A mass moves back and forth in simple harmonic motion with amplitude A and period T.(a) In terms of A, through what distance does the mass move in the time T? ?A(b) Through what distance does it move in the time 5.00T? ?A(c) In terms of T, how long does it take for the mass to move through a total distance of 2A? ?T(d) How long does it take for the mass to move through a total distance of 7A? ?T(e) If the objects undergoes simple harmonic motion with a period T. In the time 5T/2 the object moves through a total distance of 16D. In terms of D, what is the object's amplitude of motion? ?D

26 protons are in single nucleus of 5626fe

Answer:

10 m and 17.3 m

Explanation:

We can notice that the vector B represents the hypothenuse of a right triangle, in which the x-component of the vector is the side opposite to the [tex]30^{\circ}[/tex] angle, while the y-component of the vector corresponds to the side adjacent to the [tex]30^{\circ}[/tex] angle. This means that we can find the two components of the vector by using the sine and cosine function as follows:

[tex]B_x = B sin 30^{\circ} = (20 m) sin 30^{\circ}=10 m[/tex]

[tex]B_y = B cos 30^{\circ} = (20 m) cos 30^{\circ}=17.3 m[/tex]

Answer:

B) decreases

Explanation:

The density of water is maximum at a temperature of [tex]4^{\circ} C[/tex]. This means that:

- For temperatures between 0 and 4 degrees C, as the temperature increases, the density increases

- For temperatures above 4 degrees C, as the temperature increases, the density decreases

The density is related to the volume by the formula:

[tex]d=\frac{m}{V}[/tex]

where m is the mass and V the volume. Therefore, density is inversely proportional to the volume. This means that, for a constant amount of mass:

- For temperatures between 0 and 4 degrees C, as the temperature increases, the volume decreases

- For temperatures above 4 degrees C, as the temperature increases, the volume increases

Therefore, for a sample of water at 0 degrees C, if the temperature is slightly increased, the volume of the water will decreases, because its density will increase.

Final answer:

If the temperature of water slightly increases from 0 degrees Celsius, the volume decreases because of the anomalous expansion of water until it approaches 4 degrees Celsius, where it begins to expand normally again.

Explanation:

When considering a sample of water at 0 degrees Celsius, if the temperature is slightly increased, the volume of water decreases. This is due to the anomalous expansion of water, which is a unique property where water expands as it cools down from 4 degrees Celsius to 0 degrees Celsius and reaches its maximum density at 4 degrees Celsius. As the temperature rises slightly above 0 degrees Celsius, water begins to contract until it approaches 4 degrees Celsius, where it will start expanding again. This is in contrast to most substances which expand when heated.

- Two opposite charges:

The electrostatic force between two opposite charges is attractive. This means that if we have a positive charge and a negative charge, the electrostatic force between them is attractive.

- Two like charges:

The electrostatic force between two like charges is repulsive. This means that if we have two positive charges, or two negative charges, the force exerted by one charge on the other one is repulsive.

- A charge in an electric field:

The direction of the electric force on a charge in an electric field depends on the sign of the charge. In fact, we have

[tex]F=qE[/tex]

where F is the electric force, q is the charge, and E the electric field. We have the two following situations:

- If the charge is positive, q > 0, then the electric force has the same direction as the electric field (so the charge will be accelerated in the same direction as the electric field)

- If the charge is negative, q < 0, then the electric force has opposite direction to the electric field (so the charge will be accelerated in the opposite direction to the electric field)

Like charges repel each other and unlike charges attract each other. The direction of the force when a charge is placed in an electric field depends on whether the charge is positive (same direction as the field) or negative (opposite direction as the field).

The electric force that exists between two charges can be determined by their nature. If the charges are similar, that is, both are either positive or negative, then the direction of the electric force will be repulsive, and they push each other away. The electric field lines in this case would emanate away from the charges.

On the other hand, if the charges are dissimilar (that is, one positive and one negative), then the direction of the electric force is attractive—they pull towards each other. The electric field lines in this case would point from the positive charge towards the negative one.

When a charge is placed in an electric field, the direction of the force it experiences depends on its nature. If the charge is positive, it will experience a force in the direction of the field. If it's negative, the force will be in the opposite direction to the field.

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Answer:

B. 3

Explanation:

The half-life of a radioisotope is the time it needs for the mass of the radioisotope to halve with respect to its original value.

