Why is a scientific theory the most powerful explanation scientists have to offer?
O Many different scientists have added data from their own experiments to build the theory.
OA theory is the same things as a hypothesis.
O Scientific theories are usually the work of a single scientist.
O Technology is used to provide the experimental data for a scientific theory.

The particle's average speed after 10 seconds is 110 m/s, and its average velocity is zero.

When a particle is thrown upwards, its initial velocity is in the upward direction and its acceleration is in the downward direction due to gravity. The acceleration due to gravity is approximately 10 m/s² near the surface of the Earth. Therefore, the particle's velocity decreases at a rate of 10 m/s² until it reaches its highest point, where its velocity is zero. After that, the particle's velocity becomes negative and it starts to fall back to the ground.

To find the particle's average speed after 10 seconds, we need to calculate the total distance traveled by the particle in 10 seconds. The formula to calculate the distance traveled by a particle under constant acceleration is:

distance = initial velocity * time + (1/2) * acceleration * time²

Substituting the given values, we get:

distance = 60 m/s * 10 s + (1/2) * 10 m/s² * (10 s)²

distance = 600 m + 500 m

distance = 1100 m

Therefore, the average speed of the particle after 10 seconds is:

average speed = total distance / total time

average speed = 1100 m / 10 s

average speed = 110 m/s

To find the particle's average velocity after 10 seconds, we need to calculate the displacement of the particle in 10 seconds. Displacement is the change in position of the particle, which is equal to the difference between its final and initial positions. Since the particle is thrown upwards and then falls back to the ground, its displacement after 10 seconds is zero. Therefore, the average velocity of the particle after 10 seconds is also zero.

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a. The rock's gravitational potential energy, just before touching the ground is 0 J.

b. The rock's kinetic energy, just before touching the ground is 24,500 J.

The rock's gravitational potential energy, just before touching the ground is calculated as follows;

Just before touching the ground, the rock will have maximum velocity, and the kinetic energy of rock will be maximum while the gravitational potential energy will be minimum or zero.

P.E (top height) = K.E (minimum or zero height)

P.E = mgh

where;

P.E = 10 x 9.8 x 250

P.E = 24,500 J

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Answer:

Image distance: object distance + focal length or, 8.0 + 3.0 so 11.0cm

Object distance: image distance - focal length or 11.0 - 3.0 so 8.0cm

The coefficient of static friction between a 20 kg weight and a football turf is .85. Force of approximately 166.6 Newtons (N) is needed to make the weight start moving.

The force needed to make the 20 kg weight start moving can be determined using the coefficient of static friction. The equation for static friction is:

Frictional force = coefficient of static friction * normal force

The normal force is the force exerted by the surface perpendicular to the weight. In this case, it is equal to the weight of the object, which can be calculated as the mass (20 kg) multiplied by the acceleration due to gravity (9.8 m/s^2):

Normal force = mass * gravity = 20 kg * 9.8 m/s^2 = 196 N

Now, we can calculate the force needed to make the weight start moving:

Frictional force = 0.85 * 196 N ≈ 166.6 N

Therefore, a force of approximately 166.6 Newtons (N) is needed to overcome the static friction and make the 20 kg weight start moving on the football turf. This force must be applied in the opposite direction to the frictional force to initiate motion.

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Answer:

Yeah

Explanation:

Thermoproteota is a prokaryote.

Prokaryotes are unicellular

Answer:

d

Explanation:

reversing the current, reverses the poles of the electromagnet....this keeps the rotor spinning in the stationary mag field

The amount of Sertaline will you give to the patient is 30 mg that is half of the given 60mg tablet.

To give the patient a 30mg measurement of Sertraline when the accessible dose is 60mg tablets, one-half of the tablet ought to be given. The tablet ought to be part in half employing a pill cutter and after that one of the parts oughts to be managed to the persistent. This will give the persistent the precise 30mg measurements that have been requested. 

