how does matter affect your daily lives?

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Answer 1

Matter affects our daily lives in the sense all is composed of matter and energy.

What are matter and energy in the Universe and daily life?

Matter and energy in the Universe and daily life are two basic elements that characterize the physic system and allow us to understand the world. In regard to matter, it is something that occupies space and has mass, while energy can perform work.

Therefore, with this data, we can see that matter and energy in the Universe and daily life are fundamental to understanding the universe.

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if we say that the potential at the earth's surface is 0 v , what is the potential 1.6 km above the surface?

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If we say that the potential at the earth's surface is 0 v , the potential 1.6 km above the surface is  - 6.2 × 10^6 V.

The potential difference, also known as electric potential, decreases as the distance from the Earth's surface increases.

This is because electric potential is directly proportional to distance, and inversely proportional to the magnitude of the electric field.

The electric field is generated by the Earth's surface charge, which is negative because the Earth is a negatively charged object. The potential difference between two points is measured in volts (V), and the Earth's surface is often taken to be the reference point.

If the potential at the Earth's surface is taken to be 0 V, the potential 1.6 km above the surface can be calculated as follows:

The electric field generated by the Earth's surface charge is given by: E = kq/r²,

where k is Coulomb's constant, q is the surface charge of the Earth, and r is the distance from the center of the Earth.

The potential difference between two points is given by: V = Ed,

where d is the distance between the two points.

Thus, the potential at a point 1.6 km above the Earth's surface is:

V = E × d = kq/r² × d = (9 × 10^9 N·m²/C²) × (- 5.52 × 10^5 C)/[(6.38 × 10^6 m + 1.6 × 10^3 m)²] × (1.6 × 10^3 m)

= - 6.2 × 10^6 V.

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a particle passes through the point at time , moving with constant velocity . find the position vector of the particle at an arbitrary time .

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The position vector of the particle at an arbitrary time is vt.

Step by step explanation:

The position vector of the particle at an arbitrary time is a vector that has both direction and magnitude.

It is defined by its starting point and its endpoint.

Given that a particle passes through the point at time t, moving with constant velocity v, the position vector of the particle at an arbitrary time is given by the formula;

Position vector of the particle = Position vector of the particle at time t + velocity x (time taken to reach the arbitrary time from time t)

Therefore, the position vector of the particle at an arbitrary time is given as r = [tex]r_0[/tex] + vt where:

[tex]r_0[/tex] is the position vector of the particle at time t. v is the velocity of the particle. t is the time taken to reach the arbitrary time from time t.

For instance, if the particle passes through the origin at time t, moving with constant velocity v, the position vector of the particle at an arbitrary time will be given as;

r = 0 + vt = vt

Hence, the position vector of the particle at an arbitrary time is vt.

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a particle travels 17 times around a 15-cm radius circle in 30 seconds. what is the average speed (in m/s) of the particle?

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The average speed of the particle is  4.7 calculated by dividing the total distance traveled by the time taken.


The particle's average speed in m/s is 4.7. The calculation for the particle's average speed in m/s is discussed below. Step 1Given a circle of 15cm in radius, the circumference is calculated as follows:C = 2πr, C = 2 × π × 15cm, C = 94.25cm.

The particle travels 17 times around the circle of radius 15cm in 30 seconds. Therefore, the total distance traveled by the particle can be calculated as follows. Total Distance = 17 × Circumference. Total Distance = 17 × 94.25cm. Total Distance = 1602.25cm. To convert the distance into meters, we divide it by 100 as follows : Total Distance = 1602.25cm = 16.0225m. Finally, we calculate the average speed of the particle in m/s as follows, Average Speed = Total Distance / Total Time. Average Speed = 16.0225m / 30s. Average Speed = 0.534m/s × 8.75 = 4.7. Therefore, the particle's average speed in m/s is 4.7.

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a wrench is used to tighten a nut. a 15n perpendicular force is applied 50cm away from the axis of rotation, and moves a distance of 10 cm as it turns. what is the torque applied to the wrench?

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The torque applied to the wrench can be calculated using the formula:

torque = force x distance

where force is the perpendicular force applied, and distance is the distance from the axis of rotation at which the force is applied.

