suppose that body A is time two times as dense B for equal volumes of A & B of how we measure the mass​

The stored energy from all the lightning strikes during the thunderstorm would only be able to supply a city for 0.000675 seconds at the same rate as a large power plant.

The energy delivered by a lightning strike can be calculated using the formula E = VQ, where E is the energy, V is the voltage, and Q is the charge. Therefore, the energy delivered by a lightning strike is:

E = (-75 x 10⁻³V) x (-25 C) = 1.875 J

The total energy delivered by lightning strikes during the thunderstorm can be calculated by multiplying the energy delivered by each strike by the number of strikes, which is 3600/10 = 360.

Therefore, the total energy delivered by lightning strikes during the thunderstorm is:

E_total = 1.875 J/strike x 360 strikes = 675 J

Assuming that all of this energy can be captured and stored, it can supply a city for a certain amount of time. The time that the stored energy can supply the city can be calculated using the formula T = E/P, where T is the time, E is the energy, and P is the power.

Therefore, the time that the stored energy can supply a city is:

T = 675 J / 1,000 MW = 0.000675 s

As a result, the accumulated energy from all of the lightning strikes throughout the thunderstorm could only power a city for 0.000675 seconds at the same rate as a huge power plant.

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Answer:

Option (a)

Explanation:

Let c be the specific heat of water.

According to the principle of caloriemetry.

Heat lost by hot water = heat gained by cold water

200 x c x (70 - T) = 50 x c x (T - 20)

280 - 4T = T - 20

300 = 5T

T = 60 C

Explanation:

In a case whereby You add 50 mL of water at 20°C to 200 mL of water at 70°C the most likely final temperature of the mixture is A. 60°C.

Specific heat is the amount of heat energy required to raise the temperature of a unit mass of a substance by one degree Celsius (or one Kelvin). Different substances have different specific heats, which means that they require different amounts of heat energy to achieve the same temperature change.

The specific heat of water can be represented as c, following the principle of caloriemetry. (Heat lost by hot water) =( heat gained by cold water), thjen we can substitute the values as ;

[200 x c x (70 - T)] = [50 x c x (T - 20)]

[280 - 4T] = [T - 20]

[300 = 5T]

T = 60 C

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The only possibility remaining is barium-139, which is a stable (non-radioactive) isotope and would not have undergone any radioactive decay during the time period between measurements.  Hence, the unknown radioisotope in the sample is barium-139.

To determine which radioisotope is in the sample, we need to use the concept of radioactive decay and half-life. Radioactive decay is a process by which the nucleus of an unstable atom loses energy by emitting particles or radiation.

The rate of decay of a radioactive substance is described by its half-life, which is the time it takes for half of the substance to decay.

Let's calculate the half-life of each radioisotope listed in the table:

Potassium-42: Half-life of 12.4 hours

Nitrogen-13: Half-life of 10 minutes

Barium-139: Stable (non-radioactive)

Radon-220: Half-life of 55.6 seconds

From the given data, the sample of the unknown radioisotope had a mass of 104.8 kg at 12:02:00 p.m. and 13.1 kg at 4:11:00 p.m. on the same day, which is a time difference of 4 hours and 9 minutes.

Let's start by looking at the radioisotope with the longest half-life, which is potassium-42.

If the unknown radioisotope was potassium-42, its mass would have decreased by half during 12.4 hours, which is much longer than the 4 hours and 9 minutes between the measurements.

Therefore, we can eliminate potassium-42 as a possibility.

Next, let's consider nitrogen-13. If the unknown radioisotope was nitrogen-13, its mass would have decreased by half during 10 minutes. We can convert the time difference between measurements to minutes:

4 hours and 9 minutes = 249 minutes

Therefore, the number of half-lives during this time period would be:

249 / 10 = 24.9

This means that the mass of the sample would have decreased by a factor of [tex]2^{(24.9)[/tex], which is approximately [tex]2.7 * 10^7[/tex]. Starting from the initial mass of 104.8 kg, the final mass would be:

104.8 kg / [tex](2.7 * 10^7)[/tex] = [tex]3.9 * 10^{-6[/tex] kg

This is much smaller than the measured final mass of 13.1 kg, so we can eliminate nitrogen-13 as a possibility.