In this problem, the initial mass of the radioisotope at t=0 is

m0 = 50.0 mg

We see that after t = 1 min, the mass of the isotope is

m(1 min) = 25.0 mg

so, exactly half the initial mass: this means that 1 minute is exactly the half-life of this radioisotope.

So, the amount of mass left after each minute is the following:

m (1 min ) = 25.0 mg (1 half-life)

m (2 min) = 12.5 mg (2 half-lives)

m (3 min) = 6.25 mg (3 half-lives)

so, when we are left with 6.25 mg of isotope, 3 minutes have passed, which means that 3 half-lives have passed.

answer:

absorption spectrum

Answer:

The correct answer is B. Producers

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Explanation:

Answer:

Positive Z direction (out of screen)

Explanation:

Magnetic force is given by [tex]F = il \wedge B[/tex]. A quick way to gauge the components is to put your left middle finger on the direction of the current, your index on the direction of the magnetic field, and the thumb gives you the answer you want.

Answer:

positive-z direction (out of the screen)

Explanation:

As we know that length vector of the current carrying conductor is always along the direction of conventional current

So here direction of length vector will be + X direction

magnetic field is along + Y direction

now we will have

[tex]\vec F = I(\vec L \times \vec B)[/tex]

now we will have

[tex]\vec F = ILB(\hat i \times \hat j)[/tex]

[tex]\vec F = ILB\hat k[/tex]

so magnetic force will be along +Z direction (out of the screen)

First surface runoff then ground water...

Answer:44(,5 v

Explanation:

Ggthnu

Answer: at the center of curvature of the mirror on the same side of the mirror as the object

A concave mirror has a reflective surface that is curved inwards.  This type of mirrors reflects the light making it converge in a focal point, therefore they are used to focus the light.

This occurs because the light is reflected with different angles, since the normal to the surface varies from one point to another of the mirror.

Nevertheless, it is important to note the object must be within the radius of curvature of the mirror.

In addition, it is important to state clear the following:

-If the object is at a distance greater than the focal distance, a real and inverted image is formed that may be greater or less than the object.

-If the object is at a distance smaller than the focal distance, a virtual image is formed, right and larger than the object.

In this case the object is placed a great distance (it can be said at infinite) from the concave mirror, hence the image formed will be real, inverted and smaller than the object.

When we say real, it means the image is formed in the same side of the mirror as the object and the image can be seen on a screen.

Answer:

At the focal length of the mirror on the same side of the mirror as the object

Explanation: I had this question on USA test prep and I used the answer that the other guy said and it was incorrect. so the correct answer is

A)

at the focal length of the mirror on the same side of the mirror as the object

Identify a problem

Do background research

Form a hypothesis

Create and plan an experiment

Perform the experiment

Analyze the result

Create a conclusion

Communicate results

The major steps of a scientific method includes, observation, hypothesis, experiment, result and discussion and evaluation of the results and conclusion.

A scientific method includes several stages to find a solution for the scientific problem under study. The first step of the study includes an observation on the incident under study.

Based on the observations made make a scientific hypothesis which predict the solution of the problem. Later the hypothesis is tested with a well designed experiment.

So the next step is the planning of the experiment. Then perform the experiment and attain the results. Later the results are thoroughly evaluated and arrive at a conclusion.

The last step is to reveal and publish the results to communicate the important results and discussions.

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The tension in the string corresponds to the gravitational attraction between the Sun and any planet.

Answer:

Explanation:

While Steve is whirling a rubber cork, it developes a situation which is similar to the Solar System. In the rubber cork movement, there's "circular" motion which is possible thanks to the string, that is, the string works as "attraction force", allowing the system to maintain its movement.

Now, in the Solar System, the graviational force is like the string, it's the physical magnitude that maintains planets on their movements, attached to the Sun someway.

The gravitational force on a 9.0kg sphere inside the space shuttle orbiting 340km above the Earth's surface cannot be precisely determined with the information provided. However, astronauts and objects appear weightless in orbit, not because there is no gravity, but because they are in a state of continuous free-fall towards Earth, experiencing what is known as microgravity.