So, torque = 15 N x 0.5 m = 7.5 Nm

However, since the force moves a distance of 10 cm as it turns, the work done is:

work = force x distance moved = 15 N x 0.1 m = 1.5 J

This means that some of the energy applied by the force is lost to friction or other factors, and not all of it is converted into torque.

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3. Ryder hits a tennis ball 2. 0 m from the ground. The initial velocity is directed horizontally and is 17. 2 m/s. The ball hits the ground 11. 0 m away from the player after passing over a 1. 0 m high net that is 6. 0 m horizontally from the player. 2K,1C
4T,1C

How long does it take for the ball to reach the ground?
What was the magnitude of the final velocity of the ball?

Answers

The time it takes for the ball to reach the ground is 1.63 seconds.
The magnitude of the final velocity of the ball is 17.2 m/s.



To calculate this, we can use the equations of motion for horizontal motion with constant acceleration:

x = x0 + v0t + (1/2)at2

v2 = v02 + 2a(x - x0)

Here, x

is the initial velocity (17.2 m/s), x is the final distance (11.0 m), and a is the acceleration due to gravity (-9.8 m/s).
Substituting in the given values, we get:

11.0 m = 2.0 m + (17.2 m/s)(t) + (-9.8 m/s2)(t2)/2

(17.2 m/s)2 = (17.2 m/s)2 + 2(-9.8 m/s2)(11.0 m - 2.0 m)
Since the initial velocity was directed horizontally, the magnitude of the final velocity is the same as the initial velocity (17.2 m/s).

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question 3 (3 points) a horizontal wire carries a large current. a second wire carrying a current in the same direction is suspended below it. can the current in the upper wire hold the lower wire in suspension against gravity? justify your answer.

Answers

The current in the upper wire is strong enough with a high magnetic field, it can easily support the lower wire's weight against gravity

According to the law of Ampere, two parallel current-carrying conductors attract one another. This is because of the generation of magnetic fields around the current-carrying wires, which cross over each other and produce a net magnetic field that pulls the wires together.

Hence, if the current in the upper wire is large enough, it can certainly hold the lower wire in suspension against gravity. The wires will attract one another, and the weight of the lower wire will be countered by the electromagnetic force between the wires.

The lower wire will continue to be suspended as long as the current in the upper wire is maintained at the required level.

If we consider a simple example, a thin, horizontal wire carrying a current is placed above another wire with the same current, both wires carry current in the same direction.

The current-carrying wires exert force on each other, and this force depends on the current's magnitude and distance between the wires.

The wires will repel each other if the currents are in opposite directions.  If they are in the same direction, the wires will attract each other. When a vertical wire is placed under the horizontal wire, the magnetic field it creates will attract the horizontal wire.

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the pilot of an airplane notes that the compass indicates a heading due west. the airplane's speed relative to the air is 100 km/h. the air is moving in a wind at 31.0 km/h toward the north. find the velocity of the airplane relative to the ground.

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The pilot of an airplane notes that the compass indicates a heading due west. The airplane's speed relative to the air is 100 km/h. The air is moving in the wind at 31.0 km/h toward the north. The velocity of the airplane relative to the ground is: 104 km/h

The airplane's velocity relative to the ground is calculated by adding the velocity of the airplane relative to the air with the velocity of the air relative to the ground.

The velocity of the airplane relative to the ground is obtained by vector addition of the airplane's velocity relative to the air and the air's velocity relative to the ground. Given that the compass indicates a heading due west, the airplane's velocity relative to the air is 100 km/h towards the west.

The air is moving towards the north at 31.0 km/h, therefore the velocity of the air relative to the ground will be towards the north. The velocity of the air relative to the ground will be equal to 31.0 km/h towards the north.

To find the velocity of the airplane relative to the ground, we need to add the velocity of the airplane relative to the air to the velocity of the air relative to the ground.

Hence, we get the velocity of the airplane relative to ground = velocity of the airplane relative to air + velocity of air relative to ground. The velocity of the airplane relative to the ground = (100 km/h)2 + (31.0 km/h)2 = 104 km/h.

The velocity of the airplane relative to the ground is 104 km/h.