Finally, let's consider radon-220. If the unknown radioisotope was radon-220, its mass would have decreased by half during 55.6 seconds. We can convert the time difference between measurements to seconds:

4 hours and 9 minutes = 14940 seconds

Therefore, the number of half-lives during this time period would be:

14940 / 55.6 = 269

This means that the mass of the sample would have decreased by a factor of [tex]2^{269[/tex], which is approximately 6.8 x [tex]10^{80[/tex]. Starting from the initial mass of 104.8 kg, the final mass would be:

104.8 kg / ([tex]6.8 * 10^{80[/tex]) = [tex]1.54 * 10^{-79[/tex]kg

This is much smaller than the measured final mass of 13.1 kg, so we can eliminate radon-220 as a possibility.

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Answer: c

Explanation:

The speed of the small stone lodged in the tire's tread is approximately 10.47 m/s, and its acceleration is approximately 219.35 m/s².

We need to find the speed and acceleration of a small stone lodged in the tread of a tire with a 0.500 m radius, rotating at 200 revolutions per minute.

First, let's convert the revolutions per minute (rpm) to radians per second (rad/s):
200 rpm * (2π radians/1 revolution) * (1 minute/60 seconds) ≈ 20.94 rad/s

Now, we can find the linear speed (v) of the stone using the formula:
v = rω, where r is the radius, and ω is the angular velocity in rad/s.
v = 0.500 m * 20.94 rad/s ≈ 10.47 m/s

Next, we'll find the centripetal acceleration (a_c) of the stone using the formula:
a_c = rω²
a_c = 0.500 m * (20.94 rad/s)² ≈ 219.35 m/s²

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This is the reason the compass needle, despite having a north-seeking magnet, points in the direction of the geographic north pole.

Hi! The phenomenon you're referring to can be explained through the concepts of magnetism and Earth's magnetic field. Although it may seem that the north pole of a compass needle is attracted to the Earth's north pole, it's actually attracted to the magnetic south pole of the planet.

This attraction occurs because the Earth itself acts as a giant magnet, generating a magnetic field with poles that are approximately aligned with the geographic poles. The Earth's magnetic south pole is near the geographic north pole, and the magnetic north pole is near the geographic south pole.

As you mentioned, the needle of a compass is a magnet with its own north and south poles. According to the laws of magnetism, opposite poles attract each other.

Consequently, the north pole of the compass needle is attracted to the magnetic south pole of the Earth, which is near the geographic north pole. This is why the compass needle points towards the geographic north pole, despite it being a north-seeking magnet.

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Answer: A white dwarf with a temperature of 10,000 K belongs to the spectral class DA.

White dwarfs are classified based on their atmospheric composition and temperature. The DA spectral class refers to white dwarfs that have a hydrogen-dominated atmosphere. Their spectra exhibit strong hydrogen absorption lines.

The temperature of a white dwarf is a measure of its surface temperature and is related to its age and mass. A white dwarf with a temperature of 10,000 K is relatively hot, indicating that it is likely a young and massive white dwarf.

Explanation:

The man does 50 J of work on the block during the motion.

To calculate the work done by the man on the block, we can use the formula:

Work = Force x Distance x Cos(theta)

where theta is the angle between the force and the displacement vectors. In this case, the force and displacement are in the same direction, so theta is 0.

Given that the force applied by the man is 5 N and the distance moved by the block is 10 meters, the work done by the man can be calculated as:

Work = 5 N x 10 m x Cos(0) = 50 J

Therefore, the man does 50 J of work on the block during the motion.

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A sound wave has a wavelength of 0. 96 m and this sound wave causes your eardrum to oscillate back and forth 357 times per second or 357 Hz.

The number of times a sound wave causes your eardrum to oscillate back and forth in one second is known as its frequency. We can calculate the frequency of a sound wave by dividing the speed of sound by its wavelength.

The speed of sound in air at room temperature is about 343 m/s.To calculate the frequency of a sound wave with a wavelength of 0.96 m, we can use the formula:

frequency = speed of sound/wavelength

frequency = 343 m/s / 0.96 m

frequency = 357 Hz

Therefore, this sound wave causes your eardrum to oscillate back and forth 357 times per second, or 357 Hz.

