Earth orbiting the Sun The Earth is 1.5 ⋅ 10 8 km from the Sun (on average). How fast is the Earth orbiting the Sun in kilometers per second (on average)? You can assume the orbit of the Earth is a circle and that the circumference of a circle is equal to C = 2 π R where R is the radius of a circle (the distance between the center and the edge. Note that for our purposes, it is perfectly fine to assume π = 3 which allows for a pretty good approximation C = 6 R . Your answer does not need to be put into scientific notation, but if you choose to do so it will be marked correct! kilometers per second

The stress in Magapascals is 80 Magapascals

This question can be solved by applying Young's modulus.

Young's modulus : Young's modulus states that, stress  is directly proportional to strain.

Where Stress is pressure acting the body. The s.i unit of stress is N/m² or Pascal. And it can be expressed mathematically as,

From the question,

\(P = 10000/(125*10^{-6} )\)

\(P =\) \(8*10^{7}\) Pascals

\(P =\) 80 Magapascals

Hence the stress in Magapascals is 80 Magapascals

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The stress experimented by the tie-bar is 80 megapascals.

The tie-bar is under axial load. Under the assumption that force is distributed uniformly in the cross-section area, we can use the following definition of normal stress (\(\sigma\)), in megapascals:

\(\sigma = \frac{F}{A}\) (1)

Where:

\(F\) - Axial force, in meganewtons.

\(A\) - Cross-section area, in square meters.

If we know that \(F = 10\times 10^{-3}\,MN\) and \(A = 125\times 10^{-6}\,m^{2}\), then the stress experimented by the tie-bar is:

\(\sigma = \frac{10\times 10^{-3}\,MN}{125\times 10^{-6}\,m^{2}}\)

\(\sigma = 80\,MPa\)

The stress experimented by the tie-bar is 80 megapascals.

Answer:D

Explanation:D

If a product claims to be low-calorie, then it means that it must contain 40 calories or fewer per saving. Hence, option D is correct.

Our bodies need energy to keep us alive and to make sure that our organs are functioning properly.

When we eat and drink, we provide our bodies with energy. Our bodies consume that energy during routine tasks like breathing and jogging.

To keep a constant weight, the amount of energy we put into our bodies must correspond to the amount of energy we expend during routine biological functions and physical activity.

A key element of a balanced diet is striking a balance between the energy you put into your body and the energy you expend.

For instance, the more physical activity we do, the more energy we use.

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Could we have the choices? If we have those we could help you!   εїз

Answer:

Pulse transfers a single disturbance, while wave is a continuous disturbance that transfers energy.

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Answer:

They have the same acceleration.

Explanation:

50km/h to 60km/h = 10km/h

0km/h to 10 km/h = 10km/h

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Answer:

they have a same acceleration

Explanation:

50km/h to 60 km/h=10km/h

0km/h to 60 km/h=10km/h

Answer:

The velocity of the box is related to the angular velocity of the spool, which is given by the equation:

v = r * ω

where r is the radius of the spool and ω is the angular velocity of the spool. The angular velocity of the spool, in turn, is related to the torque applied to the spool by the tension in the string, which is given by the equation:

τ = I * α

where τ is the torque, I is the moment of inertia of the spool, and α is the angular acceleration of the spool.

The tension in the string is equal to the weight of the box, which is given by:

T = m * g

Putting all of these equations together, we can solve for the time it takes for the box to reach the floor. Here's how:

First, we can find the angular acceleration of the spool using the torque equation:

τ = I * α

T = m * g = τ

m * g = I * α

α = (m * g) / I

α = (35.30 kg * 9.81 m/s^2) / 4.00 kg*m^2

α = 86.53 rad/s^2

Next, we can find the angular velocity of the spool using the kinematic equation:

ω^2 = ω_0^2 + 2 * α * θ

where ω_0 is the initial angular velocity (which is zero), θ is the angle through which the spool has turned (which is equal to the distance the box has fallen divided by the radius of the spool), and ω is the final angular velocity (which is what we want to find). Solving for ω, we get:

ω^2 = 2 * α * θ

ω = sqrt(2 * α * θ)

ω = sqrt(2 * 86.53 rad/s^2 * (3.50 m / 0.10 m))

ω = 166.6 rad/s

Finally, we can find the time it takes for the box to reach the floor using the equation:

v = r * ω

v = 0.10 m * 166.6 rad/s

v = 16.66 m/s

t = d / v

t = 3.50 m / 16.66 m/s

t = 0.21 s

The diameter of the smaller end of the pipe is approximately 0.78 meters.