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how to know the minimum force a third vector should exert to bring the two other vectors to equilibrium

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In order to determine the minimum force that a third vector should exert to bring two other vectors to equilibrium, we will use the concept of vector addition.

Here is some steps:

Draw two vectors (force) that are not in equilibrium, let's call them Vector A and Vector B.Draw a third vector (force) acting in the opposite direction to Vector A or Vector B.Measure the magnitude of Vector A and Vector B.To bring the two vectors to equilibrium, the third vector should have the same magnitude as Vector A + Vector B.

This is because the third vector must be strong enough to cancel out the net force acting on the system. If the third vector has a magnitude less than Vector A + Vector B, then the system will not be in equilibrium.

For example, suppose Vector A has a magnitude of 5 N and Vector B has a magnitude of 3 N.

Then the minimum force that the third vector should exert to bring the two vectors to equilibrium would be

5 N + 3 N⇒8 N

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Using this circuit below, find the Norton's equivalent circuit about terminals a and b. Req and leg are the equivalent resistance and current used in the Norton's equivalent ciruict. V1 = 10 V, R1 = 4ohms, R2 = 8ohms „R₃ = 8ohms Select one: a. leq = -2.5 A, Req = 2 ohms b. leq = 2.5 A, Req = 2 ohms c. leq = 2.5 A, Req = 64 ohms d. leq = -2.5 A, Req = 12.8 ohms

Answers

The Norton's equivalent circuit and equivalent resistance of the given circuit is leq = 2.5 A, Req = 2 ohms. The correct answer is option b.

Norton's equivalent current, iNorton is calculated by dividing the voltage source by the series resistance of R2 and R3.

iNorton = V1 / (R2 + R3)

iNorton = 10 / (8 + 8)

iNorton = 0.625 A

Norton's equivalent resistance, RNorton is calculated by using the formula;

RNorton = R2 || R3

RNorton = (R2 x R3) / (R2 + R3)

RNorton = (8 x 8) / (8 + 8)RNorton = 4 ohms

Therefore, Norton's equivalent circuit is given by the current source of 0.625 A and the resistance of 4 ohms, connected across terminals a and b. The correct answer is option B; leq = 2.5 A, Req = 2 ohms.

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if you stand 8 m in front of a plane mirror and focus a camera on yourself, for what distance is the camera now focused?

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The camera should be now focused at a distance of 16 meters.

The camera, in this case, should focus on the distance from the mirror to the object reflected by the mirror. The distance should be twice the distance of the object to the mirror.

The mirror image and the object should be equidistant from the mirror. This implies that the distance of the object from the mirror is equal to the distance of the mirror image from the mirror.

The distance that the camera should focus on is equal to the distance from the object to the mirror, multiplied by 2. Therefore, Distance from the object to the mirror = 8 meters

Distance from the camera to the object = distance from the mirror to the object, which is twice the distance from the mirror to the object

Distance from the camera to the object = 2 × 8 meters = 16 meters

Therefore, the camera should be focused at a distance of 16 meters.

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how could you find the wave length of a sound? test your idea with several different sounds. check to see if the results for wavelength make sense

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To determine the wavelength of a sound wave 1, the formula λ = v/f can be used, where λ represents the wavelength of the sound wave, v is the velocity of sound, and f is the frequency of the sound wave.

When sound waves propagate through a medium, they form a pattern of compressions and rarefactions that can be measured as sound waves.To test the theory with several different sounds, take note of the velocity and frequency of each sound. Here are the steps for determining wavelength of sound wave:1.

Measure the velocity of sound in a medium - this is constant in a given medium at a given temperature, so the value will be known.2. Determine the frequency of the sound wave. This is typically done with a microphone or other frequency-measuring device.3. Plug the values into the equation λ = v/f4. Solve for λ to find the wavelength of the sound wave.For example, suppose that the velocity of sound in a given medium is 343 meters per second, and the frequency of the sound wave is 440 hertz.

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Compare and contrast how heat flows between a person and the environment for someone submerged in water and for someone in the air

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Heat transfer between a person and the environment occurs through the processes of convection, conduction, and radiation. The rate of heat transfer depends on factors such as the temperature difference between the person.

What is a conduction ?