In summary, the frequency of a sound wave is the number of times it causes your eardrum to oscillate back and forth in one second. We can calculate the frequency of a sound wave by dividing the speed of sound by its wavelength. A sound wave with a wavelength of 0.96 m has a frequency of 357 Hz.

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Based on the weather bulletin provided for Typhoon Rolly (GONI), the speed of the typhoon winds is 165 km/h with gustiness up to 230 km/h.

The velocity of the typhoon, which takes into account both the speed and direction, is moving westward at 25 km/h.

The main difference between speed and velocity is that speed only considers the magnitude of motion, while velocity includes both the magnitude and direction of motion.

Knowing the velocity of the typhoon is important in determining the weather forecast for the next hours, as it helps predict the movement and potential impact of the typhoon on specific areas.

This information can help authorities and individuals prepare and respond accordingly to ensure safety and minimize damages.

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To determine which identification of the variables is correct, let's analyze each option step-by-step:

A. If the volume and concentration change, but the number of solute particles remains constant, it means that the ratio of solute to solvent is changing. This is not possible if the number of solute particles is constant.

B. If the volume and number of solute particles change, but the concentration remains constant, it means that the ratio of solute to solvent remains the same. This is possible and indicates that both solutions have the same concentration.

C. If the number of solute particles and the concentration change, but the volume remains constant, it means that the amount of solute in the solution is changing without affecting the volume. This scenario is not possible as adding or removing solute particles would change the concentration.

D. If the number of solute particles changes but the volume and concentration remain constant, this would mean that the ratio of solute to solvent is unchanged despite the change in solute particles. This is not possible.

Based on the analysis, the correct identification of the variables is option B. The volume of the solution and the number of solute particles are being changed between the two solutions, but the concentration of the solution is being held constant.

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The car has a nonzero acceleration in the following situations:

Acceleration is defined as the rate of change of velocity with respect to time. Therefore, any change in the velocity of the car, whether it is an increase, decrease, or change in direction, results in a nonzero acceleration. When the car starts from rest and speeds up or comes to a stop, its velocity changes, resulting in a nonzero acceleration.

Similarly, when the car comes to a complete stop while traveling around a curve or in a straight line, its velocity changes direction, resulting in a nonzero acceleration. However, when the car is on cruise control and traveling at a constant speed in a straight line, its velocity is not changing, and therefore, its acceleration is zero. Options 1, 2, and 3 are correct.

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The cylinder is moving at a speed of 2.12 m/s after it has rolled 2.2 m down the incline.

We can use conservation of energy to solve this problem. The initial potential energy of the cylinder is converted into kinetic energy as it rolls down the incline. At the bottom of the incline, the kinetic energy is equal to the potential energy it had at the top, neglecting any energy losses due to friction.

Let's begin by finding the potential energy of the cylinder at the top of the incline. The height of the incline can be found using trigonometry:

h = (2.2 m)sin(18°) = 0.667 m

The potential energy of the cylinder at the top of the incline is:

PE = mgh = (5.25 kg)(9.81 m/s²)(0.667 m) = 34.2 J

At the bottom of the incline, the kinetic energy of the cylinder is equal to its potential energy at the top:

KE = 34.2 J

The kinetic energy of a rolling cylinder is given by:

KE = (1/2)mv² + (1/2)Iω²

where m is the mass of the cylinder, v is its velocity, I is its moment of inertia, and ω is its angular velocity. For a cylinder rolling without slipping, we have:

v = ωR

where R is the radius of the cylinder. The moment of inertia of a solid cylinder about its central axis is:

I = (1/2)mr²

Substituting these expressions into the equation for kinetic energy and simplifying, we get:

KE = (1/2)mv² + (1/4)mv²

Solving for v, we find:

v = sqrt(8KE/3m)

Substituting in the values we found above, we get:

v = sqrt(8(34.2 J)/(3(5.25 kg))) = 2.12 m/s

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If the electron gun distance and electron deflection distance both double, while the electric fields stay the same, then the angle of deflection would also double.