Sertraline is an antidepressant pharmaceutical utilized to treat depression, obsessive-compulsive disorder, freeze clutter, post-traumatic stretch clutter, social uneasiness clutter, and premenstrual dysphoric disorder. It belongs in the course of drugs known as selective serotonin reuptake inhibitors (SSRIs) and works by expanding the sum of serotonin, a natural substance within the brain, which makes a difference to preserve mental adjustment. Sertraline is as a rule taken once day by day, with or without nourishment, and ought to be taken as coordinated by a healthcare supplier. 

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The height to which the object is lifted is 0.14 meters.The problem seems to involve the concept of work, potential energy, and equilibrium.

The upward force of 442N must be balanced by the downward force of the object's weight to maintain equilibrium. Assuming the object is stationary, we can equate the upward force to the weight of the object:

442N = weight of the object

Weight = m * g, where m is the mass of the object and g is the acceleration due to gravity (9.8m/s^2).

So, 442N = m * 9.8m/s^2

Solving for m, we get m = 45.10 kg.

Now, the work done by the applied force of 32N over a distance of 2m is given by W = F * d = 32N * 2m = 64 J (Joules).

As the object is lifted, its potential energy increases by the amount of work done on it. This potential energy is given by the formula:

Potential energy = m * g * h

where h is the height to which the object is lifted.

Equating the work done on the object to the increase in its potential energy, we get:

64 J = 45.10 kg * 9.8m/s^2 * h

Solving for h, we get h = 0.14 m (rounded to two decimal places).

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The electrical change of oxygen ion is -2

An ion is an atom or molecule that has an unequal number of protons and electrons, which results in a net electrical charge. When an oxygen atom gains two electrons, it becomes negatively charged because electrons have a negative charge. This is because the number of negatively charged electrons now exceeds the number of positively charged protons in the nucleus of the atom. Therefore, the electrical charge of the oxygen ion is -2.

The electrical charge of an ion is represented by the symbol "n", which stands for the number of electrons gained or lost by the atom. In this case, the oxygen atom has gained two electrons, so n = -2. This is represented by the formula: O^2-.

In summary, when an oxygen atom gains two electrons, it becomes negatively charged and forms an oxygen ion. The electrical charge of the oxygen ion is represented by the symbol "n" and is equal to -2, which is reflected in the formula O^2-. This means that the oxygen ion has a net charge of -2, which reflects the excess of negatively charged electrons in the ion.

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The work done to lift a man and a 44-pound box up 10 meters of stairs is 11,772 J, and the power required to do this in 1 minute is 196.2 W.

a. First, we need to convert the units of the weight of the box to kilograms: 44 pounds = 20 kg. The total mass that the man needs to lift up the stairs is therefore 100 kg + 20 kg = 120 kg.

The force required to lift this mass against gravity is given by F = mg, where g is the acceleration due to gravity (9.81 m/s^2). So, F = 120 kg x 9.81 m/s^2 = 1177.2 N. The work done is then given by W = Fd, where d is the distance moved (10 meters).

Thus, W = 1177.2 N x 10 m = 11,772 J.

b. To find the power required, we need to divide the work done by the time taken. As the time is given in minutes, we need to convert it to seconds: 1 minute = 60 seconds. Therefore, the power required is P = W/t = 11,772 J / 60 s = 196.2 W.

Therefore, the work done to lift the man and the box up a flight of stairs 10 meters high is 11,772 J, and the power required to do this in only 1 minute is 196.2 W.

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The centripetal force affecting the body has to be doubled.

1. The centripetal force acting on a body moving in a circular path of radius r with speed v is given by F = mv²/r, where m is the mass of the body.

2. If the speed of the body increases to √2v while moving in the same circular path, the new centripetal force acting on the body can be calculated as follows:

  F' = m(√2v)²/r = 2mv²/r

3. Comparing the new centripetal force F' with the initial centripetal force F, we get:

  F' = 2F

4. As a result, the centripetal force acting on the body must be twice in order for the body to proceed in the same circular direction at 2v.

