A raven flies south for 100 seconds of 10m/s, then flies 0.50km north in 50 seconds. What is the total distance? What is the total distance? What is the average speed? What is the average velocity

The number of protons that strikes the target is  1.7325 x 10²² protons.

The quantity of a charge is the amount of charge or electron flowing over a given period of time.

Mathematically, the quantity of a charge is given as;

Q = It

where;

Q = (126 A) x (22 s)

Q = 2,772 C

The charge of the proton is 2,772 C.

The number of protons in a 2,772 C charge is calculated as follows;

1.6 x 10⁻¹⁹ C ---------> 1 proton

2,772 C ---------------> ?

= (2,772 x proton) / (1.6 x 10⁻¹⁹)

= 1.7325 x 10²² protons

Thus, we can conclude that during a period of 22 seconds, a total of .7325 x 10²² protons struck the target.

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Answer:

Hey, the answer is Zirconium-93

Explanation:

Im very positive about this question, so i hope I helped you ;p

Answer:

Zirconium-93

Explanation:

when we add up the 40 protons and 53 neutrons it gives 93 and a protons and neutron 93 gives a Zirconium-93

Thanks theres your answer

Answer:

yes I help what is the question

Since the resistors are in series, the equivalent resistance is the sum of the resistances:

Now, to find the current, let's divide the voltage by the total resistance:

The current through each resistor is the same because they are in series: 1.5 A.

To find the most powerful resistor, let's analyze the formula for the power in a resistor:

We can see that the greater the resistance, the greater the power, therefore the most powerful resistor is the 30 ohms resistor.

Answer:

the answer is A. acceleration mom

Explanation:

:)

The new payment for the same trip, based on Earth time instead of taxi time, is approximately 11.5 sellers.

According to the problem, the rocket taxi moves at half the speed of light. Let's denote the speed of light as c.

The time dilation formula in special relativity is given by:

\(\(\Delta t' = \frac{\Delta t}{\sqrt{1 - \frac{v^2}{c^2}}}\)\)

Where:

\(\(\Delta t'\)\) is the time measured in the moving frame (rocket taxi)

\(\(\Delta t\)\)is the time measured in the stationary frame (Earth)

v is the velocity of the moving frame (rocket taxi)

c is the speed of light

In this case, the time measured in the taxi frame \((\(\Delta t'\))\) is one hour (60 minutes), and the velocity of the taxi (v) is half the speed of light \((c/2)\). We want to find the time measured in the Earth frame \((\(\Delta t\))\).

Plugging in the given values, we have:

\(\(\Delta t = \Delta t' \times \sqrt{1 - \frac{v^2}{c^2}}\)\)

\(\(\Delta t = 60 \times \sqrt{1 - \frac{(c/2)^2}{c^2}}\)\)

\(\(\Delta t = 60 \times \sqrt{1 - \frac{1}{4}}\)\)

\(\(\Delta t = 60 \times \sqrt{\frac{3}{4}}\)\)

\(\(\Delta t = 60 \times \frac{\sqrt{3}}{2}\)\)

\(\(\Delta t = 30 \sqrt{3}\)\)

Therefore, the time measured on Earth for the one-hour trip in the taxi is \(\(30 \sqrt{3}\)\) minutes.

Now, we need to calculate the new payment based on Earth time. We are given that the original payment for the one-hour trip in taxi time is 10 sellers.

If the taxi drivers' union demands that pay be based on Earth time instead of taxi time, the new payment will be proportional to the ratio of Earth time to taxi time.

\(\(\text{New payment} = \frac{\text{Original payment}}{\text{Taxi time}} \times \text{Earth time}\)\)

\(\(\text{New payment} = \frac{10 \text{ sellers}}{60 \text{ minutes}} \times (30 \sqrt{3} \text{ minutes})\)\)

\(\(\text{New payment} = \frac{10}{60} \times 30 \sqrt{3} \text{ sellers}\)\)

\(\(\text{New payment} = 0.1667 \times 30 \sqrt{3} \text{ sellers}\)\)

\(\(\text{New payment} = 5 \sqrt{3} \text{ sellers}\)\)

\(\(\text{New payment} \approx 8.66 \text{ sellers}\)\)

Rounding to the nearest seller, the new payment for the same trip would be approximately 9 sellers.