To determine the diameter of the smaller end of the pipe, we can use the principle of conservation of mass. According to this principle, the mass flow rate of water should remain constant throughout the pipe.

The mass flow rate is given by the equation:
Mass flow rate = density of water * cross-sectional area * velocity

Since the density of the water remains constant, we can write:
Cross-sectional area1 * velocity1 = Cross-sectional area2 * velocity2

Given that the velocity1 is 13 m/s, the diameter1 is 1.2 m, and the velocity2 is 30 m/s, we can solve for the diameter2 using the equation:
(pi * (diameter1/2)^2) * velocity1 = (pi * (diameter2/2)^2) * velocity2

Simplifying the equation:
(1.2/2)^2 * 13 = (diameter2/2)^2 * 30

Calculating the equation:
(0.6)^2 * 13 = (diameter2/2)^2 * 30

0.36 * 13 = (diameter2/2)^2 * 30

4.68 = (diameter2/2)^2 * 30

Dividing both sides by 30:
0.156 = (diameter2/2)^2

Taking the square root of both sides:
0.39 = diameter2/2

Multiplying both sides by 2:
0.78 = diameter2

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The force acting on the system of two blocks connected by a rope passing over a pulley is -13.26 N.

The system of two blocks connected by a rope passing over a pulley are M1 and M2, where M1 rests on the triangle edge to the left of the pulley, which makes an angle of theta subscript 1 with the base of the triangle. The coefficient of friction between box M1 and the surface is mu subscript 1. M2 rests on the triangle edge to the right of the pulley, which makes an angle of theta subscript 2 with the base of the triangle.

The coefficient of friction between box M2 and the surface is mu subscript 2. The system sits atop a scalene triangle whose long edge forms the base. The pulley is attached to the apex of the triangle.M1 has a mass of 2.25 kg and is on an incline of angle 1=42.5∘ that has a coefficient of kinetic friction 1=0.205. M2 has a mass of 5.55 kg and is on an incline of angle 2=33.5∘ that has a coefficient of kinetic friction 2=0.105.The free-body diagram of M1 shows that the weight of M1 acts straight downwards (vertically) and the normal force acts perpendicular to the slope.

The force of friction opposes the motion and acts opposite to the direction of motion.M1 = 2.25 kgTheta subscript 1 = 42.5 degreesMu subscript 1 = 0.205g = 9.81 m/s²In the free-body diagram of M2, the normal force acts perpendicular to the incline of the slope, the weight of the object acts vertically downwards and parallel to the incline, and the force of friction opposes the motion and acts opposite to the direction of motion.M2 = 5.55 kgTheta subscript 2 = 33.5 degreesMu subscript 2 = 0.105g = 9.81 m/s²The tension in the string is the same throughout the rope. Since the masses are being pulled by the same rope, the acceleration of the objects is the same as the acceleration of the rope.

The tension in the string is directly proportional to the acceleration of the objects and the rope.A system of two blocks connected by a rope passing over a pulley has a total mass of M. The acceleration of the system is given by the formula below:a = [(m1-m2)gsin(θ1) - μ1(m1+m2)gcos(θ1)] / (m1 + m2)Where, μ1 = 0.205 is the coefficient of friction of block M1θ1 = 42.5 degrees is the angle of the incline of block M1M1 = 2.25 kg is the mass of block M1M2 = 5.55 kg is the mass of block M2g = 9.81 m/s² is the acceleration due to gravitysinθ1 = sin 42.5 = 0.67cosθ1 = cos 42.5 = 0.75The acceleration of the system is:a = [(2.25-5.55)(9.81)(0.67) - (0.205)(2.25+5.55)(9.81)(0.75)] / (2.25 + 5.55)a = -1.7 m/s² (the negative sign indicates that the system is accelerating in the opposite direction).

The force acting on the system is given by:F = MaWhere M is the total mass of the system and a is the acceleration of the system. The total mass of the system is:M = m1 + m2M = 2.25 + 5.55M = 7.8 kgThe force acting on the system is:F = 7.8(-1.7)F = -13.26 N (the negative sign indicates that the force is acting in the opposite direction).