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Answer:

Therefore, the period of vibration for a wave with a wavelength of 0.5 m moving at a speed of 6.0 m/s is approximately 0.0833 seconds.

Explanation:

To find the period of vibration, we can use the formula:

period = wavelength / speed

Given:

Wavelength (λ) = 0.5 m

Speed (v) = 6.0 m/s

Substituting the values into the formula:

period = 0.5 m / 6.0 m/s

Calculating the value:

period ≈ 0.0833 seconds

The mass of 5 cm³ of copper is 45 grams. This means that if we had a block of copper with a volume of 5 cm³, it would weigh 45 grams.

The density of copper is given as 9 g/cm³, which means that for every cubic centimeter (cm³) of copper, there is a mass of 9 grams (g). To find the mass of 5 cm³ of copper, we can use the following formula:

mass = density x volume

where mass is the mass of the object in grams, density is the density of the material in grams per cubic centimeter, and volume is the volume of the object in cubic centimeters.

Plugging in the values we have, we get:

mass = 9 g/cm³ x 5 cm³

mass = 45 g

Therefore, the mass of 5 cm³ of copper is 45 grams. This means that if we had a block of copper with a volume of 5 cm³, it would weigh 45 grams.

It is important to note that the density of a material is an important physical property that relates its mass to its volume, and is often used in calculations involving materials and objects of different shapes and sizes.

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Humans detect infrared radiation released by things using a variety of senses. The ultimate response is that people recognise it by feeling warmth on their skin, having sharper vision, seeing light radiated by the item, and having goosebumps.

1. Skin detects warmth: Objects that produce infrared radiation also emit heat. The temperature sensors in the skin sense this heat energy, allowing people to experience warmth.

2. Improved vision: Infrared radiation may pass through some things like fog or smoke. Because infrared radiation bounces off things and reaches the eyeballs, people can see more clearly in low-light circumstances.

3. Light emitted by the object: When subjected to infrared radiation, certain items emit visible light. Certain types of security cameras, for example, detect movement using infrared radiation and can generate visible light as a result.

4. Goosebumps: Goosebumps are a natural reaction to temperature fluctuations or emotional stimulation.

When people are chilly or terrified, the muscles in their skin contract, causing the hair follicles to rise up and provide a "goosebump" sensation.

This reaction can also occur when the skin senses infrared light in order to regulate body temperature.

Overall, these sensory reactions enable people to perceive and respond to infrared radiation released by objects in their surroundings.

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a) The current through R₂, R₃ and R₄ is 8/3 mA, 4/3 mA, and 40 mA respectively

The resistor with resistance R₄ is connected in series with the resistor R₁, thus the current through them is the same and 40 mA.

The parallel combination of R₂ and R₃ is connected in series with R₁ thus the current through the parallel combination is 40mA

Resistance in parallel = 5 * 10 / 5 + 10 = 10/3

Current = 40 mA

Voltage = IR according to Ohm's Law

V = 10/3 * 40

= 40/3 mV

Since voltage drop is equal in parallel combination,

40/3 = I * 10

I = 4/3 mA (R₃)

40/3 = I * 5

I = 8/3 mA (R₂)

b) The potential difference between A and B is 11.83 V

[tex]V_{AB[/tex] = 1.5 - iR

= 1.5 - 10/3 * 40

= 1.5 - 40/3

= 1.5 - 13.33

= 11.83 V

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The correct matches are Production line - Televisions, Continuous flow - Flu vaccines, Custom - Kitchen cabinets, Fixed - Submarines.

1. Production line: A production line is a manufacturing process that involves a series of workstations where individual operations are performed to create a final product. Each workstation is responsible for a specific task, and the product moves from one station to another until it is completed. The production line is used for mass production of standardized products, and it is characterized by a high level of automation and a high production rate.

2. Continuous flow: Continuous flow is a manufacturing process where raw materials or components are continuously fed into the production process, and the finished product is continuously outputted without interruption. This process is used for products that are made in large quantities and have a high demand. The continuous flow process is characterized by a high degree of automation and a low level of labor involvement.