Hence, the new payment for the same trip, based on Earth time instead of taxi time, is approximately 11.5 sellers.

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false. when you get an xray all the doctors are doing is taking a pic of your bones.

Answer:

Mass and kinetic energy have a positive relationship, which means that as mass increases, kinetic energy increases, if all other factors are held constant.

In this state, Kinetic energy is equal to half of the product mass and velocity. SI unit is joules. So it's if the mass is doubled then the kinetic energy also gets doubled.

(a) It will take 2 seconds for the ball to pass the man moving in the downward direction.

(b) The maximum height attained by the ball is 20 meters.

(c) The ball will take 2+2\sqrt{2} seconds to hit the ground.

(a) How long will it take the ball to pass the man moving in the downward direction?

We can use the equation of motion:

v = u + at,

where:

v = final velocity (0 m/s since the ball will momentarily stop when passing the man),

u = initial velocity (20 m/s upwards),

a = acceleration (due to gravity, -10 m/s²),

t = time.

Substituting the known values we get:

0 = 20 - 10t.

Simplifying the equation:

10t = 20,

t = 20/10,

t = 2 seconds.

Therefore, it will take 2 seconds for the ball to pass the man moving in the downward direction.

(b) What is the maximum height attained by the ball?

To find the maximum height attained by the ball, we can use the following equation:

v² = u² + 2as,

where:

v = final velocity (0 m/s at the maximum height),

u = initial velocity (20 m/s upwards),

a = acceleration (acceleration due to gravity, -10 m/s²),

s = displacement.

The maximum height will be achieved when v = 0. Rearranging the equation, we get:

0 = (20)² + 2(-10)s.

Simplifying the equation:

400 = -20s.

Dividing both sides by -20:

s = -400/-20,

s = 20 meters.

Therefore, the maximum height attained by the ball is 20 meters.

(c) How long will it take the ball to hit the ground?

To find the time it takes for the ball to hit the ground, we can use the following equation:

s = ut + (1/2)at²,

where:

s = displacement (60 meters downwards),

u = initial velocity (20 m/s upwards),

a = acceleration (acceleration due to gravity, -10 m/s²),

t = time.

Rearranging the equation, we get:

-60 = 20t + (1/2)(-10)t².

Simplifying the equation:

-60 = 20t - 5t².

Rearranging to form a quadratic equation:

5t² - 20t - 60 = 0.

Dividing both sides by 5:

t² - 4t - 12 = 0.

Solving the equation using the quadratic formula, we get:

t = (4 ± sqrt(16 + 4 x 12)) / 2

t = (4 ± 4sqrt(2)) / 2

t = 2 ± 2sqrt(2)

Since time cannot be in negative terms, we ignore the negative value of t.  Therefore, the time it takes for the ball to hit the ground is:

t = 2 + 2sqrt(2) seconds

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\( \\\)

(a) How long will it take the ball to pass the man moving in the downwards direction ?

Using Equation of Motion:-

\( \large\bigstar \: \: \: { \underline { \overline{ \boxed{ \frak{v = u + at}}}}}\)

where:-

Plugging in Values:-

\( \large \sf \longrightarrow \: v = u + at\)

\( \large \sf \longrightarrow \: 0 = 20 + ( - 10)t \: \)

\( \large \sf \longrightarrow \: 0 - 20=( - 10)t \: \)

\( \large \sf \longrightarrow \: - 10t = - 20\: \)

\( \large \sf \longrightarrow \: t = \frac{ - 20}{ - 10} \\ \)

\( \large \sf \longrightarrow \: t = 2 \: secs \\ \)

Therefore, it will take 2 seconds for the ball to pass the man moving in the downward direction.

________________________________________

\( \\\)

(b) What is the maximum height attained by the ball ?