The sensation of weightlessness experienced by astronauts in the space shuttle is due to the phenomenon of microgravity. While it may seem that gravitational forces are absent in space, this is not the case. The space shuttle, and everything inside it, is in a constant state of free fall towards Earth. This happens because the shuttle is moving forward at a speed that matches the rate at which it falls towards the Earth, creating a steady orbit. According to Newton's law of universal gravitation, the force of gravity decreases with the square of the distance from the center of the Earth, but it is never zero. At the altitudes where the ISS and space shuttles orbit, gravity is only slightly weaker than on Earth's surface. Therefore, the 9.0kg sphere is subject to Earth's gravity, but it appears to float because it is in free-fall along with the shuttle.

Answer:

Some work input is lost to friction

Explanation:

The efficiency of a machine is defined as:

[tex]\eta = \frac{W_{out}}{W_{in}}[/tex] (1)

where

[tex]W_{out}[/tex] is the work output

[tex]W_{in}[/tex] is the work input

Due to the law of conservation of energy, the work output can never be larger than the work input (because energy cannot be created). Moreover, in real machines part of the work input is lost due to the presence of frictions: as a result, part of the energy in input is converted into thermal energy or other forms of energy, and so the work output is smaller than the work input, and so the ratio (1) becomes less than 1, and so the efficiency is less than 100%.

Answer:

Some work input is lost to friction

During the day, the atmosphere lets through solar radiation that heats the Earth's surface. Once the surface of the planet has warmed up, this heat is returned to space in the form of infrared radiation.

However, not all infrared radiation is returned to space, because the same atmosphere (specifically carbon dioxide and water vapor present in it) prevents the release of much of this thermal radiation, reflecting part of it and returning it to the surface; thus maintaining the heat and allowing life on our planet. This phenomenon is known as the greenhouse effect.

Thus, of the solar energy that reaches the Earth by radiation, only a small percentage is reflected back into space by the surface and the atmosphere.

In fact, if the atmosphere did not exist, all that heat would completely escape into space and the Earth would cool rapidly during the night.

Therefore, the correct option is B:

(a) [tex]2.5\cdot 10^{-6}eV[/tex]

The energy of a photon is given by:

[tex]E=\frac{hc}{\lambda}[/tex]

where

[tex]h=6.63\cdot 10^{-34}Js[/tex] is the Planck constant

[tex]c=3\cdot 10^8 m/s[/tex] is the speed of light

[tex]\lambda[/tex] is the wavelength

For the microwave photon,

[tex]\lambda=50.00 cm = 0.50 m[/tex]

So the energy is

[tex]E=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{0.50 m}=4.0\cdot 10^{-25} J[/tex]

And converting into electronvolts,

[tex]E=\frac{4.0\cdot 10^{-25}J}{1.6\cdot 10^{-19} J/eV}=2.5\cdot 10^{-6}eV[/tex]

(b) [tex]2.5 eV[/tex]

For the energy of the photon, we can use the same formula:

[tex]E=\frac{hc}{\lambda}[/tex]

For the visible light photon,

[tex]\lambda=500 nm = 5 \cdot 10^{-7}m[/tex]

So the energy is

[tex]E=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{5\cdot 10^{-7} m}=4.0\cdot 10^{-19} J[/tex]

And converting into electronvolts,

[tex]E=\frac{4.0\cdot 10^{-19}J}{1.6\cdot 10^{-19} J/eV}=2.5 eV[/tex]

(c) [tex]2500 eV[/tex]

For the energy of the photon, we can use the same formula:

[tex]E=\frac{hc}{\lambda}[/tex]

For the x-ray photon,

[tex]\lambda=0.5 nm = 5 \cdot 10^{-10}m[/tex]

So the energy is

[tex]E=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{5\cdot 10^{-10} m}=4.0\cdot 10^{-16} J[/tex]

And converting into electronvolts,

[tex]E=\frac{4.0\cdot 10^{-16}J}{1.6\cdot 10^{-19} J/eV}=2500 eV[/tex]

Answer:  The new moon phase occurs when the Moon is directly between the Earth and Sun. A solar eclipse can only happen at new moon. A waxing crescent moon is when the Moon looks like crescent and the crescent increases ("waxes") in size from one day to the next. This phase is usually only seen in the west.

Answer: a new moon

Explanation:

when the moon is between the sun and the earth a new moon comes.

hope this helps btw this is correct

I got 0.0126, but it feels wrong.