3. Custom: Custom manufacturing is a production process where products are made to meet the specific needs and requirements of individual customers. This process involves the customization of products based on the customer's specifications, and it is characterized by a high level of flexibility and customization. The custom manufacturing process is often used for products that require a high degree of customization, such as furniture or clothing.

4. Fixed: Fixed manufacturing is a production process where products are made using a fixed set of specifications and processes. The fixed manufacturing process is characterized by a low level of customization and a high degree of repeatability. This process is often used for products that require a standardized manufacturing process, such as electronic components or automotive parts.

Therefore, The correct Answers are Production line - Televisions, Continuous flow - Flu vaccines, Custom - Kitchen cabinets, Fixed - Submarines.

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(a)Since the integral does not converge to 1, we can conclude that the wave function is not normalized, (b) The normalization constant for the given wave function is A = √10.

In quantum mechanics, the wave function is a mathematical description of the state of a quantum system. It encodes the probability amplitude for a particle or collection of particles to have certain properties, such as position or momentum.

a- To check if the wave function is normalized, we need to calculate the integral of the absolute value squared of the wave function over all space and check if it equals 1. Mathematically, this can be written as:

∫|Ψ(x,t)|^2dx = ∫Ψ*(x,t)Ψ(x,t)dx

where Ψ*(x,t) is the complex conjugate of the wave function.

Substituting the given wave function, we get:

∫|Ψ(x,t)|^2dx = ∫e^(−5x) e^(iwt) e^(5x) e^(−iwt) dx

= ∫1dx

= x

Since the integral does not converge to 1, we can conclude that the wave function is not normalized.

b- To normalize the wave function, we need to find the normalization constant A such that the integral of the absolute value squared of the normalized wave function over all space equals 1. Mathematically, this can be written as:

∫|AΨ(x,t)|^2dx = 1

where Ψ(x,t) is the given wave function.

Substituting the given wave function and the definition of the normalization constant A, we get:

∫|AΨ(x,t)|^2dx = ∫|A|^2 e^(−10x)dx = |A|^2 ∫e^(−10x)dx = |A|^2 * 1/10

For the integral to equal 1, we need |A|^2 * 1/10 = 1, which gives us:

|A|^2 = 10

Taking the positive square root, we get:

|A| = √10

So, the normalization constant for the given wave function is A = √10.

Therefore, The wave function is not normalised since the integral does not converge to 1, and the normalisation constant for the specified wave function is A = √10.

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To find the electric field components E(x) and E(y) at any point, we need to take the negative gradient of the potential function V.

Let's first find the partial derivative of potential function V with respect to x:

∂V/∂x = a(3x² - y²)/(x² + y²)^(5/2) - b x/(x² + y²)^(3/2)

Similarly, the partial derivative of V with respect to y is:

∂V/∂y = -2a xy/(x² + y²)^(5/2) - b y/(x² + y²)^(3/2)

Now, we can find the components of the electric field:

Ex = -∂V/∂x = -a(3x² - y²)/(x² + y²)^(5/2) + b x/(x² + y²)^(3/2)

Ey = -∂V/∂y = 2a xy/(x² + y²)^(5/2) + b y/(x² + y²)^(3/2)

Therefore, the  components of the electric field at any point (x,y) are:

Ex = -a(3x² - y²)/(x² + y²)^(5/2) + b x/(x² + y²)^(3/2)

Ey = 2a xy/(x² + y²)^(5/2) + b y/(x² + y²)^(3/2)

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Answer:

The correct answer is option B: 47.11°.