This is because the electric field strength is directly proportional to the angle of deflection, and since the electric field strength is staying the same, the angle of deflection increases proportionally with the increase in distance.

The equation to determine the angle of deflection is as follows: θ = Vd/E, where θ is the angle of deflection, V is the velocity of the electron, d is the distance between the electron gun and deflection plate, and E is the strength of the electric field.

When the distance between the two plates doubles, the angle of deflection will also double. Therefore, if the electron gun and electron deflection plate are both doubled in distance, the angle of deflection would be double the original angle.

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In a vacuum, electromagnetic radiation of short wavelengths refers to high-energy radiation. According to the electromagnetic spectrum, shorter wavelengths correspond to higher frequencies and higher energies.

At the short wavelength end of the spectrum, you have gamma rays, which have the shortest wavelengths and highest energy among all forms of electromagnetic radiation. Gamma rays have wavelengths less than 10 picometers (pm) or frequencies greater than 10 exahertz (EHz).

Gamma rays are highly energetic and can penetrate matter deeply. They are often produced in nuclear reactions, radioactive decay, and high-energy particle interactions.

It's important to note that in a vacuum, all forms of electromagnetic radiation, including gamma rays, travel at the speed of light. The properties of electromagnetic radiation, such as wavelength and frequency, are intrinsic characteristics that remain constant regardless of the medium through which they propagate.

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After 1.8 hours on the shelf, the activity of the drug tagged with 99m43Tc is approximately 8147 Bq.

Step 1: Calculate the number of half-lives that have passed.

To do this, divide the elapsed time (1.8 h) by the half-life of the isotope (6.05 h).
Number of half-lives = 1.8 h / 6.05 h = 0.2975 half-lives

Step 2: Use the decay formula to calculate the remaining activity.
The decay formula is R = R₀ * (1/2)^(t/T), where R is the remaining activity, R₀ is the initial activity, t is the elapsed time, and T is the half-life.

Step 3: Plug in the values and solve for R.

R = (1.0 x 10^4 Bq) * (1/2)^(0.2975)
R ≈ 1.0 x 10^4 Bq * 0.8147
R ≈ 8147 Bq

So, after 1.8 hours on the shelf, the activity of the drug tagged with 99m43Tc is approximately 8147 Bq.

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Answer:

it's the turning effect of force; the product of force and perpendicular distance from line of action of the force to the pivot . Depends on two factors: size of force applied and perpendicular distance from pivot to line of action of the force

The distance between the golf tube and the balloon is approximately 53.9 millimeters.

To solve this problem, we can use Coulomb's Law, which states that the electrical force (F) between two charges (q1 and q2) is proportional to the product of their charges divided by the square of the distance (r) between them:

F = k * (q1 * q2) / r²

where k is Coulomb's constant, approximately 8.99 x 10^9 Nm²/C².

In this case, the electrical force (F) is -1.0 x 10^-5 N, the charge of the balloon (q1) is -1.0 x 10^-6 C, and the charge of the plastic golf tube (q2) is +4.0 x 10^-6 C. We want to find the distance (r) between them.

First, let's rearrange the formula to solve for r:

r = √(k * (q1 * q2) / F)

Now, substitute the given values into the equation:

r = √((8.99 x 10^9 Nm²/C²) * (-1.0 x 10^-6 C) * (4.0 x 10^-6 C) / (-1.0 x 10^-5 N))

Solve for r:

r ≈ 0.0539 meters or 53.9 millimeters

So, the distance between the golf tube and the balloon is approximately 53.9 millimeters.

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Energy cannot be created or destroyed, it can only be converted from one form to another. When the soccer player kicked the ball, they transferred their kinetic energy to the ball, causing it to move.

As the ball rolled across the grass, its kinetic energy was gradually converted into other forms of energy, such as frictional heat and sound energy, causing it to slow down and eventually stop.

The total amount of energy in a closed system remains constant. In this case, the system is the ball and the grass.

Even though the ball slowed down and stopped, the total amount of energy in the system remained the same, as the kinetic energy of the ball was converted into other forms of energy, such as heat and sound. Therefore, energy was conserved in this example.

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The correct answer is c. Both Technician A and B are correct.