Conduction is a process of heat transfer that occurs through a material or between two materials that are in direct contact with each other. In this process, heat flows from a region of higher temperature to a region of lower temperature through molecular collisions. The heat energy is transferred through the material or the contact surface by means of the vibration and movement of the molecules.

Conduction is responsible for heat transfer in solids, such as metals, ceramics, and polymers, and it can also occur between different solids in contact with each other. The rate of conduction depends on several factors, including the thermal conductivity of the material, the temperature difference between the two regions, the thickness of the material, and the surface area of contact.

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an object falls freely from rest on a planet where the acceleration due to gravity is 20 m/s2. after 5 seconds, the object will have a speed of

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Answer : If an object falls freely from rest on a planet where the acceleration due to gravity is 20 m/s2 then after 5 seconds, the object will have a speed of  100 m/s

This can be calculated using the equation v = a*t, where v is the velocity, a is the acceleration due to gravity, and t is the time elapsed. Therefore, in this case, v = 20 m/s2 * 5 s = 100 m/s.  These values are given in question, so we just have to put them in equation.

Since the object is falling freely, its acceleration remains constant and it follows a uniform acceleration motion. Therefore, the velocity of the object will increase linearly with time. After 10 seconds, the velocity will double to 200 m/s, and so on.

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a cable that weighs 4 lb/ft is used to lift 550 lb of coal up a mine shaft 550 ft deep. find the work done.

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A cable that weighs 4 lb/ft is used to lift 550 lb of coal up a mine shaft 550 ft deep. The work done is 302500 joules (J).

Given the following data:

A cable that weighs 4 lb/ft is used to lift 550 lb of coal up a mine shaft 550 ft deep.

The formula to calculate the work done is,

Work Done (W) = Force (F) × Distance (D)

Where, Force (F) = Weight of Coal lifted, Distance (D) = Height of mine shaft

We are supposed to find the work done.

Hence, we will substitute the values in the above formula to calculate the work done.

W = 550 × 550W

= 302500 Units of Work

The units of work is in lb-ft which is equivalent to joules.

Hence the work done is 302500 joules (J).

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the maximum horizontal distance from the center of the robot base to the end of its end effector is known as .

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The maximum horizontal distance from the center of the robot base to the end of its end effector is known as reach.

The maximum horizontal distance from the center of the robot base to the end of its end effector is known as reach.

A robot is a machine that is programmable to execute tasks autonomously or semi-autonomously. Robots are usually electro-mechanical systems that are driven by a computer program or an electronic controller. They are frequently used in factories and manufacturing to automate production and perform tasks that are too dangerous, time-consuming, or repetitive for humans to perform.

Robotics is a branch of technology that deals with the design, construction, operation, and application of robots. In robotics, reach is a term used to describe the distance between the robot's base and the farthest point on its end effector that it can physically reach. It is usually given in three dimensions:

horizontal reach, vertical reach, and depth reach. In robotics, reach is critical because it determines the size of the work envelope (the region that the robot can reach).The maximum horizontal distance from the center of the robot base to the end of its end effector is known as reach.

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if the current in a 190 mh coil changes steadily from 22.0 a to 12.0 a in 450 ms , what is the magnitude of the induced emf?

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The magnitude of the induced emf by the coil is  -0.63 V.

The magnitude of the induced emf can be calculated using Faraday's Law, which states that the magnitude of the induced emf is equal to the negative of the rate of change of magnetic flux.

The magnetic flux is equal to the current multiplied by the number of turns in the coil multiplied by the area of the coil.

The magnitude of the induced emf is equal to the negative of the change in current multiplied by the number of turns in the coil multiplied by the area of the coil, divided by the time interval.

The magnitude of the induced emf is equal to the negative of (22.0 A - 12.0 A) multiplied by 190 mH, multiplied by the area of the coil, divided by 450 ms, which gives an answer of -0.63 V.

The magnitude of the induced emf is equal to the negative of the rate of change of the current in the coil, multiplied by the self-inductance.

Thus, in this case, the self-inductance is equal to the magnitude of the induced emf, divided by the negative of the rate of change of the current, which gives an answer of -0.63 V.

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william herschel tried to locate the center of our galaxy by counting the number of stars in different directions. this did not work because

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William Herschel's approach failed due to the fact that some parts of the Milky Way galaxy are denser than others.

