An electron in a long, organic molecule used in a dye laser behaves approximately like a quantum particle in a box with width 4.18 nm. Find the emitted photon when the electron makes a transition from the first excited state to the ground state and from the second excited state to the first excited state.

The points in the above graph that are possible long-run equilibria include B. A and B.

A and B are the equilibria. This is because long run equilibrium occurs where the aggregate demand intersects the long run aggregate supply curve. The long run equilibrium will always occur at long-run aggregate supply curve.

The other options are incorrect because they lie at the intersection of short-run aggregate supply curve and aggregate demand curve. They do not lie on long run aggregate supply curves and hence are not long-run equilibrium points.

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a) 12.1 Average.

B) It is important to record measurements to the correct number of significant figures because otherwise, the integrity of the number is compromised.

A) Determine the average length and express the answer using the correct number of significant figures:

12.2+12.1+12=36.3/3 = 12.1 Average.

B)  Based on your average length calculation in part a, discuss the importance of recording measurements to the appropriate number of significant figures:

It is important to record measurements to the correct number of significant figures because otherwise, the integrity of the number is compromised.

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Answer: the slope of the cars position-time graph

Explanation:

Just took the test

Answer:

One possible explanation students can make: The crabs cannot see the plankton they eat near the ocean floor. For the crabs to see the plankton, some color of visible light would need to reach the plankton so that it can be reflected into the crabs' eyes.

Programmable Array Logic (PAL) has a programmable AND array and a fixed OR array.

Programmable Array Logic (PAL)  is a logic device.

Hence PAL has programmable AND array and a fixed OR array.

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Answer:

a)

b is at higher potential

b)

E = 800 V/m

c)

W = -48.0 μ J

Explanation:

Given:

electric field has magnitude E and is directed in the negative x direction

potential difference between point a (at x = 0.60 m) and point b (at x = 0.90 m) = 240 V

a)

Direction of the electric field is in the negative x direction so it is away from the positive charge and the more distance away from direction, the lesser the potential gets. So b is at the higher potential.

b)

When the magnitude of electric field is constant then the potential difference between two points in the field is given by:

Vab = Ed

where

Vab is the potential difference between two points a and b in the field

E is electric field magnitude

d is the distance between two points

Compute distance between two points a and b

d = 0.90 - 0.60

d = 0.30

d = 0.3 m

Since potential difference between two points in the field is given which is 240 V

Compute E:

The above formula becomes:

Vab = Ed

E = Vab/d

  = 240/0.3

E = 800 V/m

c)

The negative charge moved from the higher potential to low so work done on the point charge by electric field is negative.

W = Fd

Electric field = E = F/q

where

E is electric field magnitude

F is electric force on q

q is point charge magnitude

The charge q is either positive or negative and when charge is positive the directions of E and F are same otherwise opposite in case of negative charge.

So to calculate the work done on the point charge by the electric field:

E = F/q

F = Eq

so

W = Fd = Eqd

Now putting the values:

E = 800 V/m

q = - 0.200 μC

d = 0.3 m

W = Eqd

W = 800 * - 0.200 * 0.3

W = 800 * - 0.2 * 10⁻⁶ * 0.3

W = - 48 * 10⁻⁶

W = - 4.8 * 10⁻⁵ J

W = -48.0 μJ

a. The point which is at a higher potential is point a.

b. The magnitude (E) of the uniform electric field is equal to 800 V/m.

c. The work done on the point charge by the electric field is equal to -48 microjoules.

Given the following data:

a. Since the uniform electric field with a magnitude (E) is directed in the negative x direction, the point at a higher potential is point a because the electric field is not close to the positive charge and it is further away from point b by 0.90 meter. Also, the potential difference of an electric field is inversely proportional to the square of the distance from a point charge.

b. To calculate the magnitude (E) of the uniform electric field:

At a constant potential difference, the magnitude (E) of this electric field is given by the formula:

\(V_{ab} = Ed\)

Where:

First of all, we would determine the distance (d).

\(d = d_b - d_a\\\\d = 0.90 -0.60\)

Distance, d = 0.30 meter.

Substituting the given parameters into the formula, we have;

\(240 = 0.3E\\\\E=\frac{240}{0.3}\)

E = 800 V/m

c. To calculate the work done on the point charge by the electric field:

Mathematically, the work done on a point charge in an electric field is given by the formula:

\(W = Fd = Eqd\\\\W = 800 \times (-0.200 \times 10^{-6}) \times 0.3\\\\W = -240 \times 0.200 \times 10^{-6}\\\\W = -48\times 10^{-6}\;Joules\)

Note: 1 microjoules = \(1\times 10^{-6}\) Joules

Work done = -48 microjoules.