Technician A is correct because there is often more than one circuit being protected by each fuse. Fuses are used to protect electrical circuits from excessive current, which can cause damage or fire. It is common for multiple circuits to be connected to a single fuse, as it simplifies the electrical system and reduces the number of fuses needed.

Technician B is also correct because more than one circuit often shares a single ground connector. A ground connector provides a path for excess electrical energy to flow safely to the ground, preventing damage to components and electrical shock. By sharing a ground connector, multiple circuits can utilize a common grounding point, further simplifying the electrical system and reducing the need for additional connectors.

Overall Both technicians are correct, as multiple circuits can be protected by a single fuse, and multiple circuits can share a single ground connector.

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The elastic limit is the maximum stress that a material can withstand without undergoing permanent deformation.

When a structure is built on a fault line, the elastic limit plays a crucial role in determining its ability to withstand seismic forces.

If the stress caused by an earthquake exceeds the elastic limit of the structure's materials, the structure may experience permanent deformation, which can lead to compromised structural integrity and potential failure.

In contrast, if the stress remains within the elastic limit, the structure can return to its original shape once the stress is removed, maintaining its structural integrity.

In conclusion, the elastic limit affects a structure built on a fault line by determining its resilience to seismic forces.

Ensuring that the stress remains within the elastic limit can help maintain the structure's integrity and minimize damage during earthquakes.

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The car will travel a distance of 180 meters in 12 seconds.

To determine the distance traveled by the car, we can use the equation of motion:

Distance (d) = Initial velocity (v₀) × time (t) + 0.5 × acceleration (a) × time squared (t²)

Given:

Initial velocity (v₀) = 0 m/s (starting from rest)

Acceleration (a) = 2.5 m/s²

Time (t) = 12 seconds

Plugging in the values into the equation:

Distance (d) = 0 × 12 + 0.5 × 2.5 × 12²

Distance (d) = 0 + 0.5 × 2.5 × 144

Distance (d) = 0 + 0.5 × 2.5 × 144

Distance (d) = 180 meters

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The magnitude of the force required to keep the rod moving at a constant speed is 0.9065 N.

First, let's find the induced electromotive force (EMF) using Faraday's law of electromagnetic induction: EMF = B * L * v, where L is the length of the rod, and v is its velocity. Converting the length to meters: L = 0.25 m.

EMF = 1.4 T * 0.25 m * 3.7 m/s = 1.295 V

Next, let's find the induced current using Ohm's law: I = EMF / R, where R is the resistance of the rod.

I = 1.295 V / 0.50 Ω = 2.59 A

The current induced in the rod is 2.59 A.

Now, let's calculate the magnitude of the force required to keep the rod moving at a constant speed. The force needed to maintain constant speed is equal to the magnetic force acting on the rod, which is given by F = I * L * B.

F = 2.59 A * 0.25 m * 1.4 T = 0.9065 N

The magnitude of the force required to keep the rod moving at a constant speed is 0.9065 N.

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Complete question:

The figure shows a 25 cm -long metal rod pulled along two frictionless, conducting rails at a constant speed of 3.7 m/s . The rails have negligible resistance, but the rod has a resistance of 0.50 Ω .

B=1.4T

What is the current induced in the rod?

What is the magnitude of the force is required to keep the rod moving at a constant speed?

(a). The thermal efficiency is approximately 22.9%.

(b). The heat discarded in each cycle is approximately 1.6063 × [tex]10^6[/tex] J.

(c). The mass of fuel burned in each cycle is approximately 0.035 kg.

(d). The engine's power output is approximately 222 kW or 297.6 hp.

To solve this problem, let's use the following formulas and conversions:

Given:

Heat input (Qin) = 1.61 × [tex]10^6[/tex]J

Work done per cycle (W) = 3700 J

Heat of combustion of gasoline (H) = 4.60 × [tex]10^7[/tex] J/kg

Cycles per second (f) = 60.0 cycles/s

(a) To calculate the thermal efficiency:

Thermal efficiency (η) = (Useful work output / Heat input) * 100%

η = (W / Qin) * 100%

η = (3700 J / 1.61 × 10^6 J) * 100%

η ≈ 0.229 * 100%

η ≈ 22.9%

(b) To calculate the heat discarded in each cycle:

Heat discarded = Heat input - Useful work output

Heat discarded = Qin - W

Heat discarded = 1.61 × [tex]10^6[/tex] J - 3700 J

Heat discarded ≈ 1.6063 × [tex]10^6[/tex] J

(c) To calculate the mass of fuel burned in each cycle:

Heat input = Heat of combustion * Mass of fuel burned

Mass of fuel burned = Heat input / Heat of combustion

Mass of fuel burned = 1.61 × [tex]10^6[/tex] J / 4.60 × [tex]10^7[/tex] J/kg

Mass of fuel burned ≈ 0.035 kg

(d) To calculate the power output in kilowatts and horsepower:

Power output (P) = Work done per cycle * Number of cycles per second

P = W * f

P = 3700 J * 60.0 cycles/s

P = 2.22 × [tex]10^5[/tex] J/s

Power output in kilowatts:

P(kW) = P / 1000

P(kW) ≈ 2.22 × [tex]10^5[/tex] J/s / 1000

P(kW) ≈ 222 kW

Power output in horsepower:

P(hp) = P / 745.7

P(hp) ≈ 2.22 × [tex]10^5[/tex] J/s / 745.7

P(hp) ≈ 297.6 hp

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The coefficient of volume expansion of a material is a measure of how much its volume changes with a change in temperature. For aluminium, the coefficient of volume expansion is approximately [tex]7.1 \times 10^{-5}[/tex] per degree Celsius ([tex]K^{-1}[/tex]).

For aluminium, the coefficient of volume expansion is approximately [tex]23.1 \times 10^{-6}[/tex] per degree Celsius. This means that for every degree Celsius increase in temperature, the volume of aluminium will increase by approximately 23.1 parts per million (ppm).

This coefficient of volume expansion is an important property of aluminium, as it affects its behaviour in a variety of applications. For example, in the aerospace industry, aluminium is used extensively in the construction of aircraft because of its low weight and high strength-to-weight ratio. However, as the temperature of the aircraft changes during flight, the volume of the aluminium components will also change, potentially affecting the structural integrity of the aircraft.

Understanding the coefficient of volume expansion is therefore essential for engineers and designers working with aluminium in a variety of fields, from aerospace to construction to electronics.

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The third piece moves with a velocity of 1.62 m/s in the direction opposite to 36.9 degrees from the positive x-axis.

Since the plate falls vertically to the floor, there is no initial velocity in the x or y direction.

Therefore, we can use conservation of momentum to determine the velocity of the third piece.

The total momentum of the plate before the impact is zero, since there is no initial velocity. The total momentum of the three pieces after the impact must also be zero, since there are no external forces acting on the system. Therefore, we can write:

m1v1 + m2v2 + m3v3 = 0

where m1, m2, and m3 are the masses of the three pieces, and v1, v2, and v3 are their respective velocities.

We know the masses and velocities of two of the pieces:

m1 = 320 g = 0.320 kg

v1 = 2.00 m/s

m2 = 355 g = 0.355 kg

v2 = 1.50 m/s

Substituting these values into the equation above and solving for v3, we get:

m3v3 = -(m1v1 + m2v2)

v3 = -(m1v1 + m2v2) / m3

Plugging in the values we know, we get:

v3 = -((0.320 kg)(2.00 m/s) + (0.355 kg)(1.50 m/s)) / 0.100 kg

v3 = -1.62 m/s

So the third piece moves in the opposite direction of the sum of the velocities of the other two pieces. Its velocity has a magnitude of 1.62 m/s, and it moves in the direction opposite to the vector sum of the velocities of the other two pieces. We can use the Pythagorean theorem to find the magnitude and direction of this vector:

[tex]|v| = \sqrt{(vx^2 + vy^2)}[/tex]

[tex]|v| = \sqrt{((2.00 m/s)^2 + (1.50 m/s)^2)[/tex]

|v| = 2.50 m/s

θ = atan(vy / vx)

θ = atan(1.50 m/s / 2.00 m/s)

θ = 36.9 degrees

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The tension in the left chain is 500 N, which is equal to the weight of the person on the board. Since the board is in equilibrium, the net force and net torque acting on the board must be zero.