A. 6.67 cm

The focal length of the lens can be found by using the lens equation:

[tex]\frac{1}{f}=\frac{1}{s}+\frac{1}{s'}[/tex]

where we have

f = focal length

s = 12 cm is the distance of the object from the lens

s' = 15 cm is the distance of the image from the lens

Solving the equation for f, we find

[tex]\frac{1}{f}=\frac{1}{12 cm}+\frac{1}{15 cm}=0.15 cm^{-1}\\f=\frac{1}{0.15 cm^{-1}}=6.67 cm[/tex]

B. Converging

According to sign convention for lenses, we have:

- Converging (convex) lenses have focal length with positive sign

- Diverging (concave) lenses have focal length with negative sign

In this case, the focal length of the lens is positive, so the lens is a converging lens.

C. -1.25

The magnification of the lens is given by

[tex]M=-\frac{s'}{s}[/tex]

where

s' = 15 cm is the distance of the image from the lens

s = 12 cm is the distance of the object from the lens

Substituting into the equation, we find

[tex]M=-\frac{15 cm}{12 cm}=-1.25[/tex]

D. Real and inverted

The magnification equation can be also rewritten as

[tex]M=\frac{y'}{y}[/tex]

where

y' is the size of the image

y is the size of the object

Re-arranging it, we have

[tex]y'=My[/tex]

Since in this case M is negative, it means that y' has opposite sign compared to y: this means that the image is inverted.

Also, the sign of s' tells us if the image is real of virtual. In fact:

- s' is positive: image is real

- s' is negative: image is virtual

In this case, s' is positive, so the image is real.

E. Virtual

In this case, the magnification is 5/9, so we have

[tex]M=\frac{5}{9}=-\frac{s'}{s}[/tex]

which can be rewritten as

[tex]s'=-M s = -\frac{5}{9}s[/tex]

which means that s' has opposite sign than s: therefore, the image is virtual.

F. 12.0 cm

From the magnification equation, we can write

[tex]s'=-Ms[/tex]

and then we can substitute it into the lens equation:

[tex]\frac{1}{f}=\frac{1}{s}+\frac{1}{s'}\\\frac{1}{f}=\frac{1}{s}+\frac{1}{-Ms}[/tex]

and we can solve for s:

[tex]\frac{1}{f}=\frac{M-1}{Ms}\\f=\frac{Ms}{M-1}\\s=\frac{f(M-1)}{M}=\frac{(-15 cm)(\frac{5}{9}-1}{\frac{5}{9}}=12.0 cm[/tex]

G. -6.67 cm

Now the image distance can be directly found by using again the magnification equation:

[tex]s'=-Ms=-\frac{5}{9}(12.0 cm)=-6.67 cm[/tex]

And the sign of s' (negative) also tells us that the image is virtual.

H. -24.0 cm

In this case, the image is twice as tall as the object, so the magnification is

M = 2

and the distance of the image from the lens is

s' = -24 cm

The problem is asking us for the image distance: however, this is already given by the problem,

s' = -24 cm

so, this is the answer. And the fact that its sign is negative tells us that the image is virtual.

The correct answers to the questions are:

To find the image distance for the question in question G is by the use of the magnification equation which is:

s1= -Ms

=> -5/9(12cm)

=>  -6.67cm.

Because the value is negative, we know that the image is virtual.

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Answer:

0.775 m

Explanation:

As the car collides with the bumper, all the kinetic energy of the car (K) is converted into elastic potential energy of the bumper (U):

[tex]U=K\\frac{1}{2}kx^2 = \frac{1}{2}mv^2[/tex]

where we have

[tex]k=2\cdot 10^6 N/m[/tex] is the spring constant of the bumper

x is the maximum compression of the bumper

[tex]m=30\cdot 10^4 kg[/tex] is the mass of the car

[tex]v=2.0 m/s[/tex] is the speed of the car

Solving for x, we find the maximum compression of the spring:

[tex]x=\sqrt{\frac{mv^2}{k}}=\sqrt{\frac{(30\cdot 10^4 kg)(2.0 m/s)^2}{2\cdot 10^6 N/m}}=0.775 m[/tex]

Answer:

0.6 m

Explanation:

The law of conservation of energy states that:

[tex]\Delta E_m=0[/tex]

The mechanical energy ([tex]E_m[/tex]) is the sum of the kinetic energy and the potential energy:

[tex]\Delta K+\Delta U=0\\K_f-K_i+U_f-U_i=0\\\frac{mv_f^2}{2}-\frac{mv_i^2}{2}+\frac{kx_f^2}{2}-\frac{kx_i^2}{2}=0[/tex]