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Answer: both environments are undergoing succession, but different kinds

Explanation:

Answer:

d

Explanation:

did it on edge 2021

Answer:

The correct answer is D C A

Answer:

10 Newtons.

Explanation:

F=MA

F=(5kg)(2 m/s^2)

10 N

Based on the information, we can infer that A. represents a Mitochondria.

Mitochondria is a term to refer to the eukaryotic cell organelles responsible for supplying most of the energy necessary for cell activity through the process called cellular respiration.

Based on the information, we can infer that the element that is labeled with the letter A is a mitochondrion because its location is that of a mitochondrion. In this case, the mitochondria is red, although in other models it can be represented with another color. In general, it is given this shape and this color to distinguish it from other elements of the cell.

Note: This question is incomplete. Here is the complete information:

Attached image

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Answer:

Therefore, the voltage required is 20 volts

Explanation:

To find the voltage that will send a current of 5 amperes through a bell circuit with a resistance of 4 ohms, we can plug in the values into the equation and solve for V: V=IR

V=(5A)(4Ω)

V=20V

Based on the information, it should be noted that the resistance R is 0.5 Ω.

When S1 and S2 are open, i leads e by 30°. In this case, the circuit consists of only the inductor (L) and the capacitor (C) in series. Therefore, the impedance of the circuit can be written as:

Z = jωL - 1/(jωC)

Since i leads e by 30°, we can express the phasor relationship as:

I = k * e^(j(ωt + θ))

Z = jωL - 1/(jωC) = j(120π)L - 1/(j(120π)C)

Re(Z) = 0

By equating the real parts, we get:

0 = 0 - 1/(120πC)

Let's assume that there is a resistance (R) in series with the inductor and capacitor. The impedance equation becomes:

Z = R + jωL - 1/(jωC)

Z = R + jωL

Im(Z) = ωL > 0

Substituting the angular frequency and rearranging the inequality, we have:

120πL > 0

L > 0

This condition implies that the inductance L must be greater than zero.

When S1 and S2 are closed, i has an amplitude of 0.5 A. In this case, the impedance is:

Z = R + jωL - 1/(jωC)

Since the amplitude of i is given as 0.5 A, we can express the phasor relationship as:

I = 0.5 * e^(j(ωt + θ))

By substituting this phasor relationship into the impedance equation, we can determine the value of R. The real part of the impedance must be equal to R:

Re(Z) = R

Since the amplitude of i is 0.5 A, the real part of the impedance must be equal to 0.5 A: 0.5 = R

Therefore, the resistance R is 0.5 Ω.

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Answer: The scenario violates the second law of thermodynamics.

Explanation: The second law states that heat cannot be converted into work without some loss of usable energy, and that the amount of usable energy in a closed system will always decrease over time. Therefore, the machine described in the scenario cannot exist because it would violate the second law by converting all of the heat into electricity without any loss of usable energy.

Answer:

The wavelengths are

     \(\lambda_1  =  614\ nm \)

     \(\lambda_2  =  687\ nm \)

\(\lambda_3 = 975\ nm \)

Explanation:

From the question we are told that

The number of slits per mm is N = 1200

The distance of the screen is \(D = 79.0 \ cm = 0.79 \ m\)

The first distance where First-order maxima is observed is \(y_1 = 58.2 cm = 0.582 \ m \)

The second distance where First-order maxima is observed is \(y_2 = 65.2 \ cm = 0.652 \ m \)

The second distance where First-order maxima is observed is \(y_2 = 92.5 \ cm = 0.925 \ m \)

Generally the distance of separation between the slits is mathematically represented as

\(d = \frac{1}{1200}= 8.33 *10^{-4} \ mm = 8.33 *10^{-7} \ m\)

Considering the first distance where First-order maxima is observed

Generally the the first distance where First-order maxima is observed is mathematically represented as

\(y_1 = \frac{\lambda_1 * D}{d}\)

=> \(\lambda_1 = \frac{0.582 * 8.33 *10^{-7} }{0.79}\)

=>     \(\lambda_1  =  6.14 *10^{-7} m \)

=>     \(\lambda_1  =  614\ nm \)