Explanation:

To calculate the angle of refraction when light passes from air to water, we can use Snell's law, which states:

n₁ * sin(θ₁) = n₂ * sin(θ₂)

where:

n₁ = refractive index of the medium of incidence (in this case, air) = 1.0

n₂ = refractive index of the medium of refraction (in this case, water) = 1.33

θ₁ = angle of incidence

θ₂ = angle of refraction (what we want to find)

Given:

θ₁ = 77º

Let's plug in the values into Snell's law and solve for θ₂:

1.0 * sin(77º) = 1.33 * sin(θ₂)

sin(77º) = (1.33 * sin(θ₂)) / 1.0

Now, isolate sin(θ₂) by multiplying both sides by 1.0:

sin(77º) * 1.0 = 1.33 * sin(θ₂)

sin(θ₂) = (sin(77º) * 1.0) / 1.33

Now, take the inverse sine (arcsin) of both sides to find θ₂:

θ₂ = arcsin((sin(77º) * 1.0) / 1.33)

Calculating the value:

θ₂ ≈ 47.11º

Therefore, the angle of refraction in the water when the angle of incidence is 77º is approximately 47.11º.

Answer:

Friction plays a crucial role in the operation of a bicycle, both in terms of its performance and maintenance. Here are some ways that friction affects moving parts on a bicycle and ways to reduce or increase friction:

Friction in the chain: The chain is one of the most critical components of a bicycle, and friction can cause significant wear and tear on it. Lubricating the chain regularly can help reduce friction and prolong its lifespan.

Friction in the brakes: The brake pads are designed to increase friction on the wheel rims to slow the bike down or bring it to a stop. Over time, the brake pads wear down, reducing their effectiveness. Replacing the pads regularly can help ensure the brakes work correctly.

Friction in the bearings: Bearings help reduce friction and keep the wheels spinning smoothly. However, if they are not adequately lubricated or are worn out, they can create significant friction. Keeping the bearings clean and lubricated is essential for reducing friction and ensuring the bike runs smoothly.

Friction in the tires: The tire pressure and the surface of the road can affect the level of friction between the tires and the ground. If the tire pressure is too low or too high, it can cause uneven wear on the tire and reduce traction. It is essential to keep the tires inflated to the recommended pressure for optimal performance.

Friction in the pedals: The pedals are where the rider applies force to move the bike forward. If the pedals are not lubricated, they can create friction and reduce the rider's efficiency. Regularly cleaning and lubricating the pedals can help reduce friction and increase the rider's power.

In summary, maintaining a bicycle involves reducing friction in some areas and increasing it in others. Regular cleaning, lubrication, and replacement of worn-out parts can help keep a bicycle running smoothly and safely

Explanation:

The tension in the guideline are 151.8 N when the model airplane is flying in a circle at a constant speed of 19.6 m/s. When the speed is increased to 39.2 m/s, the tension in the guideline increases to 283.8 N.

In Example 5, the model airplane was flying in a circle at a constant speed. The tension in the guideline was equal to the centripetal force, which was given by the equation:

Fc = mv²/r

where m = mass of the airplane, v = speed, and r = radius of the circle.

In this case, there is no upward lift on the plane, so the tension in the guideline is equal to the weight of the plane, which is given by the equation:

W = mg

where m = mass of the plane and g = acceleration due to gravity.

Set these two equations equal to each other to find an expression for the tension in the guideline:

mv²/r = mg

Solving for T:

T= mv²/r + mg

For the given values, calculate the tension in the guideline as follows:

T = (2.42 kg)(19.6 m/s)²/(18.5 m)  + (2.42 kg)(9.8 m/s²)

T = 128.1 N + 23.7 N

T = 151.8 N

Therefore, the tension in the guideline is 151.8 N when the model airplane is flying in a circle at a constant speed of 19.6 m/s. When the speed is increased to 39.2 m/s, the tension in the guideline increases to 283.8 N.

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Complete question:

Redo Example 5,assuming that there is no upward lift on the plane generated by its wings. without such lift , the guideline shapes downward due to the weight of the plane. for purposes of significant figures, use 2.42 kg for the mass of the plane , 18.5m for the length of the guideline , and 19.6 and 39.2 m/s for he speeds.