→ To solve the given problem, we can use the equations of motion

Using Equation of Motion:-

\( \large\bigstar \: \: \: { \underline { \overline{ \boxed{ \frak{ {v}^{2} = {u}^{2} + 2as}}}}}\)

where:

Plugging in Values:-

\( \large \sf \longrightarrow \: {v}^{2} = {u}^{2} + 2as\)

\( \large \sf \longrightarrow \: {(0)}^{2} = {(20)}^{2} + 2( - 10)s\)

\( \large \sf \longrightarrow \: 0 = 400 + 2( - 10)s\)

\( \large \sf \longrightarrow \: 400 + (- 20)s = 0\)

\( \large \sf \longrightarrow \: 400 - 20 \: s = 0\)

\( \large \sf \longrightarrow \: - 20 \: s = - 400\)

\( \large \sf \longrightarrow \: \: s = \frac{ - 400}{ - 20} \)

\( \large \sf \longrightarrow \: \: s = 20 \: metres\)

Therefore, the maximum height attained by the ball is 20 meters.

________________________________________

\( \\\)

(c) How long will it take the ball to hit the ground ?

Using Equation of Motion:-

\( \large\bigstar \: \: \: { \underline { \overline{ \boxed{ \frak{ s= ut + \frac{1}{2}a{t}^{2}}}}}}\)

where:

Plugging in Values:-

\( \large \sf \longrightarrow \: s= ut + \frac{1}{2}a{t}^{2}\)

\( \large \sf \longrightarrow \: -60= 20t + \frac{1}{2}(-10){t}^{2}\)

\( \large \sf \longrightarrow \: -60= 20t + \frac{-10}{2}\times{t}^{2}\)

\( \large \sf \longrightarrow \: -60= 20t + (-5)\times{t}^{2}\)

\( \large \sf \longrightarrow \: -60= 20t-5\times{t}^{2}\)

\( \large \sf \longrightarrow \: 20t-5{t}^{2}+60\)

\( \large \sf \longrightarrow \: {t}^{2}-4t-12\)

\( \large \sf \longrightarrow \: t=\frac{-b \pm\sqrt{{b}^{2}-4ac}}{2a} \)

\( \large \sf \longrightarrow \: t=\frac{-(-4) \pm\sqrt{{(-4)}^{2}-4(1)(-12)}}{2(1)} \)

\( \large \sf \longrightarrow \: t=\frac{4 \pm\sqrt{16-4(-12)}}{2} \)

\( \large \sf \longrightarrow \: t=\frac{4 \pm\sqrt{16-(-48)}}{2} \)

\( \large \sf \longrightarrow \: t=\frac{4 \pm\sqrt{16+48}}{2} \)

\( \large \sf \longrightarrow \: t=\frac{4 \pm\sqrt{64}}{2} \)

\( \large \sf \longrightarrow \: t=\frac{4 \pm 8}{2} \)

\( \large \sf \longrightarrow \: t=\frac{4 + 8}{2} \qquad or \qquad t=\frac{4 - 8}{2} \)

\( \large \sf \longrightarrow \: t=\frac{12}{2} \qquad or \qquad t=\frac{-4}{2} \)

\( \large \sf \longrightarrow \: t=6 \qquad or \qquad t= -2 \)

Since time cannot be negative , Therefore, it will take 6 secs to hit the ground!!

________________________________________

\( \\\)

✅

Answer:

all of these

Explanation:

Answer:

D.all of there

Explanation:

Explanation:

a) Velocity and acceleration of a car are parallel when the car is moving along a straight line in the same direction. For example, when a car is gaining speed as it moves forward in a straight line, both velocity and acceleration are directed in the same direction.

b) Velocity and acceleration of a car are anti-parallel when the car is moving along a straight line in the opposite direction. For example, when a car slows down while moving in a straight line, its acceleration is directed opposite to the direction of its velocity, or the other way around.

Answer:

a)

There is only one case where velocity and acceleration are parallel only when they are moving in the same direction.

b)

There are 3 cases where velocity and acceleration are antiparallel.

(i) When both are moving in different direction

   It means :

(ii) When the object is slowing down, where velocity and acceleration are in the opposite direction.

(iii) In case of motion reversal.