Considering the second distance where First-order maxima is observed

Generally the the second distance where First-order maxima is observed is mathematically represented as

\(y_2 = \frac{\lambda_2 * D}{d}\)

=> \(\lambda_2 = \frac{0.652 * 8.33 *10^{-7} }{0.79}\)

=>     \(\lambda_2  =  6.87 *10^{-7} m \)

=>     \(\lambda_2  =  687\ nm \)

Considering the third distance where First-order maxima is observed

Generally the the third distance where First-order maxima is observed is mathematically represented as

\(y_3 = \frac{\lambda_3 * D}{d}\)

=> \(\lambda_3 = \frac{0.925 * 8.33 *10^{-7} }{0.79}\)

=>     \(\lambda_3  =  9.75 *10^{-7} m \)

=>     \(\lambda_3 =  975\ nm \)

Note that the above image depicts the Movement of Lithospheric Plates. See the attached image accordingly.

The movement of lithospheric plates is a geologic process that occurs due to the movement of hot, molten material in the Earth's mantle. The lithosphere, which is the rigid outer layer of the Earth's surface, is broken up into several large plates that move relative to each other. These movements are driven by the convection of material in the mantle and the forces generated at plate boundaries.

There are three main types of plate boundaries: divergent, convergent, and transform. Divergent boundaries occur where plates move away from each other, creating new oceanic crust. Convergent boundaries occur where plates collide, causing subduction, volcanic activity, and mountain formation. Transform boundaries occur where plates slide past each other.

The movement of lithospheric plates is responsible for a range of geologic phenomena, including earthquakes, volcanic activity, and the formation of mountain ranges and ocean basins.

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Answer:

a. 200 m

b. 100 m

Explanation:

Solution:-

- We will first draw three points marked A,B and Q from left most to right most.

- We are told that the antennas at A and B radiate in phase. This means the radio-waves emitted by each antenna are synchronous in terms of ( frequency and wavelength ).

- We will denote the common wavelength of coherent sources of radio-waves ( A and B ) with λ.

- The relation between the wavelength ( λ ) and the path difference between the source and observation point ( Q ) for the case of destructive interference is:

                             AQ - BQ = n*λ/2

Where,

             n: The order of wavelength

             AQ: The distance between antenna A and point Q

             BQ: The distance between antenna B and point Q

- The point Q is positioned ( 100 + 50 ) m away from antenna A and 50 m from antenna B. Hence,

                            150 - 50 = n*λ/2

- To determine the longest wavelength ( λ ) to meet destructively at point Q with the given path difference. The order of wavelength ( n ) must be minimum ( 1 ). Therefore,

                           100 = λ/2

                           λ = 200 m  .... Answer

- The relation between the wavelength ( λ ) and the path difference between the source and observation point ( Q ) for the case of constructive interference is:

                             AQ - BQ = n*λ

Where,

             n: The order of wavelength

             AQ: The distance between antenna A and point Q

             BQ: The distance between antenna B and point Q

- The point Q is positioned ( 100 + 50 ) m away from antenna A and 50 m from antenna B. Hence,

                            150 - 50 = n*λ

- To determine the longest wavelength ( λ ) to meet constructively at point Q with the given path difference. The order of wavelength ( n ) must be minimum ( 1 ). Therefore,

                           100 = λ

                           λ = 100 m  .... Answer

The magnitude of the force on the tooth is approximately 0.022 N.

To find the magnitude and direction of the force on the tooth, we can use Hooke's Law, which states that the force exerted on an object is directly proportional to the change in length of a material when it is stretched or compressed.

First, we need to calculate the strain (ε) of the stainless-steel wire.

Strain is defined as the change in length divided by the original length:

ε = ΔL / L₀

Given that the change in length (ΔL) is 0.10 mm \((0.10 \times 10^{-3} m)\) and the unstretched length (L₀) is 3.1 cm \((3.1 \times 10^{-2} m)\), we can calculate the strain:

\(\epsilon=(0.10 \times 10^{-3} m)/(3.1 \times 10^{-2} m)=0.003225\)

Next, we can use Young's modulus (E) to calculate the stress (σ) in the wire.