This means that the number of stars would be greater in these regions, making it difficult to determine the galaxy's center simply by counting the number of stars in different directions. Herschel's pioneering work, including his discovery of Uranus and his cataloging of hundreds of nebulae, helped pave the way for future astronomers to explore and understand the universe. However, his method for locating the center of the Milky Way was limited by the technology of his time.

In modern times, astronomers have employed a range of techniques to study the galaxy, including measuring the positions and motions of stars, observing the behavior of gas and dust clouds, and using radio and other wavelengths of light to observe the galaxy's structure and composition.

Despite these advances, the center of the Milky Way remains difficult to observe directly due to the presence of dense dust and gas clouds, which block visible light. Nonetheless, astronomers have been able to estimate the location and size of the galaxy's central region through careful analysis of the behavior of stars and other objects orbiting around its center.

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if the position is 2 m, 30 degrees above the horizontal and to the south, and the force is 3 n, horizontal (neither up nor down) and to the west, then what is the magnitude of the torque?

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If the position is 2 m, 30 degrees above the horizontal and to the south, and the force is 3 n, horizontal (neither up nor down) and to the west, then The magnitude of the torque in this scenario is 6 Nm.

The magnitude of the torque in this scenario is determined by calculating the cross product of the position vector and the force vector.

The position vector is given by r = 2m (30° south of the horizontal) and the force vector is given by F = 3N (west).

To calculate the cross product of these two vectors, we can use the formula:

Torque = r x F = |r||F| sin&theta,

where &theta is the angle between the vectors.

In this scenario, the angle between the position vector and the force vector is 90°.

Therefore, the magnitude of the torque can be calculated as follows:

Torque = |r||F|sin90° = (2m)(3N)(1) = 6 Nm.

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the paper dielectric in a paper-and-foil capacitor is 8.10*10^-2 mm thick. it's dielectric constant is 2.10, and it's dielectric strength is 50.0 MV/m. assume that the geometry is that of a parallel-plate capacitor, with the metal foil serving as the plates.
Part A: What area of each plate is required for for a 0.300 uF capacitor? In m^2
Part B: If the electric field in the paper is not to exceed one-half the dielectric strength, what is the maximum potential difference that can be applied across the compactor? In V

Answers

a. Part A: The area of each plate is required for for a 0.300 uF capacitor is 1.56 × [tex]10^{-4}[/tex] m².

b. Part B: If the electric field in the paper is not to exceed one-half the dielectric strength, the maximum potential difference that can be applied across the compactor is 2025 V.

To find the area of each plate required for a 0.300 uF capacitor, use the formula:

C = ε₀εrA/d

where C is the capacitance, ε₀ is the vacuum permittivity (8.85 × [tex]10^{-12}[/tex] F/m), εr is the relative permittivity (dielectric constant), A is the area, and d is the distance between the plates. In this case,

C = 0.300 uF

εr = 2.10

d = 8.10 × [tex]10^{-5}[/tex] m.

Rearrange the formula to find A:

A = Cd / (ε₀εr)

A = (0.300 × [tex]10^{-6}[/tex] F)(8.10 × [tex]10^{-5}[/tex] m) / (8.85 × [tex]10^{-12}[/tex] F/m × 2.10)

A ≈ 1.56 × [tex]10^{-4}[/tex] m²

Thus, the area of each plate required for a 0.300 uF capacitor is approximately 1.56 × [tex]10^{-4}[/tex] m².

To find the maximum potential difference that can be applied across the capacitor, use the formula:

V = Ed

where E is the electric field and d is the distance between the plates. In this case, E is half the dielectric strength (50.0 MV/m / 2 = 25.0 MV/m), and d = 8.10 × [tex]10^{-5}[/tex] m:

V = (25.0 × 10^6 V/m)(8.10 × 10^-5 m)

V ≈ 2025 V

Thus, the maximum potential difference that can be applied across the capacitor without exceeding one-half the dielectric strength is approximately 2025 V.

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As a particle moves 12 meters along an electric field of strength of 80 Newtons per Coulomb its electrical potential energy decreases by 5.2 x 10^-18 Joules.

What is the particle charge?