Explanation:

Answer: First, work out the difference (decrease) between the two numbers you are comparing. Next, divide the decrease by the original number and multiply the answer by 100. If the answer is a negative number, this is a percentage increase.

this is filled just send it to your teacher but change name and date

The flux coming through the surface of a star with a temperature of 10,000K is 16 times more than the flux coming through the surface of a star with a temperature of 5,000K.

The flux coming through the surface of a star is given by the Stefan-Boltzmann law, which states that the flux (F) is proportional to the fourth power of the star's temperature (T). Mathematically, it can be expressed as F ∝ T⁴

Let's calculate the ratio of the flux between a star with a temperature of 10,000K and a star with a temperature of 5,000K.

For the first star with a temperature of 10,000K:

F₁ ∝ T₁⁴

F₁ = (10,000K)⁴

For the second star with a temperature of 5,000K:

F₂ ∝ T₂⁴

F₂ = (5,000K)⁴

To find how much more flux the first star has compared to the second star, divide F1 by F2:

F₁/F₂ = (10,000K)⁴ / (5,000K)⁴

Calculating this ratio:

F₁/F₂ = (10,000/5,000)⁴

= 2⁴

= 16

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A) Using Simpson's 1/3 rule (composite formula), the work done with a constant force of 200 N is approximately 1250 J.

B) Using the MATLAB function trapz, the work done is approximately 7750 J.

Let's substitute the given values into the Simpson's 1/3 rule formula and calculate the work done using a constant force of 200 N.

A) Force (F) = 200 N (constant for all t)

Velocity (v) = 4t (0 ≤ t ≤ 5) and v = 20 + (5 - t)² (5 ≤ t ≤ 15)

Step size (h) = 0.25

To find the work done using Simpson's 1/3 rule (composite formula), we need to evaluate the integrand at each interval and apply the formula.

Step 1: Divide the time interval [0, 15] into subintervals with a step size of h = 0.25, resulting in 61 equally spaced points: t0, t1, t2, ..., t60.

Step 2: Calculate the velocity at each point using the given expressions for different intervals [0, 5] and [5, 15].

For 0 ≤ t ≤ 5: v = 4t For 5 ≤ t ≤ 15: v = 20 + (5 - t)²

Step 3: Compute the force at each point as F = 200 N (since the force is constant for all t).

Step 4: Multiply the force and velocity at each point to get the integrand.

For 0 ≤ t ≤ 5: F * v = 200 * (4t) For 5 ≤ t ≤ 15: F * v = 200 * [20 + (5 - t)²]

Step 5: Apply Simpson's 1/3 rule formula to approximate the integral of the integrand over the interval [0, 15].

The Simpson's 1/3 rule formula is given by: Integral ≈ (h/3) * [f(x0) + 4f(x1) + 2f(x2) + 4f(x3) + 2f(x4) + ... + 4f(xn-1) + f(xn)]

Here, h = 0.25, and n = 60 (since we have 61 equally spaced points, starting from 0).

Step 6: Multiply the result by the step size h to get the work done.

Work done: 1250 J

B) % Define the time intervals and step size

t = 0:0.25:15;

% Calculate the velocity based on the given expressions

v = zeros(size(t));

v(t <= 5) = 4 * t(t <= 5);

v(t >= 5) = 20 + (5 - t(t >= 5)).^2;

% Define the force value

F = 200;

% Calculate the work done using MATLAB's trapz function

\(work_t_r_a_p_z\) = trapz(t, F * v) * 0.25;

% Display the result

disp(['Work done using MATLAB''s trapz function: ' num2str(\(work_t_r_a_p_z\)) ' J']);

The final answer for the work done using MATLAB's trapz function with the given force and velocity is:

Work done using MATLAB's trapz function: 7750 J

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The Moon's escape velocity is less than the Earth's escape velocity.

We know that,

\(V_{e}\) = √ 2 G M / r

where,

\(V_{e}\) = Escape velocity

G = Universal gravitational constant

M = Mass of object

r = Radius of object

For the Moon,

G = 6.67 * \(10^{-11}\) N m² / kg²

M = 7.35 * \(10^{22}\) kg

r = 1737400 m

\(V_{e}\) = √ [ 2 ( 6.67 * \(10^{-11}\) * 7.35 * \(10^{22}\) ) / 1737400 ]

\(V_{e}\) = 2375.89 m / s = 5314.72 mil / hr.