Stress is defined as the force per unit area:

σ = E * ε

Given that Young's modulus (E) for stainless-steel is 18 × 10¹⁰ N/m², we can calculate the stress:

σ = (18 × 10¹⁰ N/m²) * 0.003225 = 5.805 × 10⁸ N/m²

Now, we can find the force (F) on the tooth by multiplying the stress by the cross-sectional area (A) of the wire:

F = σ * A

The cross-sectional area (A) can be calculated using the formula for the area of a circle:

A = π * (d/2)²

Given that the diameter (d) of the wire is 0.22 mm\((0.22 \times 10^{-3} m)\), we can calculate the cross-sectional area:

\(A = \pi * (0.22 \times 10^-3 m / 2)^{2} = 3.802 \times 10^{-8} m^2\)

Finally, we can calculate the force:

\(F = (5.805 \times 10^{8} N/m^{2}) * (3.802 \times 10^-8 m^{2}) \approx 2.206 \times 10^{-2} N\)

Therefore, the magnitude of the force on the tooth is approximately 0.022 N.

Since the wire is stretched, the force is pulling the tooth in the direction opposite to the stretching.

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When filled with extra air its mass increased to 0.428kg also the top of polythene container mass connected to the perplex box of volume 1000cm³ and the number of times of air inside was 7.2 times. The density of the air filled in the polythene container is approximately 1.25 kg/m³.

The density of air filled in the polythene container can be determined by considering the change in mass and volume of the container before and after filling it with air. Given that the mass of the empty container is 0.419 kg and the mass of the container when filled with extra air is 0.428 kg, and the volume of the perplex box is 1000 cm³.

Calculate the mass of the air inside the container by subtracting the mass of the empty container from the mass of the container when filled with air:

Mass of air = Mass of filled container - Mass of empty container

= 0.428 kg - 0.419 kg

= 0.009 kg

Calculate the volume of the air inside the container using the given number of times the air inside is 7.2:

Volume of air = Volume of perplex box * Number of times air inside

= 1000 cm³ * 7.2

= 7200 cm³

Convert the volume of air to cubic meters (m³) by dividing by 1000000:

Volume of air = 7200 cm³ / 1000000

= 0.0072 m³

Calculate the density of air using the formula:

Density = Mass / Volume

Density = 0.009 kg / 0.0072 m³

≈ 1.25 kg/m³

Therefore, the density of the air filled in the polythene container is approximately 1.25 kg/m³.

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To better understand crash dynamics we have to look at "the laws of motion."

The laws of motion

The laws of motion were introduced by Sir Isaac Newton in 1687 in his book Philosophiæ Naturalis Principia Mathematica ("Mathematical Principles of Natural Philosophy"), which defined the laws of motion, or three fundamental laws that govern the movement of bodies. The laws of motion, according to Newton, govern the motion of an object or a system of objects that interact.

It defines the concepts of force and mass, and the fundamental dynamics of motion.The following are the laws of motion:Every object will remain at rest or in uniform motion in a straight line unless compelled to change its state by the action of an external force. The velocity of an object changes proportional to the force applied to it, and the acceleration of an object is proportional to both its force and its mass. For every action, there is an equal and opposite reaction.

Therefore, these laws are necessary to fully grasp crash dynamics because they explain how objects respond to outside forces that cause them to accelerate or decelerate.

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The statement "B. Gravity exists only on Earth" and the statement "E. Gravity doesn't exist between Earth and the sun"  is not true about gravity.

Gravity is a fundamental force of nature that exists in the whole universe, not just on Earth. It is a force that acts between any two objects that have mass. This means that statement "C. Gravity is a force that pulls two objects together" and "D. Gravity exists between two objects that have mass" are both true. Gravity plays a significant role in the functioning of our solar system. The sun's gravitational force acts on the planets, including Earth, keeping them in their orbits. Similarly, Earth's gravitational force attracts objects towards its center, giving weight to objects on its surface. Gravity is the force that holds Earth in orbit around the sun and is responsible for the planets' motion in the solar system. Gravity is a universal force that exists throughout the universe, acts between objects with mass, and plays a crucial role in celestial bodies' movements, including the interaction between Earth and the sun.

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Answer:

distance

Explanation:

Speed is the rate of change of position expressed as distance travelled per unit of time

Vector C = 15/4x - 13y

A vector is an object that has both a magnitude and a direction. Geometrically, we can picture a vector as a directed line segment, whose length is the magnitude of the vector and with an arrow indicating the direction. The direction of the vector is from its tail to its head.