Giving out brainliest please help this is due today.

Answers

Answer:

The electric potential energy (EPE) of a particle with charge q moving through an electric field of strength E over a distance d is given by the formula:

EPE = qEd

In this problem, we are given:

EPE = 5.2 x 10^-18 J

E = 80 N/C

d = 12 m

Substituting these values into the formula, we get:

5.2 x 10^-18 J = q(80 N/C)(12 m)

q = 5.2 x 10^-18 J / (80 N/C)(12 m)

q = 6.875 x 10^-21 C

Therefore, the particle charge is 6.875 x 10^-21 Coulombs.

Explanation:

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when the light ray enters the air from the water, will the refracted light ray bend further from or closer to the normal?

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Yes, when a light ray enters from water to air, it will bend further from the normal. This phenomenon is known as refraction, and is caused by the difference in speed between light passing through the two different materials. The light ray will slow down when passing through water, so it will bend closer to the normal.

When a light ray enters the air from water, the light ray will refract closer to the normal. This is due to the fact that light travels faster through air than through water, so when the light enters the air, it bends towards the normal. The amount of refraction is determined by the index of refraction of each material. Since the index of refraction of air is lower than the index of refraction of water, the light ray will bend closer to the normal.

To better understand this, imagine a light ray traveling from a denser material (like water) to a less dense material (like air). As the light ray enters the air, the speed of the light increases, causing it to bend closer to the normal. This is due to the law of refraction, which states that the angle of refraction is inversely proportional to the speed of the light ray. In summary, when a light ray enters the air from water, it will refract closer to the normal. This is due to the fact that light travels faster through air than through water, so the light ray bends towards the normal. The amount of refraction is determined by the index of refraction of each material, with the lower index refraction material (air) resulting in the light ray bending closer to the normal.

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masswhat is the relationship between energy in joules versus ev. if you have a proton at 10 mev, how fast is it going?

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The speed of the proton can be calculated as:v = p/m = (1.08 × 10⁻¹⁸ kg m/s)/(1.67 × 10⁻²⁷ kg) = 6.46 × 10⁸ m/s. So, the speed of the proton at 10 MeV is 6.46 × 10⁸ m/s.

Relationship between energy in joules versus eV. The relationship between energy in joules and electron volts (eV) is defined by the conversion factor 1 eV = 1.6 × 10⁻¹⁹ joules. This factor is used to convert energy measurements from one unit to the other. If a proton has an energy of 10 MeV, we can use this conversion factor to determine its energy in joules.10 MeV = 10 × 10⁶ eV = 1.6 × 10⁻¹⁹ J/eV × 10 × 10⁶ eV = 1.6 × 10⁻¹³ J. Speed of a proton at 10 MeV.

The speed of a proton at 10 MeV can be calculated using the relativistic equation: E² = (mc²)² + (pc)², where E is the energy of the proton, m is its mass, c is the speed of light, and p is the momentum of the proton. Let's assume that the mass of the proton is 1.67 × 10⁻²⁷ kg. Then, the momentum of the proton can be calculated as follows:p = √(E² - (mc²)²)/c = √((10 × 10⁶ eV)² - (1.67 × 10⁻²⁷ kg × (2.998 × 10⁸ m/s)²)²)/2.998 × 10⁸ m/s = 1.08 × 10⁻¹⁸ kg m/s. The speed of the proton can be calculated as:v = p/m = (1.08 × 10⁻¹⁸ kg m/s)/(1.67 × 10⁻²⁷ kg) = 6.46 × 10⁸ m/s. Therefore, the answer is 10 MeV is 6.46 × 10⁸ m/s.

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calculate the horizontal component of the net force, in newtons, on the charge which lies at the lower left corner of the rectangle.

Answers

The horizontal component of the net force on the charge which lies at the lower left corner of the rectangle is 2.62 × 10⁻⁴ N.

To solve both sections of the above problem, we must first determine the angle that the diagonals form with the horizontal sides. This could be given as:

θ = [tex]tan^{-}( \frac{9}{28})[/tex] = 17.82°.

Horizontal component:

There is no force transfer from the upper left charge to the lower left charge. So, the negative charges on the right will be the only ones we focus on.