The model airplane in Figure 5.6 has a mass of 0.90 kg and moves at a constant speed on a circle that is parallel to the ground. The path of the airplane and its guideline lie in the same horizontal plane, because the weight of the plane is balanced by the lift generated by its wings. Find the tension in the guideline (length = 17 m) for speeds of 19 and 38 m/s.

The emf induced in the toroid when the current decreases from 2.5 A to 1.1 A in a time of 0.15 s is 220 V.

The emf induced in a toroid can be given by the formula:

emf = -N (dΦ/dt)

where N = number of turns in the toroid and dΦ/dt = rate of change of the magnetic flux through the toroid.

To find the magnetic flux through the toroid, use the formula:

Φ = μ0 N I A / (2πr)

where μ0 = permeability of free space, I = current in the toroid, A = cross-sectional area of the toroid, and r = radius of the toroid.

Substituting the given values:

Φ = (4π x 10⁻⁷ T m/A) x (3143 turns/m) x (2.5 A) x (1.4 x 10⁻⁶ m²) / (2π x 0.073 m) = 0.0105 Wb

The rate of change of magnetic flux can be found by taking the derivative of the magnetic flux with respect to time:

dΦ/dt = -ΔΦ / Δt = -(0.0105 Wb) / (0.15 s) = -0.070 T/s

Finally, substitute the values into the formula for emf:

emf = -N (dΦ/dt) = -(3143 turns/m) x (-0.070 T/s) = 220 V

Therefore, the emf induced in the toroid when the current decreases from 2.5 A to 1.1 A in a time of 0.15 s is 220 V.

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The speed of the running shaft driving the machine, given that the main shaft to drive the machine has a pulley of diameter 140 mm is 111.4 rev/min

First, we shall list out the given parameters from the question. Details below:

The speed of the running shaft can be obtain as shown below:

S₁D₁ = S₂D₂

183 × 140 = S₂ × 230

183 × 140 = S₂ × 230

25620 = S₂ × 230

Divide both sides by 230

S₂ = 25620 / 230

S₂ = 111.4 rev/min

Thus, we can conclude that the speed of the running shaft driving the machine is 111.4 rev/min

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The spring is compressed by 0.122 m before the block comes to rest.

When the block encounters the spring, it will begin to compress the spring due to the force exerted by the spring on the block. This force will cause the block to decelerate and eventually come to a stop when all its kinetic energy has been converted into potential energy stored in the compressed spring.

We can use the conservation of energy principle to determine the compression distance of the spring.

The initial kinetic energy of the block is given by:

[tex]KE_i = (1/2) * m * v^2 = (1/2) * 0.980 kg * (1.32 m/s)^2 = 0.852 J[/tex]

This energy will be converted into potential energy stored in the compressed spring when the block comes to a stop. The potential energy stored in the spring is given by:

[tex]PE_s = (1/2) * k * x^2[/tex]

where k is the force constant of the spring and x is the compression distance of the spring.

Setting the initial kinetic energy equal to the potential energy stored in the spring, we have:

[tex]KE_i = PE_s[/tex]

0.852 J = (1/2) * 245 N/m * x^2

Solving for x, we get:

[tex]x = \sqrt{( (2 * 0.852 J) / (245 N/m) ) } = 0.122 m[/tex]

Therefore, the spring is compressed by 0.122 m before the block comes to rest.

It's important to note that this calculation assumes that the mass of the spring is negligible compared to the mass of the block, and that the spring is ideal with no damping or other losses of energy. In reality, there may be some losses due to friction or other factors, which would result in a smaller compression distance.

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Answer: B.

Explanation:

An isotope is a variation of the same element with a different number of neutrons in its nucleus. The only thing that changes is the top component, which is the total number of protons and neutrons. The bottom component remains the same (the number of protons).

The answer is "the electric potential at point B is 3750 V".

To calculate electric potential (V) at a point in an electric field due to a point charge:

V = k Q / r

Where, k is Coulomb's constant (k = 9 x [tex]10^9[/tex] N [tex]m^2[/tex] / [tex]C^2[/tex]).