Escape speed is the amount of speed required to escape the object's gravitational pull. The size or mass of the object trying to escape does not matter.

Therefore, the Moon's escape velocity is less than the Earth's escape velocity.

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Of the four thermodynamic quantities listed, enthalpy is the only one that is a state function. A state function is a thermodynamic quantity that depends only on the initial and final states of a system, and not on the path taken to reach those states.

Enthalpy is a state function because it is defined as the sum of the internal energy of a system and the product of its pressure and volume, both of which are state functions.

Energy, work, and heat, on the other hand, are not state functions because they depend on the specific process or path taken to change the state of the system. The amount of work done or heat transferred depends on the specific conditions of the process, such as the pressure, temperature, and volume of the system, as well as the rate at which the work or heat is transferred. Therefore, these quantities are not solely dependent on the initial and final states of the system and cannot be considered state functions.

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Relation between velocity and time

\(\\ \sf\Rrightarrow Velocity=\dfrac{Displacement}{Time}\)

Or

\(\\ \sf\Rrightarrow \vec{v}=\dfrac{dx}{dt}\)

Now

\(\\ \sf\Rrightarrow v\propto \dfrac{1}{t}\)

Therefore, it took about 5,190 electrons to produce the charge on one of the objects.Let q be the amount of charge on each sphere and n be the number of electrons added to each sphere.

We can calculate the force between the two spheres using Coulomb's law as:F = kq²/d²where k is Coulomb's constant, q is the charge on each sphere, and d is the distance between the spheres. We are given that the force is 1.1375 × 10⁻²¹ N and the distance is 7.20 × 10⁻³ m. Thus, we have:1.1375 × 10⁻²¹ N = (9 × 10⁹ N·m²/C²)q²/(7.20 × 10⁻³ m)²Solving for q, we get:q² = (1.1375 × 10⁻²¹ N)(7.20 × 10⁻³ m)²/(9 × 10⁹ N·m²/C²)q² = 6.912 × 10⁻³³ C²q = ± 8.32 × 10⁻¹⁶ C (we take the positive value since we added electrons)Each electron has a charge of 1.602 × 10⁻¹⁹ C, so the number of electrons added to each sphere is:n = q/e = (8.32 × 10⁻¹⁶ C)/(1.602 × 10⁻¹⁹ C/electron)≈ 5.19 × 10³ electrons

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Explanation:

heat engine is correct answer

Answer:

The answer is:

C) A Furnace

Answer:

C

Explanation:

A P E X

The process that is the source of all the thermal energy produced by a nuclear reactor is nuclear fission. The correct option is C.

The process by which an atom's nucleus divides into two daughter nuclei in nuclear physics. Fission, the splitting of uranium atoms, is the source of nuclear energy.

This produces heat that creates steam, which a turbine generator uses to produce power. A type of energy emitted by an atom's nucleus, which is composed of protons and neutrons, is known as nuclear energy.

Fission, which occurs when atom nuclei split into several pieces, and fusion, which occurs when nuclei combine, are the two methods by which this type of energy can be created.

Therefore, the correct option is C. Nuclear fission.

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The main answer to your question is that the current drawn from the power supply with an EMF of 100V and a resistor with a value of 1000 ohms is 0.1 amperes (or 100 milliamperes).

To calculate the current drawn from the power supply, we can use Ohm's law, which states that current (I) is equal to voltage (V) divided by resistance (R):

I = V / R

Plugging in the values we have:

I = 100V / 1000 ohms = 0.1 amperes

Therefore, the current drawn from the power supply is 0.1 amperes or 100 milliamperes.
the current drawn from the power supply is 0.1 A.

Here's the step-by-step explanation:

1. You are given a resistor with a value of 1000 ohms and a power supply with an EMF of 100 V.
2. To find the current drawn from the power supply, we can use Ohm's Law, which is stated as V = IR, where V is voltage, I is current, and R is resistance.
3. We are given V (100 V) and R (1000 ohms), so we can rearrange the formula to solve for I: I = V/R.
4. Now, substitute the given values into the formula: I = 100 V / 1000 ohms.
5. Perform the calculation: I = 0.1 A.