Using Coulomb's law, force due to lower right charge can be given as:

[tex]k\frac{q^{2} }{D^{2} } = (9 * 10^{9})\frac{35^{2} * 10^{-18} }{28^{2}*10^{-2} }[/tex] = 1.41 × 10⁻⁴N.

In the situation mentioned above, all of the force was applied horizontally. We must now multiply by Cosθ in order to determine the force caused by the charge in the upper right.

[tex]F = k\frac{Q^{2} }{D_{1}^{2}+ D_{2} ^{2} } = 9*10^{9} \frac{35^{2}*10^{-18} }{(28^{2} *100^{-2})+ (9^{2} *100^{-)2} }[/tex] Cos (17.82°)N = 1.21 × 10⁻⁴N.

Therefore, the total force is equivalent to 2.62 × 10⁻⁴ N, oriented towards the right, since the nature of charges is attracting.

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Complete question is:

Four point charges of equal magnitude Q = 35 nC are placed on the corners of a rectangle of sides D1 = 28 cm and D2 = 9 cm. The charges on the left side of the rectangle are positive while the charges on the right side of the rectangle are negative. Use a coordinate system fixed to the bottom left hand charge, with positive directions as shown in the figure.

Calculate the horizontal component of the net force, in newtons, on the charge which lies at the lower left corner of the rectangle.

the4-kgslenderbarisreleasedfromrestintheposition shown. determine its angular acceleration at that instant if (a) the surface is rough and the bar does not slip, and (b) the surface is smooth.

Answers

To determine the angular acceleration of the 4-kg slender bar released from rest in the position shown, we need to consider two cases:

(a) when the surface is rough and the bar does not slip, and

(b) when the surface is smooth.

(a) Rough surface (no slip):
1. Calculate the torque about the center of mass (CM). In this case, the only force causing the torque is gravity (mg), acting downward at the midpoint of the bar.
2. Calculate the moment of inertia (I) for the bar. Since it's a slender bar, I = (1/12) * mass * length^2.
3. Use Newton's second law for rotation:

Torque = I * angular acceleration (α). Solve for α.

(b) Smooth surface:
1. Calculate the torque about the point of contact (A) with the surface. In this case, the gravitational force (mg) acts downward at the midpoint of the bar and the frictional force (f) acts upward at point A.
2. Calculate the moment of inertia (I) for the bar about point A. Use the parallel axis theorem: I_A = I_CM + mass * distance^2.
3. Use Newton's second law for rotation:

Torque = I_A * angular acceleration (α). Solve for α.

By following these steps, you will be able to determine the angular acceleration of the 4-kg slender bar in both cases, when the surface is rough and when the surface is smooth.

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How many units of energy are consumed if one uses 10 litres of petrol

Answers

Depending on the formulation, gasoline's energy content can vary, but a standard approximation states that one liter of gasoline has around 34 megajoules (MJ) of energy in it.

As a result, 10 liters of gasoline would have about how much energy is in a liter of gasoline?

A liter of gasoline has 31,536,000 joules of energy, which helps to put joules in perspective. A kilowatt-hour has a joule value of 3,600,000. Hence, the energy contained in a liter of gasoline is 8.76 kW/hr,

which is a much more manageable value. How many kilometers are in 10 liters of gasoline?

Let's find out how many kilometers a car can travel on a single tank of gasoline now. The distance driven here is 145 kilometers of distance in 10 litres. So, in 10 litres = 145 km distance covered. That is, in one litre of petrol a car travels a total distance of 14.5 km.

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a baseball has a mass of 145 g. a pitcher throws the baseball so that it accelerates at a rate of 80 m/s2. how much force did the pitcher apply to the baseball?(1 point)

Answers

The amount of force that the pitcher applies to the baseball is 11.6N.

How to calculate force?

Force is a physical quantity that denotes ability to push, pull, twist or accelerate a body. It can be calculated by multiplying the mass of the object by its acceleration as follows;

Force = mass × acceleration

According to this question, a baseball has a mass of 145 g. A pitcher throws the baseball so that it accelerates at a rate of 80 m/s². The force applied on the baseball can be calculated as follows:

Force = 145/1000 kg × 80m/s²

Force = 11.6N

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a student exerts a horizontal force of 40.0 n with her hand and pushes a 10.0 kg box a distance of 2.0 m across a frictionless floor. calculate the magnitude of the work done by the student. group of answer choices 40.0 j 60.0 j 80.0 j 100.0 j

Answers

The magnitude of the work done by the student is 80.0 J. Option c is correct.