            Q is the charge in Coulombs.

            r is the distance between the point charge and the point where             potential is being calculated, in meters.

Using the values given, calculate the electric potential at point A:

[tex]V_A[/tex] = (9 x [tex]10^9[/tex] N [tex]m^2[/tex] / [tex]C^2[/tex]) x (2.5 x [tex]10^{-6 }[/tex] C) / (0.03 m)

[tex]V_A[/tex]= 7500 V

[tex]V_B[/tex] = (9 x [tex]10^9[/tex] N [tex]m^2[/tex] / [tex]C^2[/tex]) x (2.5 x [tex]10^{-6 }[/tex] C) / (0.06 m)

[tex]V_B[/tex] = 3750 V

So, the electric potential at point B is 3750 V.

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Answer:

6,063.63637 kg m/s²

Explanation:

First you have to find the acceleration

a = velocity changes / time changes

(a stands for acceleration)

so we have 52-23 = 29 m/s

we know acceleration took 11 seconds so the time changes = 11 seconds

a= 29 / 11 = 2.63636364 m / s²

and we know

F = m × a

( F stands for force and m stands for mass)

so we have 2300kg × 2.63636364 m/s = 6,063.63637 kg m/s² or 6,063.63637 Newton

hope you understand ♥️

To express the decimals as powers of ten, the answer will be a) 5 ×10-¹, b) 1.04×10-³, c) 8.4 ×10-², d) 3.6×10-⁴.

A decimal is a number with a whole and a component of fraction. Decimal numbers, which are in between integers, are used to express the numerical value of whole and partially whole quantities.

As with multiplying decimals (see Decimal Multiplication), count the base number's decimal places before multiplying by a decimal. Add the exponent to that number after that. The total number of decimal places in the response will be this.

Therefore, the answer will be

a) 0.5 = 5 ×10-¹

b) 0.001 04 = 1.04×10-³

c) 0.084 = 8.4 ×10-²

d) 0.000 36 = 3.6×10-⁴

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It takes approximately 0.00000144/45 = [tex]3.2\times 10^{-8}[/tex] seconds, or 32 nanoseconds, for the capacitor to lose 20% of its initial energy when it discharges through a 45Ω resistor.

To determine how long it takes a charged capacitor to lose 20% of its initial energy when it discharges through a resistor, we need to use the formula for the energy stored in a capacitor, which is given by:

[tex]E = (1/2)CV^2[/tex]

where E is the energy in joules, C is the capacitance in farads, and V is the voltage across the capacitor in volts. The energy stored in a capacitor is proportional to the square of the voltage across it.

When the capacitor discharges through a resistor, the voltage across it decreases over time, and the energy stored in the capacitor is converted into heat energy in the resistor. The rate at which the energy is dissipated is given by:

[tex]P = IV = V^2/R[/tex]

where P is the power in watts, I is the current in amperes, and R is the resistance in ohms.

Using the above equations, we can determine the time it takes for the capacitor to lose 20% of its initial energy as follows:

First, calculate the initial energy stored in the capacitor:

[tex]Ei = (1/2)(80\times 10^{-6})(V^2)[/tex]

Next, calculate the final energy stored in the capacitor when it has lost 20% of its initial energy:

Ef = 0.8Ei

Using the equation for power, we can find the current in the circuit:

[tex]P = IV = V^2/R[/tex]

I = V/R

We can use the formula for the rate of change of energy to find the time taken to lose 20% of the initial energy:

dE/dt [tex]= -P = -(V^2/R)[/tex]

dt/dE = [tex]-R/(V^2)[/tex]

t = -R/(V^2) * ∫ (Ei-Ef) dE

Substituting the values from steps 1-4, we can solve for the time taken:

t = -45/([tex]V^2[/tex]) * ∫ (0.8(1/2)([tex]80\times 10^{-6}[/tex])(V^2) - 0) dE

t = -45/([tex]V^2[/tex]) * [0.8(1/2)([tex]80\times 10^{-6}[/tex])V^2]

t = 0.00000144/R seconds

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