (1)/(2)(-g)t^2

=>       -532  = 71.83 t +  (1)/(2)(-9.8)t^2

Here the negative sign for the distance show that the direction is along the negative y-axis

 =>   4.9t^2 - 71.83t - 532 = 0

Solving this using quadratic formula we obtain that

    t =  20.06 \  s

Generally the x-component of the velocity is  

     u_x  =  u  *  cos (\theta)

=>    u_x  =   92.3  *  cos (51.1)

=>   u_x  =   57.96 \ m/s

Generally the distance travel in the horizontal  direction is    

     D =  u_x  *  t

=>   D =  57.96  *   20.06

=>    D =  1162.7 \  m

Generally the angle of the velocity vector relative to the ground is mathematically represented as

       \beta  =  tan ^(-1)[(v_y)/(v_x ) ]

Here v_y is the final  velocity of the package along the vertical  axis and this is mathematically represented as  

     v_y  =  u_y  -   gt

=>  v_y  =  71.83  -    9.8 *  20.06

=>  v_y  =  -130.05 \  m/s  

and  v_x is the final  velocity of the package which is equivalent to the initial velocity u_x

So

       \beta  =  tan ^(-1)[-130.05}{57.96 } ]

       \beta =- 65.55^o

The negative direction show that it is moving towards the south east direction

   

If the absolute pressure of gas is 550.280 kPa, its gauge pressure is

Answers

pressure absolute = pressure gage + pressure atmosphere

Answer:

650.280

Explanation: 100kpa + 550.280kpa

An electron is a subatomic particle (m = 9.11 x 10-31 kg) that is subject to electric forces. An electron moving in the +x direction accelerates from an initial velocity of +6.18 x 105 m/s to a final velocity of 2.59 x 106 m/s while traveling a distance of 0.0708 m. The electron's acceleration is due to two electric forces parallel to the x axis: = 8.87 x 10-17 N, and , which points in the -x direction. Find the magnitudes of (a) the net force acting on the electron and (b) the electric force .

Answers

Complete question:

An electron is a subatomic particle (m = 9.11 x 10-31 kg) that is subject to electric forces. An electron moving in the +x direction accelerates from an initial velocity of +6.18 x 105 m/s to a final velocity of 2.59 x 106 m/s while traveling a distance of 0.0708 m. The electron's acceleration is due to two electric forces parallel to the x axis: F₁ = 8.87 x 10-17 N, and , which points in the -x direction. Find the magnitudes of (a) the net force acting on the electron and (b) the electric force F₂.

Answer:

(a) The net force of the electron, ∑F = 4.07 x 10⁻¹⁷ N

(b) the electric force, F₂ = 4.8 x 10⁻¹⁷ N

Explanation:

Given;

initial velocity of the electron, v_0 = +6.18 x 10⁵ m/s

final velocity of the electron, v_f = 2.59 x 10⁶ m/s

the distance traveled by the electron, d = 0.0708 m

The first electric force, F_1 = 8.87*10^(-17) \ N

(a) The net force of the electron is given as;

∑F = F₁ - F₂ = ma

where;

a is the acceleration of the electron

a = (v_f^2 -v_0^2)/(2d) \n\na = ((2.59*10^6)^2 -(6.18*10^5)^2)/(2(0.0708))\n\na = 4.468*10^(13) \ m/s^2

∑F = ma = (9.11 x 10⁻³¹ kg)(4.468 x 10¹³)

∑F = 4.07 x 10⁻¹⁷ N

(b) the electric force, F₂ is given as;

∑F = F₁ - F₂

F₂ = F₁ - ∑F

F₂ = 8.87 x 10⁻¹⁷ - 4.07 x 10⁻¹⁷

F₂ = 4.8 x 10⁻¹⁷ N

Final answer:

The problem involves calculating the acceleration of an electron, then using Newton's second law to find the net force on the electron. This is used to find the magnitude of a second electric force acting on the electron.

Explanation:

First, we can calculate the acceleration of the electron using the formula a = Δv/Δt, where 'a' is acceleration, 'Δv' is the change in velocity, and 'Δt' is the change in time. In this case, Δv = vf - vi = 2.59 x 106 m/s - 6.18 x 105 m/s = 1.972 x 106 m/s. The time taken by the electron to travel 0.0708 m can be found using the equation d = vi t + 0.5 a t₂. We use these values to get Δt which we use to find 'a'.

Next, let's use Newton's second law F = ma to find the net force acting on the electron. The only forces acting on the electron are electric forces, and we know one them is 8.87 x 10-17 N. If we designate this known force as F₁ then the total force F total = F₁ + F₂ where F₂ is the unknown electric force.

Finally, we can find F₂ = F total - F₁. This gives the magnitude of the second electric force.

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A bicycle racer is going downhill at 11.0 m/s when, to his horror, one of his 2.25 kg wheels comes off when he is 75.0 m above the foot of the hill. We can model the wheel as a thin-walled cylinder 85.0 cm in diameter and neglect the small mass of the spokes. (a) How fast is the wheel moving when it reaches the bottom of hill if it rolled without slipping all the way down? (b) How much total kinetic energy does it have when it reaches bottom of hill?

Answers

Answer:

a.) Speed V = 29.3 m/s

b.) K.E = 1931.6 J

Explanation: Please find the attached files for the solution

Final answer:

The wheel's speed at the bottom of the hill can be found through the conservation of energy equation considering both translational and rotational kinetic energy, while the total kinetic energy at the bottom of the hill is a sum of translational and rotational kinetic energy.

Explanation:

These two questions address the physics concepts of conservation of energy, kinetic energy, and rotational motion. To answer the first question, (a) How fast is the wheel moving when it reaches the bottom of the hill if it rolled without slipping all the way down?, we need to consider the potential energy the wheel has at the top of the hill is completely converted into kinetic energy at the bottom. This includes both translational and rotational kinetic energy. Solving for the final velocity, vf, which would be the speed of the wheel, we get vf = sqrt((2*g*h)/(1+I/(m*r^2))), where g is the acceleration due to gravity, h is the height of the hill, I is the moment of inertia of the wheel, m is the mass of the wheel, and r is the radius of the wheel.

For the second question, (b) How much total kinetic energy does it have when it reaches bottom of the hill?, we use the formula for total kinetic energy at the bottom of the hill, K= 0.5*m*v^2+0.5*I*(v/r)^2. Substituting the value of v found in the first part we find the kinetic energy which we can use the formula provided in the reference information.

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