Two examples of vectors are those that represent force and velocity. Both force and velocity are in a particular direction. The magnitude of the vector would indicate the strength of the force or the speed associated with the velocity.

From the question;

Vector A = 3x + 5y

Vector B = -3x + 6y

2A + 7B + 4C = 0

Substituting the vectors A and B

2(3x + 5y) + 7(-3x + 6y) + 4C = 0

6x +10y - 21x + 42y +4C = 0

Taking like terms

-15x +52y + 4C = 0

4C = 15x - 52y

Dividing both sides by 4

Vector C = 15/4x - 13y

In conclusion, our Vector C is derived by substituting in the equation given.

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The energy stored in the circuit at any time t is given by \(U = (1/2)L*I^{2} + (1/2)Q^{2} /C = (1/2)L*(V_{0} /R)^{2} *e^{(-2t/(R*C))} + (1/2)C*V_{0} ^{2} *(1 - e^{(-2t/(R*C)})).\)The units are in seconds.

The total energy stored in the circuit can be calculated using the formula: U = (1/2)L*I² + (1/2)Q²/C, where L is the inductance, I is the current, Q is the charge on the capacitor, and C is the capacitance.

Initially, the capacitor is uncharged, so the second term is zero.

Therefore, the initial energy stored in the circuit is U₀ = (1/2)L*I₀², where I₀ is the initial current, which is zero.

When the magnetic field is switched on, a current begins to flow in the solenoid.

This current increases until it reaches its maximum value, given by I = V/R, where V is the voltage across the solenoid and R is its resistance.

Since the solenoid is connected in series with the capacitor, the voltage across the solenoid is equal to the voltage across the capacitor, which is given by V = Q/C, where Q is the charge on the capacitor.

The charge on the capacitor is given by Q = C*V, where V is the voltage across the capacitor at any time t.

Therefore, we have I = V/R = Q/(R*C) = dQ/dt*(1/R*C), where dQ/dt is the rate of change of charge on the capacitor.

This is a first-order linear differential equation, which can be solved to give \(Q(t) = Q_{0} *(1 - e^{(-t/(R*C)}))\), where Q₀ is the maximum charge on the capacitor, given by Q₀ = C*V₀, where V₀ is the voltage across the capacitor at t=0.

The current in the solenoid is given by I(t) = \(dQ/dt*(1/R*C) = (V_{0} /R)*e^{(-t/(R*C)}).\)

The energy stored in the circuit at any time t is given by\(U = (1/2)L*I^{2} + (1/2)Q^{2} /C = (1/2)L*(V_{0} /R)^{2} *e^{(-2t/(R*C))} + (1/2)C*V_{0} ^{2} *(1 - e^{(-2t/(R*C)})).\)

The time t at which the energy stored in the circuit drops to 10% of its initial value can be found by solving the equation U(t) = U₀/10, or equivalently, \((1/2)L*(V_{0} /R)^{2} *e^{(-2t/(R*C)}) + (1/2)C*V_{0} /R)^{2}*(1 - e^{(-2t/(R*C)})) = (1/20)L*I_{0} /R)^{2}.\)

This equation can be solved numerically using a computer program, or graphically by plotting U(t) and U₀/10 versus t on the same axes and finding their intersection point.

The solution is t = 1.74 ms.

The units are in seconds.

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The car is moving at approximately 12.5 meters per second.

The kinetic energy (KE) of an object can be calculated using the formula:

KE = 1/2 * m * \(v^2\)

where

KE = kinetic energy,

m =Mass of the object, and

v = velocity.

In this case, we are given the mass (m) of the car as 1600 kg and the kinetic energy (KE) as 125,000 J. To find the velocity .

Substituting the  values , we have:

125,000 J = 1/2 * 1600 kg *\(v^2\)

Now, we can solve for v by rearranging the equation:

\(v^2\) = (2 * 125,000 J) / 1600 kg

\(v^2\) = 156.25 \(m^2/s^2\)

Taking the square root, we find:

v = √156.25\(m^2/s^2\)

v ≈ 12.5 m/s

Therefore, the car is moving at approximately 12.5 meters per second.

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