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The equipotential surfaces associated with a point charge are concentric shells.

Charged material experiences a force when it is exposed to an electromagnetic field due to the physical property of electric charge. You can have a positive or negative electric charge.

The equipotential surface is the region where all points with the same potential are located. On the equipotential surface, a charge can be moved from one point to another without exerting any effort.

Thus,a point charge's related equipotential surfaces are concentric shells.

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The work done by gas is 0.753 J and change in internal energy is 4.247J

So we are given that mass is 2kg , radius 1 cm and the amount of heat is 5 cm

The piston raised by 2.4cm

As we know that Work done is PΔV

Where ΔV is change in volume

Therefore ΔV =  πr^2 h = π x (.01)^2 x .024 =7.53×10^(-6)m^3

Here pressure is 10^5 pa

So W = \(10^5\times7.53\times10^-(6)\)

Therefore W = 0.753 J

Now coming to change in internal energy

Change in Internal Energy = Heat Added - Energy lost in work

∴ 5J - 0.753 J = 4.247J

Hence the change in internal energy is 4.247 J

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Answer:

A: Linear

Explanation:

Formula for Resistance of a metallic conductor is given as;

R = R_o(1 + αΔT)

Where;

R_o is the original resistance

R is the final resistance after change in temperature

ΔT is change in temperature

α is coefficient of linear expansion

Now, from the formula given, we can see that the change in resistance is directly proportional to the change in temperature.

Thus, the higher the final temperature, the more the change in resistance and the lower the final temperature, the lesser the change in resistance.

Thus, for temperature less than zero, since the change is resistance is directly proportional to the temperature, it means it follows a linear relationship.

To convert one system of unit to another  of derived quantities is not a use of dimensional analysis.

Checking for consistency in the dimensions on both sides of an equation entails looking at the dimensions of the physical quantities involved in a problem, such as length, mass, time, electric charge, and temperature.

The core tenet of dimensional analysis is that physical quantities, such as length, mass, and time, may be described in terms of their basic dimensions.

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All of the options listed are included in the uses of dimensional analysis.

Dimensional analysis is a powerful tool used in physics to:

The dimensional analysis involves analyzing the dimensions of physical quantities and using them to establish relationships between them.

By using the principles of dimensional analysis, we can simplify complex physical problems and gain insights into the behavior of physical systems.

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Answer:

M v^2 / R = centripetal force

For Red: M v^2 / R = 5 * 9^2 / 1.5 = 270

For Blue M v^2 / R = 270 = 25 * 1.8^2 / Rb

So Rb = 25 * 1.8^2 / 270 = .3 m

1.Unit vector in the direction BA: BA/|BA| = (2/3, 2/3, 1/3)

2.The angular velocity (ω) of the body is given as 3 radians per second.

3.Without the position of point P given, it is not possible to write the position vector of P.

4.Left-handed rotation refers to the direction of rotation where the rotation follows the left-hand rule.

5.Position vector of point A: (4, 1, 1)

Position vector of point B: (2, -1, 0)

The direction vector BA = (-2, -2, -1)

1.To determine the unit vector pointing in the direction BA, we subtract the coordinates of point B from the coordinates of point A and normalize the resulting vector.

The direction vector BA is given by:

BA = (4 - 2, 1 - (-1), 1 - 0) = (2, 2, 1)

To obtain the unit vector in the direction of BA, we divide the direction vector by its magnitude:

|BA| = √(2^2 + 2^2 + 1^2) = √(4 + 4 + 1) = √9 = 3

Unit vector in the direction BA: BA/|BA| = (2/3, 2/3, 1/3)

2.The angular velocity (ω) of the body is given as 3 radians per second.

3.Without the position of point P given, it is not possible to write the position vector of P. Please provide the position of point P to proceed with the calculation.

4.Left-handed rotation refers to the direction of rotation where the rotation follows the left-hand rule. In a left-handed coordinate system, if you curl the fingers of your left hand in the direction of rotation, your thumb will point in the direction of the rotation axis. It is the opposite direction to a right-handed rotation.

5.The position vectors of points A and B are:

Position vector of point A: (4, 1, 1)

Position vector of point B: (2, -1, 0)

The direction vector BA can be obtained by subtracting the coordinates of point A from the coordinates of point B:

BA = (2 - 4, -1 - 1, 0 - 1) = (-2, -2, -1)

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Answer:

8 N

Explanation:

The net force formula is F = m×a

a = 2 m/s^2, m2 = 4 kg

So, now we can find F:

F = 4 × 2 = 8 N

It takes 30 s from launch to the top of the rocket's trajectory.