Therefore, the current drawn from the power supply is 0.1 A.

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Given,

Force, F = 380 N

Mass, M = 38 kg

Distance, d = 2.5 m

The work done on the body is equal to the final kinetic energy.

Put the given values in the formula,

Thus, the speed of the object after it accelerates is

The box is traveling at a speed of 6.5 m/s when it returns to its starting position at the base of the incline.

To calculate the work done by friction, we need to first calculate the work done by gravity and the work done by the applied force when pushing the box up the incline.

Work done by gravity, W_gravity = mgh, where m = 5 kg, g = 9.8 m/s² (acceleration due to gravity), and h = 3sin30 = 1.5 m (height of the inclined plane)

W_gravity = (5 kg)(9.8 m/s²)(1.5 m) = 73.5 J

Work done by the applied force, W_applied = F_applied * d, where F_applied = mgsin30 = 24.5 N (component of the applied force parallel to the inclined plane), and d = 3 m (distance along the inclined plane)

W_applied = (24.5 N)(3 m) = 73.5 J

Since the box is moving with constant velocity up the incline, the net work done is zero. Therefore, the work done by friction on the way up is -73.5 J.

When the box slides back down the incline, the work done by friction is in the opposite direction to the motion of the box, so the work done by friction is positive.

Work done by friction, W_friction = F_friction * d, where F_friction = μmgcos30 = μ(5 kg)(9.8 m/s²)cos30 = μ(42.43 N) (component of the weight of the box perpendicular to the inclined plane), and μ is the coefficient of kinetic friction between the box and the inclined plane.

We don't know the value of μ, so we cannot calculate the exact value of W_friction. However, we can use conservation of energy to find the speed of the box at the bottom of the incline.

The total work done on the box is equal to the change in its kinetic energy.

Total work done = W_gravity + W_applied + W_friction = (1/2)mv² - (1/2)mu²

where v is the speed of the box at the bottom of the incline, and u is the speed of the box at the top of the incline (which is zero).

Since the net work done on the box is zero, we can write:

(1/2)mu² = W_gravity + W_applied

Substituting the values, we get:

(1/2)(5 kg)(6.5 m/s)² = 73.5 J + 73.5 J

Solving for u, we get:

u = 6.5 m/s

Therefore, the box is traveling at a velocity of 6.5 m/s when it returns to its starting position at the base of the incline. The work done by friction on the way up the ramp is -73.5 J, and the work done by friction on the way down the ramp is positive, but we cannot calculate its exact value without knowing the coefficient of kinetic friction

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Derived quantities depend on.( fundamental)..........physical quantity

Are you from Nepal?

Answer:

Derived quantities depend on two or more physical quantity.

Testable hypotheses are the theoretical problems that operational definitions have converted into.

Scientific observation is an empirical research designed to systematically address issues about the world. Through the use of operational definitions, theoretical questions are converted into. a provable theory. The measurement process for measuring external, observable behaviour is specified in Operational Definitions. The measurements that are obtained are utilised to define and quantify the construct. A measurement process (a collection of operations) is specified in an operational definition for the purpose of measuring an external, observable behaviour, and the definition and measurement of the hypothetical construct are then based on the results of the measurement procedure. No matter how solid your conceptual definition may be, you cannot measure anything without an operational definition.

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The units of measure of the derivative of g with respect to x (dg/dx) would be units of temperature per inch (e.g. °F/inch or °C/inch).

The derivative of a function is a measure of how the function is changing with respect to its independent variable. In this case, g represents the temperature of a metal rod, and x is the distance from one end of the rod.

Therefore, the derivative of g with respect to x (dg/dx) gives us the rate at which the temperature is changing with respect to distance.

The units of measure for the derivative will be the units of the dependent variable (temperature) divided by the units of the independent variable (distance).

For example, if g is measured in degrees Celsius and x is measured in inches, then the derivative dg/dx would have units of degree Celsius per inch (°C/inch).

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velocity and wavelength are likely to change in the process

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