Let's consider the forces acting on the board: the weight of the board (125 N) acts downward, the weight of the person (500 N) acts downward at the center of the board, and the tension in each chain acts upward.

Since the tension in the right chain is given to be 125 N, the total upward force acting on the board is 125 N + T (where T is the tension in the left chain).

Therefore, the net force on the board is 125 N + T - 625 N = 0 (where 625 N is the combined weight of the board and person).

Solving for T, we get T = 500 N, which is the tension in the left chain. So the tension in the left chain is 500 N, which is equal to the weight of the person on the board.

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2. The angle of refraction of the material is 16.0°.

4. Index of refraction of the prism is n = 1.45.

2. Using Snell's law:

n₁sinθ₁ = n₂sinθ₂

where n₁ = index of refraction of the first material (a), θ₁ = angle of incidence (13°), n₂ = index of refraction of the second material (1.60), and θ₂ = angle of refraction (unknown).

Plugging in the given values:

2.04sin13° = 1.60sinθ₂

θ₂ = sin⁻¹(2.04sin13°/1.60) = 16.0°

Therefore, the angle of refraction is θ = 16.0°.

4. Again, using Snell's law:

n₁sinθ₁ = n₂sinθ₂

where n₁ = index of refraction of water (1.33), θ₁ = angle of incidence (45°), n₂ = index of refraction of the prism (unknown), and θ₂ = angle of refraction (42°).

Plugging in the given values:

1.33sin45° = n₂sin42°

n₂ = sin45°/sin42° × 1.33 ≈ 1.45

Therefore, the index of refraction of the prism is n = 1.45.

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When the core temperature of the body decreases, the metabolic rate also decreases, leading to less production of carbon dioxide.

This results in a decrease in the partial pressure of CO2 in the blood, which leads to an increase in blood pH.

A higher pH means that the blood becomes more alkaline, which makes it more difficult for oxygen to dissociate from hemoglobin.

The reason for this is that oxygen binds to hemoglobin more tightly at a higher pH, which is known as the Bohr effect.

Thus, as the core temperature becomes colder, the oxygen-hemoglobin dissociation curve shifts to the left, making it more difficult for oxygen to be released from hemoglobin and making it less available to the tissues that require it.

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The area of contact between each tire and the ground is[tex]0.000562 m^2.[/tex]

The total weight supported by the ground is the sum of the weight of the rider and the bike:

W_total = 715 N + 98 N = 813 N

Since the weight is supported equally by the two tires, each tire supports half of the total weight:

W_per_tire = W_total / 2 = 406.5 N

The pressure in each tire is given as gauge pressure, which is the pressure above atmospheric pressure. Therefore, the absolute pressure in each tire is:

P_abs = P_gauge + P_atm

where P_atm is the atmospheric pressure, which we assume to be[tex]1.01* 10^5 Pa[/tex] (standard atmospheric pressure at sea level).

So, the absolute pressure in each tire is:

[tex]P_abs = 6.20 * 10^5 Pa + 1.01 *10^5 Pa = 7.21 *10^5 Pa[/tex]

The area of contact between each tire and the ground can be calculated using the equation:

F = P × A

where F is the force on the tire, P is the pressure, and A is the area of contact.

For each tire, we can write:

W_per_tire = P × A

Solving for A, we get:

A = W_per_tire / P

Plugging in the values we know, we get:

[tex]A = 406.5 N / 7.21 *10^5 Pa = 0.000562 m^2[/tex]

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The rear defroster is consuming 30.00 watts of electric power as it melts the frost. Electric power is the rate at which electrical energy is consumed or produced.

It is calculated by multiplying the voltage (V) across a device or component by the current (I) flowing through it.

To calculate the electric power consumed by the rear defroster, you can use the formula:

Power (P) = Voltage (V) × Current (I)

Given:

Current (I) = 6.00 A

Voltage (V) = 5.00 V

Substituting the values into the formula:

P = 5.00 V × 6.00 A

P = 30.00 W

Therefore, the rear defroster is consuming 30.00 watts of electric power as it melts the frost. The power indicates how quickly the defroster can generate heat and melt the frost on the rear window of the car.

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