[tex]U_i[/tex] is zero since the spring is not initially compressed and [tex]K_f[/tex] is zero since all kinetic energy becomes potentital energy:

[tex]\frac{kx_f^2}{2}=\frac{mv_i^2}{2}[/tex]

Finally, we solve for x and replace the given values:

[tex]x_f^2=\frac{mv_i^2}{k}\\x_f=\sqrt{\frac{mv_i^2}{k}}\\x_f=\sqrt{\frac{(30*10^4kg)(2\frac{m}{s})^2}{2*10^6\frac{N}{m}}}\\x_f=0.6 m[/tex]

To find the distances and times at which the two cars pass each other, we can use equations of motion. The car passes the Porsche at 20 meters and 2 seconds after the light turns green. The Porsche passes the car at 60 meters and 5 seconds after the light turns green.

To solve this problem, we will use equations of motion to find the distances and times at which the two cars pass each other. Given that the Porsche starts from rest and accelerates at 3 m/s², we can use the equation x = xo + vot + ½at² to find the distance it travels before the other car reaches it. Similarly, for the other car which is traveling at a constant speed of 10 m/s, we can use the equation x = Ut. By solving these equations simultaneously, we can find the distances and times at which the two cars pass each other.

a. The distance at which you pass the Porsche can be found by setting the distances traveled by both cars equal to each other: 10t = 15 + 0.5(3)(t²). By solving this equation, we find that you pass the Porsche at a distance of 20 m from the stop line and at a time of 2 seconds after the light turns green.

b. To find the distance at which the Porsche passes you, we need to find the time at which the Porsche catches up to you. We can do this by setting the equation for the Porsche's distance equal to your distance: 15 + 0.5(3)(t²) = 10t. Solving this equation gives us a time of 5 seconds after the light turns green. Plugging this time into the equation for your distance, we find that the Porsche passes you at a distance of 60 meters from the stop line.

c. To determine if either of you will be pulled over for speeding, we need to compare your speeds to the speed limit of 50 km/h. Your speed is given as 36 km/h (10 m/s), which is less than the speed limit. The Porsche's speed can be found by taking the derivative of its distance equation with respect to time: v = at. Plugging in the time at which it passes you, we find that its speed is 30 m/s, which is also less than the speed limit. Therefore, neither of you will be pulled over for speeding.

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You pass the Porsche 16.65 meters beyond the stop line at 3.33 seconds, while the Porsche catches up 66.7 meters from the stop line at 6.67 seconds. Only the Porsche would be pulled over for speeding. The primary topic is the analysis of two moving vehicles using physics.

Your car's constant speed is [tex]36 km/h[/tex] or [tex]10 m/s[/tex] . The Porsche accelerates from rest at [tex]3 m/s^2[/tex].

Calculate the time it takes for the Porsche to reach your speed:
Using the equation: [tex]v = u + at[/tex],
where[tex]v = 10 m/s[/tex],[tex]u = 0[/tex], and[tex]a = 3 m/s^2[/tex], we get [tex]t = v/a = 10/3 \approx 3.33 seconds[/tex].

Now, find the distance the Porsche covers in this time with the equation: [tex]s = ut + 0.5at^2[/tex], where[tex]u = 0[/tex], [tex]a = 3 m/s^2[/tex] and [tex]t = 3.33 seconds[/tex],
thus [tex]s = 0.5 \times 3 \times 3.332 \approx 16.65 meters[/tex].

Your car covers [tex]3.33 \times 10 = 33.3 meters[/tex] in this same time.

Hence, you pass the Porsche [tex]33.3 - 16.65 = 16.65 meters[/tex] beyond the stop line.

To find when the Porsche catches up, consider the distance equations for both vehicles:

Your distance: [tex]d_{you}(t) = 10t[/tex].

Porsche’s distance: dporsche(t) = 0.5 * 3 * t2.

Set the distances equal to solve for [tex]t: 10t = 1.5t2[/tex],  leading to [tex]t = 0[/tex] or [tex]t = 10/1.5 = 6.67 seconds[/tex].

The Porsche catches up [tex]0.5 \times 3 \times 6.672 = 66.7[/tex]meters from the stop line.

Determine both cars' speeds when the Porsche passes you again:

Your speed:[tex]10 m/s = 36 km/h[/tex], under the limit.

Porsche’s speed: [tex]v = u + at = 3 \times 6.67 = 20.01 m/s \approx 72 km/h[/tex], over the limit.