The total time of flight is calculated to be equal to 45 s.

Trajectory is the path that any object with mass in motion follows through space as function of time.

As we know, Vf = Vi + at

d = Vi × t + (1/2) × \(\mathrm{at^2}\)

Vf = final velocity  ; Vi = initial velocity  ; a = acceleration  ; t = time  d = distance

As, F = ma

So, F = mg

ma = mg

a = g

Vf = Vi + at

Vf = 0 + (\(\mathrm{9.81 m/s^2}\)) × 15 s

Vf = 147.15 m/s

So the velocity of the rocket at the instant the rocket motor shuts down is 147.15 m/s.

Vf = Vi + at

0 = 147.15 m/s + (\(\mathrm{-9.81 m/s^2}\)) × t

t = 15 s - (147.15 m/s)/(\(\mathrm{-9.81 m/s^2}\))

t = 30 s

So it takes 30 s from launch to the top of the rocket's trajectory.

total time of flight = 15 s (accelerating upwards) + 30 s (reaching maximum height and falling back to ground)

Total time of flight = 45 s

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Answer:

V = a t      velocity after time t

t = 33.1 / 9.80 = 3.38 sec   (time ball had been falling)

S = 1/2 a t^2 = 55.9 m

So the ball had been falling  for 7 * 8 = 56 m  

The child was 7 floors from the top

Since he was on the eight floor the floors below him were

7 * 8 + 12 = 68 m     total floors below child

68 + 56 = 124 m      total height of building

Total floors in building = 7 + 7 + 1 = 15 floors

PE at top = KE at bottom

KE = m g h = .125 * 9.80 * 124 = 152 Joules

Answer:

The viscosity of the fluid is 1.16 N-s/m²

Explanation:

Given that,

Area = 0.6 m²

Angle = 30°

Velocity = 0.36 m/s

Thickness = 1.8 mm

Weight = 280 N

We need to calculate the viscosity of the fluid

Using balance equation

\(w\sin\theta=\dfrac{\mu\times v}{t}\times A\)

Put the value in the equation

\(280\sin30=\dfrac{\mu\times0.36}{1.8\times10^{-3}}\times(0.6)\)

\(140=\mu\times120\)

\(\mu=1.16\ N-s/m^2\)

Hence. The viscosity of the fluid is 1.16 N-s/m².

The mass of the object is approximately 0.457 kg.

The mass of the object attached to the trolley can be calculated using the principle of conservation of momentum. Since the two trolleys couple together and move as a single system after the collision, the total momentum before and after the collision should be the same. Given the mass of one trolley is 0.80 kg and the initial speed is 1.1 m/s, the momentum before the collision is 0.80 kg * 1.1 m/s = 0.88 kg·m/s. After the collision, the total mass is the sum of the two trolleys, and the final speed is 0.70 m/s.

Using the momentum equation, the mass of the object can be calculated as follows:

Total momentum before collision = Total momentum after collision

0.88 kg·m/s = (0.80 kg + mass of the object) * 0.70 m/s

Solving for the mass of the object, we get:

0.88 kg·m/s = (0.80 kg + mass of the object) * 0.70 m/s

0.88 kg·m/s = 0.56 kg + 0.70 kg * mass of the object

0.88 kg·m/s - 0.56 kg = 0.70 kg * mass of the object

0.32 kg = 0.70 kg * mass of the object

Dividing both sides by 0.70 kg, we find:

mass of the object = 0.32 kg / 0.70 kg = 0.457 kg

The two trolleys collide and couple together, the total momentum before the collision is equal to the total momentum after the collision according to the principle of conservation of momentum.

The momentum of an object is defined as the product of its mass and velocity. In this case, the mass of one trolley is known (0.80 kg) and the initial speed is given (1.1 m/s), allowing us to calculate the momentum before the collision.

After the collision, the two trolleys move together at a new speed (0.70 m/s). By setting the initial momentum equal to the final momentum and solving for the unknown mass of the object, we can find its value.

In the calculation, we subtract the masses of the two trolleys from the total mass in order to isolate the mass of the object.

Dividing the difference in momentum by the product of the known mass and the new speed, we obtain the mass of the object. In this case, the mass of the object is approximately 0.457 kg.

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The initial velocity of the lander, at t = 0 and just before it reaches the lunar surface, is 64 m/s. This is because the velocity of the lander is given by c = 64 m/s.