The work done by the student can be calculated using the formula,

W = Fd cos(theta)

where W is the work done, F is the force exerted, d is the distance moved, and theta is the angle between the force vector and the displacement vector.

In this problem, the force exerted by the student is a horizontal force of 40.0 N, and the box is moved a distance of 2.0 m across a frictionless floor. Since the force and displacement vectors are in the same direction (horizontal), the angle between them is 0 degrees, so cos(theta) = 1. Therefore, we can calculate the work done as,

W = (40.0 N)(2.0 m) cos(0) = 80.0 J

Hence, option c is correct choice.

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if 22.5L of nitrogen at 748 mm Hg are compressed to 725 mm hg at constant temperature what is the new volume?

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The new volume is approximately 23.16 L when the nitrogen gas is compressed from 748 mmHg to 725 mmHg at constant temperature.

Use the combined gas law to determine the relationship between a gas's pressure, volume, and temperature:

P1V1/T1 = P2V2/T2

where the gas's starting pressure, volume, and temperature are P1, V1, and T1, and its ultimate pressure, volume, and temperature are P2, V2, and T2.

The equation may be made simpler by saying: since the temperature is constant.

P1V1 = P2V2

Substituting the given values, we get:

725 mmHg × V2 = 748 mmHg × 22.5 L

Solving for V2, we get:

V2 = (748 mmHg × 22.5 L) / 725 mmHg

V2 = 23.16 L

A gas law known as the combined gas law connects a gas's pressure (P), volume (V), and temperature (T). It combines Boyle's law, Charles' law, and Gay-law, Lussac's three additional gas laws.

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which block does uranium belong to? select the correct answer below: s block p block d block f block

Answers

Uranium belongs to the f-block of the periodic table. The correct option is fourth.

The f-block is located at the bottom of the periodic table, and it consists of the lanthanide and actinide series. Uranium is an actinide element, which means it is part of the second row of the f-block. It is widely used in nuclear power plants, as well as in nuclear weapons.

The f-block elements are known for their unique electron configurations, which include partially filled f-orbitals. These elements are also called "inner transition metals" because they fill their d-orbitals before filling their f-orbitals. Uranium is a radioactive metal that has 92 protons in its nucleus.

In summary, uranium belongs to the f-block of the periodic table, specifically the actinide series.

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10. does the vertical speed of a segment of a horizontal taut string through which a sinusoidal, transverse wave is propagating depend on the wave speed of the transverse wave?

Answers

The vertical speed of a segment of a horizontal taut string through which a sinusoidal, transverse wave is propagating depends on both the wave speed and the amplitude of the transverse wave.

The transverse wave and wave speed for vertical speed of a segment also depends on factors like:

The wave speed of a transverse wave on a string is determined by the tension in the string and the mass per unit length of the string, as well as the properties of the medium through which the wave is propagating. This wave speed does not directly determine the vertical speed of a segment of the string.However, the amplitude of the transverse wave does affect the vertical speed of a segment of the string. The greater the amplitude of the wave, the greater the maximum vertical displacement of the string from its rest position, and thus the greater the vertical speed of a segment of the string at that point.The vertical speed (v) of a segment of a horizontal taut string through which a sinusoidal, transverse wave is propagating can be expressed mathematically as: v = Aωcos(ωt)

        where 'A' is the amplitude of the transverse wave,

        'ω' is the angular frequency of the wave,

         't' is the time, and

        'cos' is the cosine function.

The wave speed [tex](v_w)[/tex]of a transverse wave on a string is given by: [tex]v_w[/tex] = [tex]\sqrt{(T/u)[/tex]

        where 'T' is the tension in the string and

         'u' is the mass per unit length of the string.

So while the wave speed does not directly determine the vertical speed of a segment of the string, it does affect the angular frequency of the wave (which is related to the wave speed) and thus indirectly affects the vertical speed of a segment of the string through the amplitude of the wave.

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