Thus, only the Porsche would be pulled over for speeding.

Answer:

-10.9 rad/s²

Explanation:

ω² = ω₀² + 2α(θ - θ₀)

Given:

ω = 13.5 rad/s

ω₀ = 22.0 rad/s

θ - θ₀ = 13.8 rad

(13.5)² = (22.0)² + 2α (13.8)

α = -10.9 rad/s²

Use the formula

[tex]{\omega_f}^2-{\omega_i}^2=2\alpha\Delta\theta[/tex]

[tex]\implies\left(13.5\dfrac{\rm rad}{\rm s}\right)^2-\left(22.0\dfrac{\rm rad}{\rm s}\right)^2=2\alpha(13.8\,\mathrm{rad})[/tex]

[tex]\implies\alpha=-10.9\dfrac{\rm rad}{\mathrm s^2}[/tex]

so the magnitude is 10.9 rad/s^2.

In order to be able to calculate an answer, we must assume that the boat, with her on it, was motionless in the water until she jumped off of it.

I'll make that assumption, and then I'll go ahead and answer the question that I just invented:

-- Before she jumped off of the boat, she had no momentum and the boat had no momentum.  

-- The SUM of (her momentum) + (the boat's momentum) was zero.

-- Momentum is conserved, so the SUM of (her momentum) + (the boat's momentum) has to still be zero after she jumps off of the boat.

-- Momentum = (mass) · (speed in some direction)

The boat's momentum after the jump = (42kg) · (1.5 m/s) that way ==>

The boat's momentum after the jump = 63 kg-m/s that way ==>

Her momentum after the jump has to be 63 kg-m/s this way <==

Her momentum = 63 kg-m/s = (59 kg) · (her speed this way <== )

Divide each side by (59 kg):

Her speed this way <== = (63 kg-m/s <==) / (59 kg)

Her speed = (63/59) · (m/s this way <== )

Her speed = 1.07 m/s opposite to the direction the boat is moving.

= = = = = = = = = =

A casual but striking observation:

Our 'student' is carrying 17 kg more mass than the boat she sails !  

The mind boggles at the implied zaftigkeit.  

To find the speed of the physics student, we can use the principle of conservation of momentum. By setting up an equation using the initial and final velocities of both the student and the sailboat, we can solve for the final velocity of the student. The answer is approximately 0.038 m/s.

To solve this problem, we can use the principle of conservation of momentum. The initial momentum of the system, which includes the student and the laser sailboat, is equal to the final momentum of the system. The momentum of an object is calculated by multiplying its mass by its velocity. We know the mass and velocity of the laser sailboat after the student jumps off, so we can use that information to find the speed of the physics student.

We can set up the equation as follows:

(mass of student)(initial velocity of student) + (mass of sailboat)(initial velocity of sailboat) = (mass of student)(final velocity of student) + (mass of sailboat)(final velocity of sailboat)

Plugging in the given values, we have:

(59 kg)(0 m/s) + (42 kg)(0 m/s) = (59 kg)(final velocity of student) + (42 kg)(1.5 m/s)

Simplifying the equation, we find that the final velocity of the student is approximately 0.038 m/s.

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The speed of a wave is given by:

[tex]v=f.\lambda[/tex]  (1)

Where [tex]f[/tex] is the frequency and  [tex]\lambda[/tex] the wavelength.

In the case of light, its speed is:

[tex]c=f.\lambda[/tex] (2)

On the other hand, the described situation is known as Refraction,   a phenomenon in which the light changes its direction when passing through a medium with a refractive index different from the other medium.  

In this context, the Refractive index [tex]n[/tex] is a number that describes how fast light propagates through a medium or material, and is defined as the relation between the speed of light in vacuum ([tex]c=3(10)^{8}m/s[/tex]) and the speed of light [tex]v[/tex] in the second medium:

[tex]n=\frac{c}{v}[/tex] (3)

In addition, as the light changes its direction, its wavelength changes as well:

[tex]n=\frac{\lambda_{air}}{\lambda_{glass}}[/tex] (4)

Knowing this, let's begin with the answers:

From equation (2) we can find [tex]f[/tex]:

[tex]f=\frac{c}{\lambda}[/tex]  (5)

Knowing that [tex]1nm=(10)^{-9}m[/tex]:

[tex]f=\frac{3(10)^{8}m/s}{632.8(10)^{-9}m}[/tex]  

[tex]f=4.74(10)^{14}Hz}[/tex]     (6)   >>>Frequency of the helium-neon laser light

We already know the wavelength of the light in air [tex]\lambda_{air}[/tex] and the index of refraction of the glass.