Velocity is a vector quantity that describes the rate of change of an object's position with respect to time. It is the magnitude of an object's displacement per unit of time, expressed as a speed in a specific direction.

Velocity is calculated by dividing the change in position (displacement) by the time interval in which the displacement occurs. It has units of distance per time (e.g. meters per second). The direction of velocity is the same as the direction of displacement, and its magnitude represents the speed of an object's motion.

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Answer:

paying too much on the black market instead of getting a prescription

Explanation:

i just took the quiz

Answer:

Paying too much on the black market instead of getting a prescription

Explanation:

The rest of the options are risks associated with using legal drugs without medical supervision.

Answer:

B

Explanation:

A tractor is the heaviest

Answer:

The wood block reaches a height of 4.249 meters above its starting point.

Explanation:

The block represents a non-conservative system, since friction between wood block and the ramp is dissipating energy. The final height that block can reach is determined by Principle of Energy Conservation and Work-Energy Theorem. Let suppose that initial height has a value of zero and please notice that maximum height reached by the block is when its speed is zero.

\(\frac{1}{2}\cdot m \cdot v^{2} = m \cdot g\cdot h + \mu_{k}\cdot m\cdot g\cdot s \cdot \sin \theta\)

\(\frac{1}{2}\cdot v^{2} = g\cdot h + \mu_{k}\cdot g\cdot \left(\frac{h}{\sin \theta} \right)\cdot \sin \theta\)

\(\frac{1}{2}\cdot v^{2} = g\cdot h +\mu_{k}\cdot g\cdot h\)

\(\frac{1}{2}\cdot v^{2} = (1 +\mu_{k})\cdot g\cdot h\)

\(h = \frac{v^{2}}{2\cdot (1 + \mu_{k})\cdot g}\) (1)

Where:

\(h\) - Maximum height of the wood block, in meters.

\(v\) - Initial speed of the block, in meters per second.

\(\mu_{k}\) - Kinetic coefficient of friction, no unit.

\(g\) - Gravitational acceleration, in meters per square second.

\(m\) - Mass, in kilograms.

\(s\) - Distance travelled by the wood block along the wooden ramp, in meters.

\(\theta\) - Inclination of the wooden ramp, in sexagesimal degrees.

If we know that \(v = 10\,\frac{m}{s}\), \(\mu_{k} = 0.20\) and \(g = 9.807\,\frac{m}{s^{2}}\), then the height reached by the block above its starting point is:

\(h = \frac{\left(10\,\frac{m}{s} \right)^{2}}{2\cdot (1+0.20)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}\)

\(h = 4.249\,m\)

The wood block reaches a height of 4.249 meters above its starting point.

Answer:

huh

Explanation:

Explanation:

w= mg

weight on earth= 50 * 9.8= 490 Newtons

weight on mars,= 50 * 3.7= 185 Newtons

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Answer:

Different colour lights (RBG) uses additive properties to bring light of a certain color to our eyes.

Explanation:

:))

The classification of the phase changes include the following:

A gas and a liquid= Condensation and evaporation.

A gas and a solid= sublimation and deposition

A liquid and a solid= freezing.

Phase change is defined as the ability of matter to change from one form to another ( solid, liquid or gas) due to changes in its surrounding environment.

The classification of 0hase changes include the following:

Therefore, in the classification of pase change of matter, A gas and a liquid= Condensation and evaporation, a gas and a solid= sublimation and deposition, a liquid and a solid= freezing.

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Answer:

18 m/s

Explanation:

Range of a projectile on level ground is:

R = v₀² sin(2θ) / g

14.3 m = v₀² sin(2×13°) / 9.8 m/s²

v₀ = 17.9 m/s

Rounded to two significant figures, the launch speed was 18 m/s.

If the bullet is launched at an angle of 50 degrees above ground level, the target will be struck. The angle remains the same. The launch angle obtained is 50 degrees.

Given:

The initial shot was fired at an angle of 50 degrees above ground.

The projectile's starting velocity (v) and magnitude of velocity will remain constant if it is shot at the same pace.

Let the angle of the projectile is x,

The horizontal component of velocity can be calculated as follows:

\(v(x) = v * cos(x)\)

We can write:

since the horizontal part of velocity remains constant:

\(v(x1) = v(x2)\)

\(cos(50) = cos(x)\)

\(50 = x\)

Therefore, if the projectile is launched at the same speed at a 50° angle above ground level, it will strike the target.

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The correct response is given when the angle is asked a question.