So, we only have to find the wavelength in glass [tex]\lambda_{glass}[/tex] from equation (4):

[tex]\lambda_{glass}=\frac{\lambda_{air}}{n}[/tex]

[tex]\lambda_{glass}=\frac{632.8(10)^{-9}m}{1.48}[/tex]

[tex]\lambda_{glass}=427(10)^{-9}m=427nm[/tex]   (7)   >>>Wavelength of the helium-neon laser light in glass

From equation (3) we can find the speed [tex]v[/tex]of this light in glass:

[tex]v=\frac{c}{n}[/tex]

[tex]v=\frac{3(10)^{8}m/s}{1.48}[/tex]

[tex]v=2.027(10)^{8}m/s[/tex]   (8)  >>>Speed of the helium-neon laser light in glass

The frequency of the red helium-neon laser light is approximately 4.74 x 10^14 Hz. When it travels in glass (with index of refraction 1.48), its wavelength shortens to 427.6 nm, and its speed reduces to about 203 Mm/s.

The wavelength of the red helium-neon laser light in air is given as 632.8 nm.

(a) Frequency: To calculate the frequency of the light, we can use the formula: frequency = speed of light / wavelength. Given the speed of light (c) is approximately 3.00 x 10^8 m/s, we can convert the wavelength from nm to m (1 nm = 1 x 10^-9 m), giving us 632.8 x 10^-9 m. Therefore, frequency (f) = 3.00 x 10^8 m/s / 632.8 x 10^-9 m = 4.74 x 10^14 Hz.

(b) Wavelength in Glass: The wavelength of light is smaller in a medium as compared to vacuum. We use the formula: wavelength in medium = wavelength in vacuum / index of refraction. Substituting the values: wavelength in glass = 632.8 nm / 1.48 = 427.6 nm.

(c) Speed in Glass: The speed of light in a medium is also dependent on refractive index, calculated with the formula: speed in medium = speed in vacuum / index of refraction. Therefore, speed in glass = 3.00 x 10^8 m/s / 1.48 ≈ 2.03 x 10^8 m/s, which is 203 Mm/s when converted to Mm/s.

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Answer:

increases

Explanation:

In a circuit, the power supplied by the source is given by:

[tex]P=VI[/tex] (1)

where V is the voltage of the source and I is the current in the circuit.

Using Ohm's law, we can rewrite the current as:

[tex]I=\frac{V}{R}[/tex]

where R is the equivalent resistance of the circuit. Substituting into (1), we can rewrite the power as

[tex]P= \frac{V^2}{R}[/tex] (2)

so we see that the power is inversely proportional to the resistance.

For resistors added in parallel, the equivalent resistance is given by the formula

[tex]\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+...+\frac{1}{R_n}[/tex]

so we see that when adding new resistors in parallel, the term [tex]\frac{1}{R}[/tex] increases, so the equivalent resistance R will decreases. As a result, the power supplied by the source (given by eq.(2)) wil increase.

As more resistors are added in parallel across a constant voltage source the power will; increase

Formula for Poweer across a circuit with constant voltage is;

P = I²R

Where;

1/R = 1/R1 + 1/R2 + 1/R3....

Where R is total resistance.

This means that the more resistors you add in parallel, the greater the value of R.

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The puck starts with velocity vector

[tex]\vec v_0=\left(2.35\dfrac{\rm m}{\rm s}\right)(\cos(-22^\circ)\,\vec\imath+\sin(-22^\circ)\,\vec\jmath)=(2.18\,\vec\imath-0.880\,\vec\jmath)\dfrac{\rm m}{\rm s}[/tex]

Its velocity at time [tex]t[/tex] is

[tex]\vec v=\vec v_0+\vec at[/tex]

Over the 0.215 s interval, the velocity changes to

[tex]\vec v=\left(6.42\dfrac{\rm m}{\rm s}\right)(\cos50.0^\circ\,\vec\imath+\sin50.0^\circ\,\vec\jmath)=(4.13\,\vec\imath+4.92\,\vec\jmath)\dfrac{\rm m}{\rm s}[/tex]

Then the acceleration must have been

[tex]\vec v=\vec v_0+(0.215\,\mathrm s)\vec a\implies\vec a=\dfrac{\vec v-\vec v_0}{0.215\,\rm s}=(9.06\,\vec\imath+27.0\,\vec\jmath)\dfrac{\rm m}{\mathrm s^2}[/tex]

which has a direction of about [tex]71.4^\circ[/tex].

The direction of the acceleration is determined by the direction of the change in velocity. This would be calculated by subtracting the initial velocity vector from the final velocity vector. However, the calculation would involve complex trigonometric functions.

In order to find the direction of the acceleration, we need to calculate the direction of the change in velocity and that direction will be the direction of the acceleration.

To calculate the change in velocity, we subtract the initial velocity from the final velocity: (6.42 m/s, 50.0°) - (2.35 m/s, -22°). We then calculate the angle of this vector which represents the change in velocity, and hence the direction of acceleration.

However, this calculation is not straightforward because it involves vector operations and would require the use of trigonometric functions to solve. This is due to the fact that velocity is a vector, meaning it has both a magnitude and a direction. Consequently, this becomes a multi-step process involving trigonometry and physics.

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a. 12,500 turns

The transformer equation states that

[tex]\frac{N_p}{V_p}=\frac{N_s}{V_s}[/tex]

where

Np is the number of turns in the primary coil

Ns is the number of turns in the secondary coil

Vp is the voltage in the primary coil

Vs is the voltage in the secondary coil

For the transformer in the problem,

Vp = 15,000 V

Vs = 120 V

Ns = 100

So we can find Np by rearranging the equation:

[tex]N_p = V_p \frac{N_s}{V_s}=(15,000 V)\frac{100}{120 V}=12,500[/tex]

b.  2 A

For an ideal transformer, the output power is equal to the input power:

[tex]P_i = P_o\\V_p I_p = V_s I_s[/tex]

where

[tex]V_p = 15,000 V[/tex] is the voltage in the primary coil

[tex]I_p[/tex] is the current in the primary coil

[tex]V_s = 120 V[/tex] is the voltage in the secondary coil

[tex]I_s = 250 A[/tex] is the current in the secondary coil

Solvign the formula for Ip, we find:

[tex]I_p = \frac{V_s I_s}{V_p}=\frac{(120 V)(250 A)}{15,000 V}=2 A[/tex]

The primary coil of the transformer has 12500 turns, and the current in the 15,000 V line from the electrical substation is 2 A.

The number of turns in the primary coil of the transformer can be found using the transformer equation, which states that the ratio of the secondary voltage to the primary voltage equals the ratio of the number of loops in the secondary coil to the number of loops in the primary coil.

Therefore, if the secondary coil has 100 turns, and the voltages are 15000 V (primary) and 120 V (secondary), we can set up the equation (15000 V / 120 V) = (x turns / 100 turns), where x is the number of turns in the primary coil. Solving for x gives us x = 12500 turns.

The power input to the primary coil equals the power output from the secondary coil, since no energy is lost in an ideal transformer. Power (P) equals voltage (V) times current (I), so Pin = Pout, or (15,000 V * Iin) = (120 V * 250 A). Solving for Iin, the current in the 15,000 V line from the substation, gives us Iin = 2 A.

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Answer:

Duration of the total trip

Explanation:

Acceleration is defined as the rate of change of velocity of an object:

[tex]a=\frac{\Delta v}{\Delta t}[/tex]

where

[tex]\Delta v[/tex] is the change in velocity

[tex]\Delta t[/tex] is the duration of the change in velocity

Acceleration is measured in meters per squared seconds, [tex]m/s^2[/tex]. Also, we should note that acceleration is a vector quantity, so in order to completely define acceleration we also need to specify the direction.

Answer: Time

Explanation: Took the test and got it right! :)

Final answer:

Low pH indicates a high hydrogen-ion concentration, showcasing that the environment is acidic.

Explanation:

Low pH means that the hydrogen-ion concentration is high. The pH scale is a logarithmic representation of hydrogen ion concentration, where a low pH number indicates a higher concentration of H+ ions, creating an acidic environment. In contrast, a high pH suggests a lower concentration of hydrogen ions and a basic or alkaline condition. For example, a pH of 3 is ten times more acidic than a pH of 4 due to the logarithmic scale of pH, where each whole number change represents a tenfold change in ion concentration.